Unit 1 Theory of Numbers
Natural Numbers The numbers 1, 2, 3, …, which are used in counting are called natural numbers or positive integers. Basic Properties of Natural Numbers In the system of natural numbers, we have two ‘operations’ addition and multiplication with the following properties. Let x,, y z denote arbitrary natural numbers, then 1. x+ y is a natural number i.e., the sum of two natural number is again a natural number. 2. Commutative law of addition x+ y = y + x 3. Associative law of addition x+()() y + z = x + y + z 4. x⋅ y is a natural number i.e., product of two natural numbers is a natural number. 5. Commutative law of multiplication x⋅ y = y ⋅ x 6. Associative law of multiplication x⋅()() y ⋅ z = x ⋅ y ⋅ z 7. Existence of multiplicative identity a⋅1 = a 8. Distributive law x() y+ z = xy + xz
Divisibility of Integers An integerx ≠ 0 divides y, if there exists an integer a such that y= ax and thus we write as x| y ( x divides y). If x does not divides y, we write as x\| y (x does not divides y ) [This can also be stated as y is divisible by x or x is a divisor of y or y is a multiple of x]. Properties of Divisibility 1. x | y and y | z ⇒ x | z 2. x | y and x | z ⇒x| ( ky± lz ) for all k, l∈ z. z = set of all integers. 3. x | y and y | x ⇒ x= ± y 4. x | y, where x>0, y > 0 ⇒x≤ y 5. x | y ⇒ x | yz for any integer z. 6. x | y iff nx | ny, where n ≠ 0. 2 Indian National Mathematics Olympiad
Test of Divisibility Divisibility by certain special numbers can be determined without actually carrying out the process of division. The following theorem summarizes the result : A positive integer N is divisible by • 2 if and only if the last digit (unit's digit) is even. • 4 if and only if the number formed by last two digits is divisible by 4. • 8 if and only if the number formed by the last three digits is divisible by 8. • 3 if and only if the sum of all the digits is divisible by 3. • 9 if and only if the sum of all the digits is divisible by 9. • 5 if and only if the last digit is either 0 or 5. • 25 if and only if the number formed by the last two digits is divisible by 25. • 125 if and only if the number formed by the last three digits is divisible by 125. • 11 if and only if the difference between the sum of digits in the odd places (starting from right) and sum of the digits in the even places (starting from the right) is a multiple of 11. Division Algorithm For any two natural numbers a and b, there exists unique numbers q and r called respectively quotient and remainder, a= bq + r, where 0 ≤r < b. Common Divisor If a number ‘c’ divides any two numbers a and b i.e., if c | a and c | b, then c is known as a common divisor of a and b. Greatest Common Divisor If a number d divides a and b and is divisible by all the common divisors of a and b, then d is known as the greatest common divisor (GCD) of a and b or HCF of a and b. The GCD of numbers a and b is the unique positive integer d with the following two properties. (i) d | a and d | b (ii) If c | a and c | b; then c | d We write it as (,)a b= d For example, (,);12 15= 3 (,)7 8= 1 Note 1. (,)(,)a b= b a 2. If a|b; then (,)a b= a Properties of GCD 1. If (,)b c= g and d is a common divisor of b and c, then d is a divisor of g. 2. For any m > 0, (mb , mc )= m ( b , c ) b c 1 3. If d |b and d |c and d > 0, then ,(,) = b c d d d b c 4. If (,)b c= g, then , = 1 g g 5. If (,)b c= g, then there exists two integers x and y such that g= xb + yc 6. If (,)a b = 1 and (,)a c = 1, then (,)a bc = 1 7. If a| bc = 1 and (,),a b = 1 then a | c = 1 Theory of Numbers 3
For example a=6,, b = 21 c = 10 6| 21× 10 but (,)6 21= 3 and (,)6 10= 2 and 6 divides neither 21 nor 10 LCM of two integers a, b is the smallest positive integer divisible by both a and b and it is denoted by [a, b]. The Euclidean algorithm can be used to find the GCD of two integers as well as representing GCD as in 5th property. Consider 2 numbers 18, 28 28= 1. 8 + 10 ;18= 110. + 8 10= 1 ⋅ 8 + 2 ; 8= 4 ⋅ 2 + 0 (,)18 28= 2 (,).18 28= 2 = 10 − 1 8 =10 −(.) 18 − 110 =210.. − 118 =2(.). 28 − 118 − 118 =228.. − 318 =228.() + − 3 18
Note The representation in 5th property is not unique. In fact we can represent (,)a b as xa+ yb in infinite number of ways where x, y∈ Z Z = set of all integers. In above example, 252 is LCM of 18 and 28. 252= 9. 28 252= 1418. (,).()18 28= 2 28 + − 3 ⋅ 18 =228.() + 252KK + − 318 − 252 =()()2 + 9KK 28 + − 3 − 14 18 where K is any integer. Unit 1 is called unit in the set of positive integers. Prime A positive integer P is said to be prime, if (i) P > 1 (ii) P has no divisors except 1 and P i.e., A number which has exactly two different factors, itself and one, is called a prime number. Thus, 2, 3, 5, 7, 11, ... are primes. 2 is the only even number which is prime. All other primes being odd. But the converse is not true i.e., every odd number need not be prime. Composite Every number (greater than one) which is not prime is called composite number. i.e., a number which has more than two different factors is called composite. For example 18 is a composite number because 2, 3, 6, 9 are divisors of 18 other than 1 and 18. We can also define a composite number as : A natural number n is said to be composite, if there exists integers l and m such that n = lm, where 1 Remark G A prime number P can be written as a product only in one way namely P.1 . G A composite number n can also be written as n.1. But composite number can be written in one more way also as mentioned above. G A composite number has at least three factors. Note 1 is neither prime nor composite. Twin Primes A pair of numbers is said to be twin primes, if they differ by 2. e.g., 3, 5 are twin primes. Perfect Number A number n is said to be perfect if the sum of all divisors of n (including n) is equal to 2n. For example 28 is a perfect number because divisors of 28 are 1, 2, 4, 7, 14, 28. Sum of divisors of n ()=28 = 1 + 2 + 4 + 7 +14 + 28 =56 = 2n Coprime Integers Two numbers a and b are said to be coprime, if 1 is only common divisors of a and b. i.e., if GCD of a and b = 1 i.e., if (,)a b = 1 e.g., (4, 5) = 1, (8, 9) = 1. Theorem 1 If a= qb + r , then (,)(,)a b= b r . Proof Let (,)a b= d and (,)b r= e Q (,)a b= d ∴ d | a and d | b ∴ d | a and d| qb ∴ d | (a –qb) i.e., d |r []Qa− qb = r ∴ d is common divisor of b and r. ∴ d |e [Qe is the GCD of b and r] …(i) Again Q (b, r) = e e e ∴ e |b and e |r ∴ and bq r ∴ e|bq + r i.e., e|a []Qa= bq + r ∴ e is a common divisor of a and b. ∴ e|d [Qd is the GCD of a and b] …(ii) From Eqs. (i) and (ii), we have d= e i.e., (,)(,)a b= b r Remark The above result can also be stated as : GCD of a and b is same as GCD of b and r, where r is remainder obtained on dividing a by b. Corollary If (,)a b = 1, then (,)(,)b r= a b = 1 i.e., if a is coprime to b, then r is coprime to b. r is remainder obtained on dividing a by b. Theory of Numbers 5 Theorem 2 If d is the greatest common divisor of a and b, then there exists integers x and y such that d= xa + yb and d is the least positive value of xa+ yb. Proof Case I By successive application of division algorithm to numbers a and b. K Let r1,,, r 2 rn be successive remainders. = + < < Therefore, a bq1 r 10, r 1 b (Dividing a by b) = + < < b r1 q 2 r 20, r 2 r 1 (Dividing b by remainder r1 ) = + < < r1 r 2 q 3 r 3, 0 r 3 r 2 (Dividing r1 by remainder r2 ) M M M = + < < rn−2 r n − 1 qn r n , 0 rn rn − 1 = + < < rn −1 rn qn + 1 r n + 1, 0 rn + 1 rn > > K Since r1 r 2 is a set of decreasing integers, this process must terminate after a finite number of step. i.e., remainder must be zero after some stage. = ∴ = + = So let rn + 1 0 rn −1 rn qn + 10 rn qn + 1 ∴ = = rn | rn − 1 (,)rn − 1 rn r n [If a | b, then ()]a, b a = = = = K = = Now, ()()(,)(,)a,, b b r1 r 1 r 2 r 2 r 3 (,)rn − 1 rn rn …(i) i.e., GCD of a and b is rn . = − = + = = − From first of above equations r1 a bq 1 ax 1 by 1, where x11,, y 1 q = − = − Putting the value of r1 a bq 1 in r2 b r 1 q 2 = − = − − = − + r2 b r 1 q 2 b() a bq 1 q 2 b aq2 bq 1 q 2 = − + + = + = − aq2 b1( q 1 q 2 ) ax2 by 2 , where x2 q 2 = + y21 q 1 q 2 = + = + Similarly, r3 ax 3 by 3 and so on rn ax n by n = + or rn ax by = = where xn x and yn y = = = + i.e., GCD of a and b rn can be expressed as (,)a b d ax by [By Eq. (i)] Case II (,)a b= d (Q d | a and d | b) ∴ d| ( ax+ yb ) for all values of x and y. ∴ ∃ an integer k such that xa+ yb = kd …(ii) But least value of k is 1. Putting k = 1 in Eq. (ii), least value of xa+ yb is d. Corollary If a and b are coprime integers i.e., if (,),a b = 1 then there exists integers x and y such that ax+ by = 1 a b Example 1 If (,)a b= d, then , =1. d d Solution Q (,)a b= d ∴ d| a and d| b [By definition of GCD]. ∴ = There exists integers a1and b1 such that a da1 …(i) = b db1 …(ii) Again Q (,)a b= d 6 Indian National Mathematics Olympiad There exists integers x and y such that ax+ by = d [By Theorem 2] Putting values of a and b from Eqs. (i) and (ii) + = da1 x db 1 y d. + = or a1 x b 1 y 1 = (,)a1 b 1 1 [By corollary theorem 1] a b = Q =a = b or , 1 FromEq.(),(), ia1, fromEq. ii b 1 d d d d Remark = = = = (,)a b d and a a1 d, b b 1 d, then a1 and b1 are coprime i.e., (,)a1 b 1 1 Example 2 If a|bc and (,),a b =1 then a|c. Solution Q a|bc ∴ There exists an integer d such that bc= ad …(i) Q (,)a b =1 ∴ There exist integers m and n such that am+ bn =1 …(ii) Multiplying both sides of Eq.(ii) by c, acm+ bcn = c …(iii) Putting bc= ad from Eq. (i) in Eq. (iii) acm+ adn = c a() cm+ dn = c ∴ a|c Note If a|bc and (,)a b =1, then a|c. This result is also known as Gauss Theorem. Example 3 Prove that every two consecutive integers are coprime. Solution Let n and n + 1 be two consecutive integers. Let (,)n n+1 = d ∴ d|n and d|n + 1 d|( n+1 ) − n or d |1 ∴ d =1 ∴ (,)n n +1 = 1 i.e., n and ()n + 1 are relatively prime. Example 4 Show that GCD of a+ b and a− b is either 1 or 2, if (,)a b =1. Solution Let (,)a+ b a − b = d ∴ d|( a+ b ) and d|( a− b ) ∴ d|( a+ b + a − b ) and d|( a+ b ) − ( a − b ) or d| 2 a and d| 2 b. i.e., d is a common divisor of 2a and 2b. ∴ d|(2 a , 2 b ) [By definition of GCD] Theory of Numbers 7 i.e., d|2 ( a , b ) [Q(,)(,)]ma mb= m a b But (,)a b =1 ∴ d | 2 ∴ d =1 or d = 2 . Example 5 Find GCD of 858 and 325 and express it in the form m 858+ n 325. Solution 858= 325. 2 + 208 …(i) Dividing 858 by 325 325= 2081. + 117 …(ii) Dividing 325 by 208 208= 1171. + 91 …(iii) Dividing 208 by 117 117= 911. + 26 …(iv) Dividing 117 by 91 91= 26. 3 + 13 …(v) Dividing 91 by 26 26= 13. 2 ∴ GCD of 858 and 325 is d =13 From Eq. (v), d =13 = 91 − 26 3. Substituting the value of 26 from Eq. (iv) ⇒ 91− 3(.) 117 − 911 =91 − 3117.. + 3 91 =4.. 91 − 3117 Substituting the value of 91 from Eq. (iii) =4208(). − 117 − 3117 =4.. 208 − 7117 Substituting the value of 117 from Eq. (ii) =4.() 208 − 7 325 − 2081 =4... 208 − 7 325 + 7 208 =11208.. − 7 325 =11(.). 858 − 325 2 − 7 325 [Putting the value of 208 from Eq. (i)] =11858... − 22 325 − 7 325 =11858.. − 29 325 =m858 + n. 325 where m=11, n = − 29 Example 6 If a and b are relatively prime, then any common divisor of ac and b is a divisor of c. Solution a and b are relatively prime ∴ ∃ integers x and y such that ax+ by =1 …(i) Let d be any common divisor of ac and b. Q d| ac, ∴ ∃ an integer m such that ac= dm …(ii) Q d| b , ∴ ∃ an integer n such that b= dn …(iii) Multiplying both sides of Eq. (i) by c acx+ bcy = c …(iv) 8 Indian National Mathematics Olympiad Putting the values of ac and b from Eqs. (ii) and (iii) in Eq. (iv). dmx+ dncy = c or d() mx+ ncy = c ∴ d | c Example 7 If a and b are any two odd primes, show that ()a2− b 2 is composite. Solution a2− b 2 =()() a − b a + b Q a.0 and b are odd primes. So, let a=2 k + 1 b=2 k ′ + 1 ∴ a− b =2 k + 1 − 2 k ′ − 1 =2k − 2 k ′ =2(k − k ′ ) is even a+ b =2 k + 1 + 2 k ′ + 1 =2k + 2 k ′ + 2 =2()k + k ′ + 1 is even ∴ Neither ()a− b nor ()a+ b is equal to 1. ∴ Neither of the two divisors ()a− b and ()a+ b of ()a2− b 2 is equal to 1. ∴ ()a2− b 2 is composite. [Q Out of the two divisors of a prime number p, one must be equal to 1] Example 8 If a|c, b|c and (,)a b =1, then ab|c. Solution Q a|c ∴There exists an integers d such that c= ad …(i) Q b|c ∴ There exists an integer e such that c= be Q (,)a b =1, therefore there exist integers m, n such that am+ bn =1 …(ii) Multiplying both sides by c acm+ bcn = c …(iii) Putting c = be from Eq. (ii) in acm and c= ad from Eq. (i) in bcn, Eq. (iii) becomes abem+ band = c ab() em+ dn = c ∴ ab| c Example 9 Ifa2− b 2 is a prime number, show thata2− b 2 = a + b, wherea, b are natural numbers. Solution a2− b 2 =()() a − b a + b …(i) Q ()a2− b 2 is a prime number ∴ One of the two factors =1 Q a− b =1 [Qa− b < a + b] Theory of Numbers 9 Q The only divisor of a prime number are 1 and itself. Eq. (i) becomes a2− b 2 =1()a+ b or a2− b 2 = a + b e.g., 32− 2 2 = 5 (which is prime) ⇒ 32− 2 2 = 3 + 2, 3 2, ∈N. Example 10 Prove that an integer is divisible by 9 if and only if the sum of its digits is divisible by 9. = K Solution Let a an a3 a 2 a 1 be an integer [Note a is not the product of a1,, a 2 a 3,…, an but a1,,, a 2 a 3 …,an are digits in the value of a. For example 368 is not the product of 3, 6 and 8 rather 3, 6, 8 are digits in value of 368 =8 + 6 × 10 + 3 × ()] 10 2 = a an …a3 a 2 a 1 = +1 +2 +3 + K + n − 1 a1 ()()()10 a2 10 a3 10 a4 ( )10 an = + + + + K a110 a 2 100 a 3 1000 a 4 = + + + + + + + a1()() a 29 a 2 a 3 99 a 3 ()...a4999 a 4 = + + + +K + + + + K ()a1 a 2 a 3 a 4 ()9a2 99 a 3 999 a 4 …(i) = + + + + or a S9(...) a2 11 a 3 111 a 4 = + + + + K where S a1 a 2 a 3 a 4 is the sum of digits in the value of a ∴ − = + + + a S9(...) a2 11 a 3 111 a 4 ∴ 9|(a− S ) …(ii) Case I a is divisible by 9 i.e., 9|a …(iii) ∴ 9| [()]a− a − S [From Eqs. (ii) and (iii)] i.e., 9|S i.e., sum of digits is divisible by 9. Case II S (sum of digits) is divisible by 9 i.e., 9|S …(iv) From Eqs. (ii) and (iv), 9|[()]a− S + S i.e., 9|a i.e., the integer a is divisible by 9. Theorem 3 Prove that the product of any r consecutive numbers is divisible byr ! . Proof Let = + +K + − Pn n( n1 )( n 2 ) ( n r 1 ) …(i) be the product of r consecutives integers beginning with n. We shall prove the theorem by Induction method. = = For r1, Pn n is divisible by 1! for all n. ∴ The theorem is true for r = 1 i.e., the product of 1 (consecutive) integer is divisible by 1!. Let us assume the theorem to be true for the product of ()r − 1 consecutive integers. 10 Indian National Mathematics Olympiad i.e., every product of ()r − 1 consecutive integers is divisible by ()!r − 1 . Changing n to n + 1 in Eq. (i) ∴ = + +K + Pn + 1 ( n1 )( n 2 ) ( n r ) Multiplying both sides by n = + +K + nPn + 1 n( n1 )( n 2 ) ( n r ) =n( n +1 )( n + 2 )K ( n + r − 1 )( n + r ) = + nPn + 1 () n r Pn = + nPn rP n − = ⋅ or n() Pn + 1 Pn r P n − = ⋅ Pn or P+ Pn r n 1 n + +K + − = ⋅ n( n1 )( n 2 ) ( n r 1 ) r [Using value of Pn ] n − = + +K + − or Pn + 1 Pn r( n1 )( n 2 ) ( n r 1 ) − = or Pn + 1 Pn r Product of ()r − 1 consecutive integers. = rP …(i) Where P denotes the product of ()r − 1 consecutive integers. But the product P of ()r − 1 consecutive integers is divisible by ()!r − 1 . [By assumption] ∴ P= k( r − )!1 ∴ − = − = − = Eq. (i) becomes Pn + 1 Pn rk()! r 1 kr( r )1 ! k()! r − ∀ i.e., r! | ( Pn + 1 Pn ), n Put n = 1 ∴ − r!| ( P2 P 1 ) = ⋅ ⋅K = But P1 1 2 3 r r ! is divisible by r ! i.e., r1!| P 1 ∴ − + r!| ( P2 P 1 ) P 1 i.e., r!| P2 Put n = 2, ∴ − r!| ( P3 P 2 ) But r!| P2 ∴ − + r!| ( P3 P 2 ) P 2 i.e., r!| P3 and so on. Generalising we can say that r!| Pn for all n. n Corollary Cr is an integer − −K − + − n = n ! = n( n1 )( n 2 ) ( n r 1 )( n r )! Cr r!( n− r )! r!()! n− r − −K − + = n( n1 )( n 2 ) ( n r 1 ) r ! the product ofr consecutive integers = = An integer r ! (QThe product of r consecutive integer is divisible by r !) . Note Therefore the product of two consecutive integers is divisible by 2!;= 2 the product of any three consecutive integers is divisible by 3= 6! and so on. Theory of Numbers 11 Example 1 Prove that product of two odd numbers of the form 4n + 1 is of the form ()4n + 1 . Solution Let a=4 k + 1, b=4 k ′ + 1 be two numbers of the form ()4n + 1 ∴ ab=()()4 k + 1 4 k ′ + 1 =16kk ′ + 4 k + 4 k ′ + 1 =4() 4kk ′ + k + k ′ + 1 =4l + 1 (where l=4 kk ′ + k + k ′) Which is in form ()4n + 1 . Example 2 Prove that square of each odd number is of the form 8j + 1 . Solution Let n=2 m + 1 be an odd number . n2=()2 m + 1 2 = 4 m 2 + 4 m + 1 =4m() m + 1 + 1 Now, m() m + 1 being product of two consecutive integers, is divisible by 2! = 2 ∴ m() m+1 = 2 j ⇒ n2 =4() 2 j + 1 =8j + 1 Example 3(a) Show that sum of an integer and its square is even. Solution Let n be any integer. So, we have to prove that n2 + n is even. ⇒ n2 + n = n() n + 1 which is product of two consecutive numbers n and n + 1 and hence divisible by 2! = 2 Hence, n2 + n is an even number. Example 3(b) If n is an integer. Prove that product n() n 2 −1 is multiple of 6. Solution n()()() n2 −1 = n n − 1 n + 1 =()()n −1 n n + 1 Which being the product of three consecutive integers is divisible by 3! = 6 ∴ n() n 2 −1 is divisible by 6. i.., e n() n 2 −1 is a multiple of 6. Note If n is a multiple of p, we shall write n= M() p . Example 4 If r is an integer, show that r()() r2 −1 3 r + 2 is divisible by 24. Solution r()()()()() r2 −1 3 r + 2 = r r − 1 r + 1 3 r + 2 =()()()r −1 r r + 1{} 3 r + 2 − 4 =3()()()()()r − 1 r r + 1 r + 2 − 4 r − 1 r r + 1 ()()()r−1 r r + 1 r + 2 being the product of four consecutive integers is divisible by 4! = 24 ∴ 3()()()r− 1 r r + 1 r + 2 is also divisible by 24. 12 Indian National Mathematics Olympiad Again ()()r−1 r r + 1 is divisible by 3! = 6 ∴ 4()()r− 1 r r + 1 is also divisible by 4⋅ 6 = 24 ∴ r()() r2 −1 3 r + 2 =3()()()()r − 1 r r + 1 r + 2 − 4 r − 1 r() r + 1 is also divisible by 24. Example 5 If m, n are positive integers, show that (m+ n )! is divisible by m!! n . + ⋅ ⋅KK + + + Solution (m n )! = 1 2 3m()()() m 1 m 2 m n m!! n m!! n + +K + = m!()()() m1 m 2 m n m!! n + +K + = ()()()m1 m 2 m n n! = The product ofn consecutive integers n! = An integer ∴ (m+ n )! is divisible by m!! n . Example 6 If ()4x− y is a multiple of 3, show that 4x2+ 7 xy − 2 y 2 is divisible by 9. Solution Q 4x− y is a multiple of 3. ∴ 4x− y = 3 m ∴ y=4 x − 3 m On putting value of y in 4x2+ 7 xy − 2 y 2 =4x2 + 7 x()() 4 x − 3 m − 2 4 x − 3 m 2 =4x2 + 28 x 2 − 21 xm −2() 16x2 + 9 m 2 − 24 xm =4x2 + 28 x 2 − 21 xm − 32 x 2 − 18 m 2 + 48 xm =27mx − 18 m2 =9m() 3 x − 2 m ∴ 4x2− 7 x − 2 y 2 is divisible by 9. Example 7 If n is an integer, prove that n()() n+1 2 n + 1 is divisible by 6. Solution n()() n+1 2 n + 1 =n()[()()] n +1 n + 2 + n − 1 =n()()()() n +1 n + 2 + n n + 1 n − 1 =n()()()() n +1 n + 2 + n − 1 n n + 1 Each of the two ()()n−1 n n + 1 and ()()n+2 n + 1 n being the product of three consecutive integers is divisible by 3= 6! ∴ n()() n+1 2 n + 1 is also divisible by 6. Example 8 Prove that 4 does not divide ()m2 + 2 for any integer m. Solution Q m is an integer. ∴ Either m is even or m is odd. Case I m is even So let m= 2 k Theory of Numbers 13 ∴ m2+2 =() 2 k 2 + 2 =4k 2 + 2 =2() 2k 2 + 1 =(2 × an odd integer) Which is not divisible by 4. Case II m is odd Let m=2 k + 1 ∴ m2+2 =() 2 k + 1 2 + 2 =4k2 + 1 + 4 k + 2 =()2 + 4k + 4 k 2 + 1. Which being an odd integer is not divisible by 4. Note Two important formulae. 1. If n is either even or odd, − − − − xn− y n =()( x − y x n1 + x n 2 y +xn3 y 2 +K + y n 1) 2. If n is odd, − − − − xn+ y n =()( x + y x n1 − x n 2 y +xn3 y 2 −K + y n 1) Example 9 Prove that 8n− 3 n is divisible by 5. − − − Solution 8n− 3 n =()(. 8 − 3 8 n1 + 8 n 2 3 + K + 3n 1) − − − or 8n− 3 n = 5(.) 8 n1 + 8 n 2 3K + 3 n 1 ∴ 8n− 3 n is divisible by 5. Example 10 If p >1 and 2p − 1 is prime, then prove that p is prime. Solution If possible, let p be not prime ∴ p is composite. (Q p >1) ∴ p= mn, where m >1 and n >1 ∴ 2p− 1 = 2 mn − 1 =(m )2 n − 1 n Putting 2m =a =an −1 n , where a =2m > 2 − − − =()()a −1 an1 + a n 2 +K + 1 n 1 Now, each of the two factors on RHS is greater than 1. ∴ 2p − 1 is composite. But it is contrary to given ∴ p must be prime. Remark p G But converse is not true i.e., when p is prime, 2− 1 need not be prime. 11 G For example, p =11 is prime but 2− 1 is divisible by 23 and hence is not prime. Theorem 4 The number of primes is infinite. Proof If possible suppose that the numbers of prime is finite. ∴ ∃ the greatest prime say q. Let b denote the product of these primes 2, 3, 5, …, q. i.e., let b=2 ⋅ 3 ⋅ 5 K q …(i) let a= b + 1 …(ii) Surely, a ≠ 1 (Qa= b +1 > 1) 14 Indian National Mathematics Olympiad ∴ The number a must have a prime say factor p i.e., p| a. Now, p is one of the primes 2, 3, 5, 7 … q (Because according to our assumption 2, 3, 5, 7, … q are the only primes). ∴ p| b (Qb= 235..K q ) Now p| a and p| b ∴ p| a− b or p |1 [Qfrom Eq. (ii), a− b = 1] p = 1 (which is impossible) (Q1 is not prime) So our supposition is false. ∴ The number of primes is infinite. Theorem 5 Fundamental Theorem of Arithmetic each natural number greater than 1 can be expressed as a product of primes in one and only one way (except for the order of the factors). Example 1 Every natural number other than 1 admits of a prime factor. Solution Suppose that n ≠1 is a natural number. If n itself is a prime number, the example is proved in as much as the prime number n is a factor of itself. If n is composite, then n must have factors other than 1 and n. Let l be the least of these factors of n other than 1 and n. i.e., 1 Example 2 Show that every odd prime can be put either in the form 4k + 1 or 4k + 3 (i.e., 4k − 1), where k is a positive integer. Solution Let n be any odd prime. If we divide any n by 4, we get n=4 k + r where 0≤r < 4 i.e., r = 0,,, 1 2 3 ∴ Either n= 4 k or n=4 k + 1 or n=4 k + 2 or n=4 k + 3 Clearly, 4n is never prime and 4n+ 2 = 2() 2 n + 1 cannot be prime unless n = 0 (Q 4 and 2 can not be factors of an odd prime) Theory of Numbers 15 ∴ An odd prime n is either of the form 4k + 1 or 4k + 3. But 4k+ 3 = 4() k + 1 − 4 + 3 = 4 k ′ − 1 (where k′ = k + 1) ∴ An odd prime n is either of the form 4k + 1 or ()4k + 3 i.e., 4k′ − 1. Note 1. Every number of the form 4k + 3 is of the form 4k − 1 and conversely. 2. Every number of the form 4k + 1 or 4k − 1 is not necessarily prime. 3. The above result should be committed to memory. Example 3 Show that there are infinitely many primes of the form 4n + 3. Solution If possible, let number of primes of form ()4n + 3 be finite. These primes are 3, 7, 11,…, q (put n = 0, 1, 2, …) Let q be the greatest of these primes of the form ()4n + 3 . Let a= 3,,,, 7 11 K q be the product of all primes of the form ()4n + 3 . Let b=4 a − 1 …(i) ⇒ b >1 [Qa ≥ 3,∴b=4 a − 1, b ≥11] ∴ By fundamental theorem b can be expressed as a product of primes say K p1.. p 2 p 3 pr . = K i.e., b p1. p 2 pr …(ii) Now, b=4 a − 1 is odd and hence 2 can't be a factor of b. ∴ None of the prime factors in RHS of Eq. (ii) is 2. i.e., Every prime factor in RHS of Eq. (ii) is odd. ∴ K + + Each of p1,,, p 2 pr is of the form ()4n 1 or ()4n 3 . K + Again, all p1,,, p 2 pr can't be of the form ()4n 1 . [Q If it were so, then b (their product) will also be of the form ()]4n + 1 . But this is contrary to Eq. (i) as b=4 a − 1 is of the form ()4n + 3 . ∴ K + At least one of p1, p 2 pr (say p) is (a prime factor of b) of the form ()4n 3 i.e., p|b. Also p|a [Q p is one prime of the form ()4n + 3 and a is product of all such primes] . ∴ p|4a ∴ p|(4 a− b ) ∴ p |1 [Q from Eq. (i), 4a− b = 1] Which is impossible [Q p being prime >1 ] ∴ Our supposition is wrong. ∴ Number of primes of the form ()4n + 3 is infinite. Theorem 6 The number of divisors of a composite number n : If n is a composite number of the α α α = 1 2 k order n p1. p 2 …pk , then the number of divisors denoted by d() n is α+ α +K α + (11 )( 2 1 ) (k 1 ) Proof Let n be any composite number, let d() n denote the number of divisors of composite number n by Fundamental Theorem of Arithmetic. n can be expressed as the product of the powers of primes. α α α = 1 2 K k K α α α n p1. p 2 pk , where p1 ,p2 ,, pk are distinct primes and 1 , 2…, k are non negative integers. 16 Indian National Mathematics Olympiad α α Q 1 2 3 K 1 p1 is a prime number, therefore, the only divisors ofp1 are 1, p1, p1 ,,, p1 p1 α 1 α= + The number of these divisors of p1 1 1 α 2 α= + Similarly, the number of divisors ofp2 2 1 α k α= + The number of divisors of pk k 1 α α α =1 ⋅ 2 K k =α + α+K α + Therefore, the total number of divisors of n p1 p 2 pk ()()()1 1 2 1k 1 α Q i ≤ ≤ [ Every divisor of pi ()1 i k is a divisor of n] =α + α + α + i.e., d( n ) (11 )( 2 1 )…()k 1 . α α α =1 ⋅ 2 ⋅ 3 Note Let n p1 p 2 p 3 [,,p1 p 2 p 3 are distinct prime numbers] Let d() n denotes number of divisor . 1. If n is a perfect square then d() n is odd Q ( all the di are even) 2. If n is not a perfect square then, d() n is even. [The number of ways of writing n are the product of two factors.] d() n + 1 If n is a perfect square, then number of ways are equal to . 2 d() n If n is not a perfect square, then number of ways are equal to . 2 Theorem 7 The sum of the divisors of any composite number n is denoted by σ()n which is equal to α+1 α +1 α + 1 p 1− 1 p 2 − 1 p k − 1 1 2 K k . − − − p1 1 p2 1 pk 1 Proof Let n be any composite number and let σ()n is sum of positive divisors of n. By Fundamental Theorem of Arithmetic n can be expressed as the product of the powers of primes. α α α ∴ = 1 2 K k n p1. p 2 pk K α αK α [where p1,,, p 2 pk are distinct primes and 1,,, 2 k are non-negative integers] Q p1 is prime number α α ∴ 1 2 K 1 divisors of p1 are only 1,,p1 p1 ,, p1 α + 1 α α 1[]p − 1 Sum of these divisors of p1 =1 + p + p2 +K + p 1 = 1 1 1 1 1 p − 1 n − Q = = =α + = a( r )1 RHS is a Geometric Progression with a 1, r p1, n 1 1 and Sn r − 1 α + 1 α ()p 2 − 1 Similarly sum of divisors ofp 2 = 2 2 − p2 1 α + 1 α ()p k − 1 Sum of divisors of p k = k k − pk 1 α α α ∴ σ = =1 ⋅ 2 k ()n Sum of divisors of n p1 p 2 ... pk α+1 α +1 α + 1 ()p 1− 1 ()p 2 − 1 ()p k − 1 = 1 . 2 K k . − − − ()p1 1 ()p2 1 (pk 1) α Q i ≤ ≤ [ Every divisor of pi ()1 i k is divisor of n] Theory of Numbers 17 Note If dk () n denotes the sum of k th power of divisor of n, then α+ α + α + ()pk ()()11 −1 ()pk 2 1 −1 ()pk ()m 1 −1 d() n = 1 ⋅ 2 K m k k − k − k − ()p1 1 ()p2 1 ()pm 1 Example 1 Find the sum of the cubes of divisor of 12. Solution 12= 22 × 3 3()() 2+ 1 − 3 1 + 1 − = 2 1 × 3 1 = dk ()12 2044 23 − 1 33 − 1 Example 2 Find the number of divisor of 600. Solution 600= 23 × 3 1 × 5 2 α= α = α = 13,, 2 1 3 2 =α + α + α + Number of divisors ()()()11 2 1 3 1 =()()()3 + 1 1 + 1 2 + 1 =4 × 2 × 3 = 24 Greatest Integer Function Greatest integer function is also known as Bracket Function. If x is any real number, then the largest integer which does not exceed x is called the integral part of x and will be denoted by []x . The function which associates with each real number x, the integer []x is often called the bracket function. For example, [],[],3= 3 − 4 = − 4 [ . ]37= 3 5 [.],,−42 = − 5 = 1 [−π ] = − 4 3 Note 1. []x is the largest integer ≤ x. 2. If a and b are positive integer, such that a= qb + r, 0 ≤r < b a r r Then, =q + , where 0< < 1 b b b ∴ a = q b a i.e., is the quotient in the division of a by b. b Properties of Greatest Integer Function (1) [][]x≤ x < x + 1 and x−1 < [ x ] ≤ x , 0≤x,[] 0 ≤ x − x < 1 (2) If x≥0,[] x = Σ 1 1 ≤i ≤ x (3) [][],x+ m = x + m If m is an integer. (4) [][][][][]x+ y ≤ x + y ≤ x + y + 1 (5) [][],x+ − x = 0 If x is an integer = − 1 otherwise. 18 Indian National Mathematics Olympiad []x x (6) = , if m is a positive integer. m m (7) −[] −x is the least integer greater than or equal to x. This is denoted as ()x (read as ceiling x) For example, (.),(.)25= 3 − 25 = − 2 (8) [.]x + 05 is the nearest integer to x. If x is midway between two integers, [.]x + 05 represents the larger of the two integers. n (9) The number of positive integers less than or equal to n and divisible by m is given by . m (10) If p is a prime number and e is the largest exponent of p such that ∞ e n p| n ! , then e = Σ i = 1 pi []a a Theorem 1 If a is real number, c is natural number, then = c c Proof Let []a= n i.e., n is largest integer ≤ a ∴ a= n + r, 0≤r < 1 …(i) []a n Let = = m c c n ∴ =m + s, where 0≤s < 1 c ∴ n= mc + cs, where 0 ≤cs < c …(ii) []a n Now, LHS = − = = m c c a n+ r RHS = = [Putting value of a from Eq. (i)] c c Putting the value of n from Eq. (ii). + + = mc cs r c cs+ r =m + c From Eq. (ii), cs ≤()c − 1 and r < 1 Adding cs+ r < c −1 + 1 ⇒ cs+ r < c cs+ r ∴ < 1 c cs+ r ∴ RHS =m + = m c ∴ LHS = RHS. Theory of Numbers 19 Theorem 2 For every positive real number x x + 1 + = []x 2 2 Proof First suppose that x=2 m + y , where m is an integer and 0≤y < 1. x [],x= 2 m = m 2 x+ 1 2 m+ 1 + y = = m 2 2 1 1 + y Since ≤ < 1 2 2 x x + 1 ∴ + = []x 2 2 Next, let x=(),2 m + 1 + y where m is an integer and 0≤y < 1. x x+ 1 ()2 m+ 2 + y Then, = m, = =m + 1 2 2 2 []x=2 m + 1 x x + 1 ∴ + = []x 2 2 The desired equality holds for all possible values of x. Example 1 Find the highest power of 3 contained in 1000!. Solution p=3, n = 1000 n = 1000 = 1 = 333 333 p 3 3 n = 333 = = []111 111 p2 3 n = 111 = = []37 37 p3 3 n = 37 = 1 = 12 12 p4 3 3 n = 12 = = []4 4 p5 3 n = 4 = 1 = 1 1 p6 3 3 n = 1 = 0 p7 3 ∴ Highest power of 3 contained in 1000! n n n n n n n = + + + + + + p p2 p 3 p 4 p5 p6 p 7 =333 + 111 + 37 + 12 + 4 + 1 + 0 = 498 20 Indian National Mathematics Olympiad n n + 1 2n Theorem 3 If n and k are positive integers and k is greater than 1, then + ≤ k k k Proof Let n= qk + r, q and r are integers and 0≤r ≤ k − 1 n r n + 1r + 1 Then, =q + , =q + , k k k k 2n 2r =2q + k k n n + 1 2n (i) r< k − 1, then = q,,= q ≥ 2q k k k The desired result is immediate. (ii) r= k − 1, then n n + 1 = q, =q + 1, k k 2n 2()k − 1 =2q + =2q + 1 k k + ∴ n + n 1 = 2n k k k From which the desired result is immediate. Theorem 4 If n be any positive integer, then show that n+ 1 n+ 2 n+ 4 n + 8 + + + +K = n 2 4 8 16 Proof We know that, x x + 1 []x = + 2 2 n n n n Applying above formula ton,,,,, K 2 4 8 16 n n + 1 []n = + 2 2 n n (/)n 2+ 1 = + 2 4 2 n n (/)n 4+ 1 = + 4 8 2 n n (/)n 8+ 1 = + 8 16 2 n n n Adding corresponding sides and cancelling out the terms ,,,… from both sides, we have 2 4 8 n+ 1 n+ 2 n + 4 n = + + + K 2 4 8 []n= n Theory of Numbers 21 Theorem 5 For every real number x 1 2 n − 1 []x+ x + +x + +K +x + = []nx n n n Proof Let x=[], x + y where 0≤y < 1 Let p be an integer such that p−1 ≤ ny < p (This is always possible because given a real number, we can always find two consecutive integers between which the number lies). k k Now, x + =[]x + y + n n k p−1 + k p+ k Also, y + lies between and n n n p−1 + k So long as < 1, n i.e., k< n −( p − )1 k y + is less than 1 and consequently n k x + = []x n k i.e., x + = []x for k = 0, 1, …, n− p n k But x + =[x ] + 1, for k= n − p + 1,…, n − 1 n 1 n − 1 ∴ []x+ x + +K +x + n n =[x ] +K + [ x ]( n − p + 1 times) +([x ] +1 ) + ([ x ] + 1 ) +K ( p − 1 ) times =n[ x ] + ( p − )1 …(i) Also, [][[]]nx= n x + ny =n[ x ] + ( p − )1 Since p−1 ≤ ny < p …(ii) From Eqs. (i) and (ii) 1 2 n − 1 []x+ x + +x + + K +x + = []nx n n n Theorem 6 The highest power of a prime number p contained in n ! is given by n n n k( n !) = + + + K p p 2 p 3 Proof Letk( n !)denote the highest power of p contained inn !n !is the product of the factors 1, 2, 3, …, n. The factors in n ! which will be divisible by p are n p,,,2 p 3 p K p p n n ∴ k( n !)= + k ! …(i) p p 22 Indian National Mathematics Olympiad n Changing n to in Eq. (i) p n n n k !! = + k …(ii) p p 2 p 2 Putting the value from Eq. (ii) in Eq. (i) n n n k( n !)= + + k ! p p 2 p 2 n n n Continuing this process, we get k( n !) = + + + K p p 2 p 3 This process must end after a finite number of steps. Congruences If a and b are two integers and m is a positive integer, then a is said to be congruent to b modulo m, if m divides a− b denoted by m| ( a− b ). In notation form we express it as a≡ b mod m or a− b ≡ 0 mod m. Note 1. a≡ b mod m, then m|( a− b ) or ()a− b is a multiple of m. 2. If m|( a− b ) [m does not divides ()],a− b then a is said to be incongruent to b mod m and this fact is expressed as a is not congruent to b mod m. 3. If m| a, then a ≡ 0 mod m For example : (i) 13≡ 1 mod 4 (Q 4 | ())13− 1 = 12 (ii) 4≡ − 1 mod 5 (Q 5|( 4− ( − 1 )) = 5 ) (iii) 12≡ 0 mod 4 (Q 4|12) (iv) 17 is not congruent to 3 mod 5 (Q 5| ( 17− 3 )) Theorem 7 If a≡ b mod m, then (i) a+ c = b + c mod m (ii) ac≡ bc mod m, where c is any integer. Proof (i)Qa≡ b mod m m | (a− b) ⇒ m | {()()a+ c − b + c }[Qa+ c −()] b + c = a − b a+ c ≡ b + c mod m (ii)Qa≡ b mod m m| ( a− b ) m| c ( a− b ) m| ( ac− bc ) ∴ ac≡ bc mod m Note The converse of theorem 15 (ii) is not true. Theorem states that if a≡ b mod m, then ac≡ bc mod m. i.e., a congruence can always be multiplied by an integer But the converse is not true i.e.,. It is not always possible to cancel a common factor from a congruence. For example : 16≡ 8 mod 4 [Q 4|16 – 8] But if we cancel the common factor 8 from numbers 16 and 8, we get 2≡ 1 mod 4 which is a false result because 4|( 2− 1 ) Theory of Numbers 23 Theorem 8 If a≡ b mod m and c≡ d mod m, then (i) a+ c ≡ b + d mod m (ii) a− c ≡ b − d mod m (iii) ac≡ bd mod m Proof (i)Qa≡ b mod m and c≡ d mod m ∴ m| ( a− b ) and m| ( c− d ) m| (( a− b ) + ( c − d )) or m| (( a+ c ) − ( b + d )) ∴ a+ c ≡ b + d mod m (ii) a≡ b mod m and c≡ d mod m ∴ m| ( a− b ) and m| ( c− d ) ∴ m| ( a− b ) − ( c − d ) or m| ( a− c ) − ( b − d ) a− c ≡ b − d mod m (iii)Qa≡ b mod m and c≡ d mod m m| ( a− b ) and m| ( c− d ) ∴ There exists integer h and k such that a− b = mh and c− d = mk a= b + mh and c= d + mk Multiplying the two equations, we get ac=( b + mh )( d + mk ) =bd + mbk + mhd + m2 hk ac− bd = m() bk + hd + mhk ∴ By definition of divisibility m| ( ac− bd ) or ac≡ bd mod m Corollary If a≡ b mod m, then a2≡ b 2 mod m Q a≡ b mod m and again a≡ b mod m Multiplying the two congruence a2≡ b 2 mod m Theorem 9 (i) Prove that a≡ a mod m i.e., every integer is congruent to itself. (ii) If a≡ b mod m, then prove that b≡ a mod m (iii) If a≡ b mod m, b≡ c mod m prove that a≡ c mod m Proof (i) We know that m|0 ()m ≠ 0 m| ( a− a ) a≡ a mod m [by definition of congruence] (ii) Let a≡ b mod m m| ( a− b ) ∴ m|− ( a − b ) or m| ( b− a ) ∴ b≡ a mod m (iii) Let a≡ b mod m and b≡ c mod m ∴ m| ( a− b ) and m| ( b− c ) ∴ m| ( a− b ) + ( b − c ) or m| ( a− c ) ∴ a≡ c mod m 24 Indian National Mathematics Olympiad Theorem 10 If a≡ b mod m, then ak≡ b k mod m for every positive integer k. Proof We know that, − − − − ak− b k =( a − b )( a k1 + a k 2 b + a k 3 b 2 + K + b k 1 ) …(i) But a≡ b mod m ⇒m| ( a− b ) ∴ There exists an integer t such that a− b = mt …(ii) Putting this value of ()a− b from Eq. (ii) in Eq. (i) − − − ak− b k = mt( a k1 + a k 2 b + K b k 1 ) ∴ m| ( ak− b k ) ∴ ak≡ b k mod m ≡ =n + n−1 + n − 2 +K + + Theorem 11 If a b mod m and f() x p0 x p 1 x p2 x pn − 1 x pn is an integral rational function of an indeterminate x with integral coefficients, then f()() a≡ f b mod m Q =n + n−1 + n − 2 + K + + Proof f() x p0 x p 1 x p2 x pn − 1 x pn Putting x= a ∴ =n + n−1 + n − 2 + K + + f() a p0 a p 1 a p2 a pn − 1 a pn …(i) Putting x= b ∴ =n + n−1 + n − 2 + K + + f() b p0 b p 1 b p2 b pn − 1 b pn …(ii) Subtract Eq. (i) from Eq. (ii), we get − =n − n + n−1 − n − 1 + K + − f()()()() a f b p0 a b p 1 a b pn − 1 () a b = −n−1 + n − 2 +K + n − 1 + p0 ( a b )( a a b b ) −n−2 + n − 3 +K + n − 2 +K + − p1 ( a b )( a a b b ) pn − 1 () a b − = −n−1 + n − 2 + K +n−2 + n −2 + n − 3 +K + n − 2 or f()()()[( a f b a b p0 a a b b)() p1 a a b b +K + pn − 1 ] …(iii) =()a − b t (say) But a≡ b mod m (given) m| ( a− b ) ∴ ∃ an integer k such that a− b = mk Putting this value of a− b = mk in Eq. (iii) f()() a− f b = mkt ∴ m| f ( a )− f ( b ) ⇒ f()() a≡ f b mod m Theorem 12 Fermat Theorem If p is prime, then ()()a+ bp = a p + b p mod p. Proof Expanding by binomial theorem +p = p + p p−1 + p p − 2 2 + K +p p− 1 + p ()a b a C1 a b C2 a b Cp − 1 ab b p − 1 +p = p + p + Σ p p− r r or ()()a b a b Cr a b …(i) r = 1 Theory of Numbers 25 p = p ! ≤ ≤ − Now, Cr ;()1r p 1 r!( p− r )! But p !..= 123 … p is divisible by p p is coprime to r ! Q p is coprime to 1, 2, 3, … r (Qr< p, p is prime) ∴ p is co prime to their product = r ! Also for the same reason p is coprime to ()!p− r ∴ p = p ! Cr is divisible by p. r!( p− r )! ∴ ∃ an integer kr such that p = Cr pk r p Putting this value of Cr in Eq. (i) p − 1 +p − p + p = Σ p− r r ()()a b a b p kr a b r =1 which is divisible by p. ∴ ()()a+ bp ≡ a p + b p mod p Generalization If p is a prime number, prove that + + +K + p ()a1 a 2 a 3 an ≡p + p + p +K + p ()a1 a 2 a 3 an mod p + + +K +p = + p ()()a1 a 2 a 3 an a 1 b 1 = + +K + where b1 a 2 a 3 an ≡p + p ()a1 b 1 mod p. ≡p + + +K + p [()]a1 a2 a 3 an mod p ≡p + + p [()]a1 a2 c 2 mod p = + +K + ≡p + p + p where c2 a 3 a 4 an ()a1 a 2 c 2 mod p continuing like this, we get + + +K + p ≡p + p +K + p ()a1 a 2 a 3 an ()a1 a 2 an mod p Theorem 13 If p prime number, then ap = a mod p Proof We know that, + + +K + p ≡p + p + p K + p ()a1 a 2 a 3 an (a1 a 2 a 3 an ) mod p …(i) = = =K = = Putting a1 a 2 a 3 an 1 in Eq. (i) ()1+ 1 + 1 +K + 1 p ≡()1p + 1 p + 1 p +K + 1 p mod p or np ≡()1 + 1 + 1 +K + 1 mod p or np ≡ n mod p for every natural number n. Replacing n by a. ap = a mod p 26 Indian National Mathematics Olympiad Theorem 14 Fermat Little Theorem − If p is a prime number and (,)a p ≡ 1, prove that ap 1 = 1 mod p. Proof As p is prime. ∴ ap = a mod p Cancelling a from both sides. [∴a is coprime to p] − We have ap 1 ≡ 1 mod p. =n − n − n + n − n − + −n− 2 n n + − n− 1 n Theorem 15 n!()()... n C1 n1 C 2 n 2 ()()1CCn − 2 2 1 n − 1 Proof Expanding by binomial theorem x− n = nx − n()() n−1 x + n n − 2 x−K +n −n− 2 2 x + n − n−1 x + − n ()e1 e C1 e C2 e Cn − 2 ()()1 e Cn − 1 1 e ( )1 …(i) 2 3 θ θ θ θ We know that e =1 + + + + K 1!!! 2 3 θ Using this expansion of e , Eq. (i) becomes n x x2 x 3 x n 1 + + + +KK + + − 1 1!!!! 2 3 n 2 n − − 2 − n = +nx +() nx +KK +() nx + −n + ()n1 x+ [( n1 ) x ] +K + [(n1 ) x ] + K 1 C1 1 1! 2! n ! 1! 2! n ! − − 2 − n +n + ()n2 x+ [( n2 ) x ] +KK+ [(n2 ) x ] + C2 1 1! 2! n ! 2 n +K +n −n − 2 + 2x+()2 x + KK+(x )2 + Cn − 2 ()1 1 1! 2! n 2 n + −n − 1n +x + x +KK + x + + − n ()1Cn − 1 1 ( )1 1!!! 2 n Comparing coefficient of x n on both sides nn ()n − 1n ()n − 2 n 1 = − nC + nC −K + n ! 1n ! 2 n ! − n − n1 n n− 2 n 2 ()1 Cn − 1 ()−1 C − + n 2 n ! n ! Multiplying both sides by n ! =n − n − n + n − n −K + −n− 2 n n + − n− 1 n n!()() n C1 n1 C 2 n 2 ()()1CCn − 2 2 1 n − 1 Theorem 16 Wilson Theorem If p is prime, then ()!p −1 + 1 ≡ 0 mod p Proof Case I when p = 2 ()!2− 1 + 1 ≡ 0 mod 2 [Putting p = 2 in ()!p −1 + 1 ≡ 0 mod p ] ⇒ 1! + 1 ≡ 0 mod 2 ⇒ 2≡ 0 mod 2 which is true ∴ Wilson theorem is true for p = 2 . Theory of Numbers 27 Case II If p is an odd prime. =n − n − n + n − n − + −n− 2 n n + − n− 1 n n!()()... n C1 n1 C 2 n 2 ()()1CCn − 2 2 1 n − 1 Put n= p − 1 on both sides ∴ − = −p−1 − p − 1 − p − 1 ()!()()p1 p 1 C1 p 2 +p−1 − p −1 + + − p − 3 p −1 p − 1 + − p−2 p − 1 C2 ()..() p 3 1 Cp − 32 ( )1 Cp − 2 …(i) Q p is prime. ∴ p is coprime to all numbers < p. i.e., p is coprime to p−1,,, p − 2 p − 3 … 2, 1. ∴ Putting a= p −1,,, p − 2 p − 3 … 2 in Fermat Theorem − ap 1 ≡ mod p − ()p −1p 1 ≡ 1 mod p − or ()p −1p 1 − 1 = M(p) − ()()p−1p 1 = M p + 1 − Similarly, ()()p−2p 1 = M p + 1 − ()()p−3p 1 = M p + 1 ………………… ………………… ………………… − 2p 1 =M() p + 1 − − − Putting these values of (),()p−1p1 p − 2 p 1, …, 2p 1 in Eq. (i). ∴ − = + −p − 1 + +p − 1 + + + − p−3 p − 1 ()![()][()]p1 M p 1 C1 M p 1 C2[()]..() M p1 1 Cp − 3 + + − p−2 p − 1 [()]()M p1 1 Cp − 2 p − 1 − = + −p−1 + p − 1 − K + −p − 3 + − p−2 p − 1 or ()!()p1 M p 1 C1 C2 ()()1CCp − 3 1 p − 2 − Adding and Subtracting ()−1 p 1 in RHS. − = +p−1 − p − 1 +p−1 − p −1 +K + − p −3 p − 3 (p1 )! M ( p ) [( 1 ) C1 CCC2 3 ()1 p − 3 + − p−2 p − 1 + −p−1 − − p − 1 ( )1 Cp − 2 ()]()1 1 − − ()!()()()p−1 = M p + 1 − 1p1 − − 1 p 1 − ∴ ()!()()p−1 = M p + 0 − − 1 p 1 =M( p ) − 1 Qp is odd. ∴p − 1 is even. − ∴ ()−1p 1 = 1 or ()!()p−1 + 1 = M p ()!p −1 + 1 is divisible by p. ∴ ()!p −1 + 1 ≡ 0 mod p. Theorem 17 Converse of Wilson Theorem If p > 1 and ()!p −1 + 1 ≡ 0 mod p, then p is a prime number. Proof If possible let p be not prime. ∴ p is composite (Qp > 1). = < < < < < ≤ − < ≤ − So let p p1 p 2, where (,)1p1 p 1 p 2 p or 1p1 p 1,1p2 p 1 28 Indian National Mathematics Olympiad < ≤ − Now, 1p1 p 1 ∴ − − p1 is one of the factors in the value of ()!p 1 and therefore p1 divides ()!p 1 . = Also p p1 p 2 …(i) ⇒ p1 | p …(ii) But ()!p −1 + 1 ≡ 0 mod p ∴ p| ( p −1 )! + 1 …(iii) From Eqs. (ii) and (iii) − + p1 | ( p 1 )! 1 …(iv) From Eqs. (iv) and (i) − + − − p1 | ( p1 )! 1 ( p 1 )! i.e., p1 1| Q > But this is impossible. ( p1 1) ∴ p is a prime number. Euler’s Function Definition : The number of integers≤ n and coprime to n is called Euler's function for n and is denoted by φ ().n Examples φ( )1 = 1 [Q1 is the only integer ≤ 1 and coprime to 1]. φ( )2 = 1 [Q1 is the only integer < 2 and coprime to 2]. φ( )8 = 4 [Q1, 3, 5, 7 are the only four integers < 8 and coprime to 8]. Remark G If p is a prime number, then 1, 2, 3, …()p −1 are all less than p and coprime to p and are()p −1 in total. ∴ φ()p = p −1 1 1 1 Theorem 18 Prove that φ()n = n 1 − 1 − K 1 − where p,,..., p p are distinct prime 1 2 r p1 p 2 pr factors of n. Q K Proof p1,,, p 2 pr are distinct prime factors of n. k k ∴ = 1 2 K kr n p1. p 2 pr k k k k ∴ φ = φ 1 2 kr = φ1 φ 2 K φ kr ()(....)n p1 p 2 pr ()()()p1 p 2 pr Q K φ = φ φ [ p1,,, p 2 pr are distinct primes and hence are coprime to each other and ()()()ab a b , if a and b are coprime to each other.] k 1 k 1 1 =p 11 − p 2 1 − Kp kr 1 − 1 2 r p1 p2 pr k k 1 1 1 =p1. p 2 K p kr 1 − 1 − K 1 − 1 2 r p1 p 2 pr 1 1 1 k k =n 1 − 1 − K 1 − [.]QKn= p1 p 2 p kr 1 2 r p1 p 2 pr Theory of Numbers 29 1 Theorem 19 Prove that φ()pk = p k 1 − , where p is prime. p − Proof Number of integers from 1 to p k which are not coprime to p k are p.1, p.2, p.3,… p. p k 1. − Total number of such integers, which are not coprime to pk= p k 1. ∴ φ()p k = Number of integers coprime to p k and < p k. − =pk − p k1 = p k (/)1 − 1 p Remark If a and b are coprime to each other, then φ()()()ab = φ a φ b . Example 1 Find the number of positive integers ≤ 3600 that coprime to 3600. Solution n =3600 = 24 × 3 2 × 5 2 φ()()()n = φ3600 = φ 24 × 3 2 × 5 2 1 1 1 1 1 1 =n1 − 1 − 1 − =3600 1 − 1 − 1 − p1 p 2 p 3 2 3 5 = = = [Here p1 2, p23,] p 3 5 1 2 4 =3600 × × × 2 3 5 ∴ φ()3600 = 960 Example 2 If m > 2, show that φ ()m is even. Solution If (,)a m =1, then (,)m− a m =1 ∴ Integers coprime to m occur in pairs of type a and m− a . ∴ φ ()m is even. Example 3 For what values of m is φ ()m odd. Solution If m>2, φ ( m ) is even. φ( )1 = 1 φ( )2 = 1 Only for m=1, m = 2 φ ()m is odd. 10n− 1 10 n − 1 Concept Let a = = 10− 1 9 10n − 1 We can express any a of the form in terms of perfect square. 9 10n − 1 a = ⇒ 9a = 10n − 1 9 9a + 1 = 10n Let b=9 a + 1 c=8 a + 1 Now, consider 4ab+ c =4a() 9 a + 1 + 8 a + 1 =36a2 + 12 a + 1 =()6a + 1 2 30 Indian National Mathematics Olympiad Verification ()6× 1 + 12 = 7 2 = 49 =()6 × 11 + 12 = 67 2 ()6× 111 + 12 = 667 2 Now, consider ()a−1 b + c =(a −1 )( 9 a + 1 ) + 8 a + 1 =9a2 + a − 9 a − 1 + 8 a + 1 =9a2 = () 3 a 2 Verification a = 1 ⇒32 a = 11 ⇒ ( )33 2 Now, consider ()16ab+ c 16a() 9 a+ 1 + 8 a + 1 ()12a + 1 2 This is also a perfect square. Concept Prove that every number of the sequence 49, 4489, 44489, 4448889 is a perfect square. If there are n fours and ()n − 1 eight and one 9. Let us denote 444889 as 43 8 2 9. Consider 667 written as 62 7 We know 444889 = ()667 2. ∴ 2 = We develop ()6nm −1 7 4n 8n − 1 9 If this is true then 2 2 2 n − ×n + n + 2 = 6() 10 1 + = 6 10 3 = 2. 10 1 ()6n − 1 7 1 9 9 9 2n+ n + 40 40+ 1 = 410.. 410 1 = n−1 n − 1 = 4n 8− 9 9 9 n 1 Example 1 Let n be the natural number. If 2n + 1and 3n + 1are perfect square. Then prove that n is divided by 40. Solution 40= 23 × 5. It is sufficient to prove that n is divisible by 8 and 5. Let 2n+ 1 = x 2 …(i) and 3n+ 1 = y 2 …(ii) ⇒ x 2 is odd. ⇒ x is odd. Let x=2 a + 1 ()()2n+ 1 = 2 a + 1 2 2n+ 1 = 4 a2 + 4 a + 1 n=2 a2 + a ⇒ n is even. If n is even ⇒ 3n + 1 is odd ⇒ y 2 is odd ⇒ y is odd Let y=2 b + 1 Subtract Eq. (i) from Eq. (ii), we get n= y2 − x 2 …(iii) Theory of Numbers 31 ⇒ n=()()2 b + 12 − 2 a + 1 2 We know square difference of odd number is always divisible by 8. ∴ n is divisible by 8. …(iv) If we eliminate n between 1 and 2 3x2− 2 y 2 = 1 Since square of odd number ends with 1, 5 or 9 ∴ 3x 2 ends with 3, 5 or 4, 7 ⇒ 2y 2 ends with 2, 0, 8 ⇒ x 2 ends with 1 and y 2 ends with 1 ⇒ n= y2 − x 2 [from Eq. (iii)] = 0 ∴ It is divisible by 5. Example 2 Prove that there are infinitely many squares in the sequence 1, 3, 6, 10, 15, 21, 28,… Solution SupposeTn is a square n() n + 1 LetT of the above sequence be n 2 n( n + )1 ⇒ T = n 2 = 2 If it is a square thenTn () m n() n + 1 ⇒ = ()m 2 2 ⇒ n()() n+1 = 2 m 2 Also,T4n n + 1() is also a square. 4n()[()()] n+ 1 4 n n + 1 + 1 ∴ T + = 4n() n 1 2 4()[] 2m2 4 n 2 + 4 n + 1 = =4m2() 2 n + 1 2 2 T4n n + 1()is also a perfect square. ∴ = perfect squares areT1 1 = T8 36 is a perfect square. T288 is also a perfect square. Example 3 If N =123 × 3 4 × 5 2, find the total number of even factor of N. Solution If N =123 × 3 4 × 5 2 Then, N =26 × 3 7 × 5 2 ∴ Total Number of factors are =()()()6 + 1 7 + 1 2 + 1 = 7 × 8 × 3 =168 In above factors, some of these are odd multiple and some are even. The odd multiples are formed only with combination of 35 and 55. So total number of odd multiples is ()()7+ 1 2 + 1 = 24 ∴ Even multiples =168 − 24 = 144 32 Indian National Mathematics Olympiad Example 4 Show that n2 −3 n − 19 is not a multiple of 289 for any integer n. Solution Suppose172|n 2 − 3 n − 19 Since n2 −3 n − 19 =()() n + 7 n − 10 + 51 17|(n+ 7 )( n − 10 ); ∴ 172 |(n+ 7 )( n − 10 ) (Q n+7 ≡ n − 10 (mod 17 )) ⇒ 172|(n 2 − 3 n − 19 ) − ( n + 7 )( n + 10 ) i.e.,172 | 51 which is a contradiction Consequently n2 −3 n − 19 is not a multiple of 289. Example 5 Determine all integers n such that n4− n 2 + 64 is the square of an integer. Solution Since n4− n 2 +64 > n 4 − 2 n 2 + 1 =()n 2 −1 2 for some non negative integer k, ()()n4− n 2 +64 = n 2 + k 2 = n 4 + 2 n 2 k + k 2 64 − k 2 i.e., n 2 = from which we find that the possible values 64, 1, 0 for n 2 are 2k + 1 obtained when k = 0,, 7 8 respectively. Hence, n ∈(,,)0 ± 1 ± 8 Example 6 Let a,,, b c d, e be consecutive positive integers such that b+ c + d is a perfect square and a+ b + c + d + e is a perfect cube. Find smallest possible value of c. Solution a, b, c, d, e are consecutive positive integerb+ c + d = 3 c anda+ b + c + d + e = 5 c Now, 3|3c ⇒ 32 | 3c (Q 3c is a square) ⇒ 3|c ⇒ 3|5c ⇒ 33 | 5c (Q 5c is a cube) ⇒ Also 5 5| c ⇒ 53 | 5c ⇒ 52 |c ∴ 33 5 2|c i.e., 675|c ∴ 675 being a possible value of c is the smallest of such numbers. Example 7 If 11+ 11 11a 2 + 1 is an odd integer where a is a rational number. Prove that a is perfect square. Solution Let λ =11 + 11 11a 2 + 1 Then, ()()λ −112 = 11 2 11a 2 + 1 Simplifying, we get λ() λ −22 = 113a 2 r Putting |a |= ,r, s∈ N such that (,)r s =1 gives λ() λ −22s2 = 11 3 r 2. s Since112|s because otherwise 11 would divide r,11|λ. Writing11µ λ= , we get µ() µ −2s2 = 11 r 2 Since 11|s for otherwise we would have 11|r. It follows that s =1. Thus we have µ() µ −2 = 11r 2. Since µ − 2 and µ are consecutive odd integers they are relatively prime. If11|µ − 2, then µ is a square of form11n + 2 which is not possible. ∴11|µ and hence µ =11n 2 for some n∈ N. Thus, we have λ=11 µ = 112n 2 Theory of Numbers 33 Example 8 Determine all pairs of positive integers (,)m n for which 2m+ 3 n is a perfect square. Solution Let 2m+ 3 n = k 2 Since ()−1m ≡ 2 m ≡k 2 ≡ 1 (mod 3) (Q 3 |k ) m is even, say 2p. Now, ()k − 2p (k + 2p ) = 3n + ⇒k −2p = 1 and k +2p = 3 n ⇒ 2p1 + 1 = 3 n + Since ()−1n ≡ 3 n (mod 4) =2p 1 + 1 ≡ 1, n is even, say 2q. + Now ()()3q− 1 3 q + 1 = 2 p 1 ⇒ 3q − 1 = 2 ⇒ 3q = 3 ⇒q =1 and hence p = 2 So, we have only one solution (4, 2). Example 9 Determine the set of integers n for which n2 +19 n + 92 is a square. Solution Let n2+19 n + 92 = m 2, m is a non negative integer. Then, n2+19 n + 92 − m 2 = 0 1 Solving for n, we get n=() −19 ± 4 m2 − 7 2 ∴ 4m2 − 7 is a square i.e., 4m2− 7 = p 2 Where p∈ N ∴ ()()2m− p 2 m + p = 7 ∴ 2m+ p being positive therefore (2m+p) is 7 and 2m− p = 1 Hence, 4m = 8 ⇒m = 2 Thus, we have n2 +19 n + 92 = 4 ⇒ n2 +19 n + 88 = 0 ⇒ ()()n+8 n + 11 = 0 ⇒ n = − 8 or –11 Example 10 Find n, if 2200− 2 192 ⋅ 31 + 2n is a perfect square. Solution 2200− 2 192 31 + 2n = 2 192() 2 8 − 31 + 2 n =2192() 256 − 31 + 2n =2192 ⋅ 225 + 2n ∴ For some m∈ N 2n =m2 − 2 192 ⋅ 225 =m2 −()2 96 ⋅ 15 2 =()()m −296 ⋅ 15 m + 2 96 ⋅ 15 α α+ β So, m =296 ⋅ 15 = 2 and m +296 ⋅ 15 = 2 for some non negative integers α, β . +αβ α α β Hence, 297 ⋅ 15 = 2 − 2 =2() 2 − 1 α β ⇒ 2= 297 and 2− 1 = 15. i.e., α = 97 and β = 4 ∴ n =2α + β = 198 Example 11 Find the number of values of n for which 211+ 2 8 + 2n is a perfect square. Solution We can write 211+ 2 8 + 2n as 28() 2 3 + 1 + 2n ⇒ 28 ⋅ 9 + 2n − ⇒ 2n() 28 n ⋅ 9 + 1 34 Indian National Mathematics Olympiad − Note that for any k < 8, 2k() 28 k 9+ 1 is not a square, when k is odd, 2k is not a square and in the other case, the second factor is not a square. Hence n ≥ 8. Now write − 211+ 2 8 + 2n as 28() 9+ 2n 8 . Then the problem is to find the number of non negative integers k such that 9+ 2k is a square. 9+ 2k = t 2 ⇒ 2k =()()t − 3 t + 3 + ⇒t −3 = 2p and t +3 = 2p q for some non negative integers p and q. ∴2p() 2 q − 1 = 6 implying p =1 from which it follows that t = 5. Hence, there is a unique solution. Example 12 Find all positive integers n for which n 2 + 96 is a perfect square. Solution Suppose m is a positive integer, such that n2+96 = m 2 Then, m2− n 2 =96 ⇒()()m− n m + n = 96 sincem− n < m + n andm− n,m+ n must be both even [asm+ n =( m − n ) + n .2 Thereforem− n,m+ n must be both odd or both even; also if both of them are odd, then the product cannot be even. ∴ Only possibilities are m− n =2, m + n = 48 ⇒m=25, n = 23 m− n = 4,m+ n = 24 ⇒m=14, n = 10 m− n =6, m + n = 16 ⇒m=11, n = 5 m− n = 8,m+ n =12 ⇒m=10, n = 2 Example 13 Give with justification, a natural number n for which 39+ 3 12 + 3 15 + 3n is a perfect cube . − Solution 39+ 3 12 + 3 15 + 3n = 3 9( 1 + 3 4 + 3 6 + 3n 9 ) − =()()()33 3{ 1 + 3 ⋅ 3 2 + 3 3 2 2 + 3 2 3 + 3n 9 − 3() 32 2} − =()()33 3 1 + 3 2 3 provided 3n 9− 3 5 = 0 − = ()270 3 provided 3n 9= 3 5 i.e., provided n =14 So, given number is a perfect cube when n =14 Example 14 Prove that 2p+ 3 p is not a perfect power if p is a prime number. Solution If p =2, 2p + 3 p = 22 + 3 2 = 13 (not a perfect power) Let now p be a prime > 2.. x+ a divides xp+ a p, whenever p is odd [factor theorem] ∴ 2p+ 3 p is divisible by 2+ 3 = 5. We shall show that 2p+ 3 p is not divisible by 52. − − − − xp+3 p =()( x + 3 x p1 − 3 x p 2 +32x p 3 +K +()) − 3 p 1 When x = − 3, then − − − xp1−3 x p 2 +K +() − 3 p 1 − − − =()()() −3p1 − 3 − 3 p 2 +K + − 3 p 1 − = p3p 1 Theory of Numbers 35 Showing that x + 3 does not divide − − − − xp1−3 x p 2 + 3 2 x p 3 +K +() − 3 p 1 Consequently ()x + 3 2 does not divide xp + 3 p. So, ()2+ 3 2 does not divide 2p+ 3 p. Since 2p+ 3 p is a multiple of 5 but is not a multiple of 52. ∴ it cannot be a perfect power. Example 15 A 4 digit number has the following properties (I) It is a perfect square (II) its first 2 digit are equal to each other (III) its last two digit are equal to each other.Find all such four digit number. Solution We want to find positive integers x and y such that1≤x ≤ 9,0≤y ≤ 9 and xxyy is a perfect square. Since,102= 100100, 2 = 10000. It follows that xxyy must be the square of a 2 digit number. Suppose that ()ab2 = xxyy. The number xxyy is clearly a multiple of 11. Since, it is a perfect square it must be a multiple of112 i.e., 121. ∴It must be of the form 121× 1121,, × 4121 × 9,121× 16,121× 25, 121× 36,121× 49,121× 64,121× 81 Out of these121× 64 i.e., 7744 is of the form xxyy, we conclude that 7744 is the desired number. Example 16 Show that for any integer n, the number n4−20 n 2 + 4 is not a prime number. Solution n4−20 n 2 + 4 =() n 4 − 4 n 2 + 4 − 16 n 2 =()n2 −2 2 − 16 n 2 =()()n2 −4 n − 2 n 2 + 4 n − 2 …(i) Note It can be easily seen that none of the factors n2 −4 n − 2,n2 +4 n − 2 can have the value ± 1, whatever integral value n may have. Here four cases arises. 4± 28 (i) n2 −4 n − 2 = 1 ⇒n = 2 4± 20 (ii) n2 −4 n − 2 = − 1 ⇒n = 2 −4 ± 28 (iii) n2 +4 n − 2 = 1 ⇒n = 2 −4 ± 20 (iv) n2 +4 n − 2 = − 1 ⇒n = 2 From the above four cases, we find that whatever integral value n may have, n4−20 n 2 + 4 is the product of the integers n2 −4 n − 2 and n2 +4 n − 2 neither of which equals ± 1.. Example 17 Prove that the product of four consecutive natural numbers cannot be a perfect cube. Solution Consider the product P= n( n + )1(n+2 )( n + 3 ), where n is a natural number. If possible, that P is a perfect cube = k 3 Two cases arises. Case I If n is odd. n,(),() n+1 n + 3 are all prime to n + 2 Now, we know that every common divisor of n+ p and n+ q must divide q− p. 36 Indian National Mathematics Olympiad ∴ n + 2 and n()() n+1 n + 3 are relatively prime. Since, their product is a perfect cube, each of them must be a perfect cube. Since, n3 < n()()() n +1 n + 3 < n + 3 3 ∴ n()()() n+1 n + 3 = n + 1 3 or ()n + 2 3 As n()() n+1 n + 3 and ()n + 2 3 are relatively prime, so second possibility ruled out. Also n()()() n+1 n + 3 = n + 1 3 ⇒n =1. Since P = 24, when n =1 which is not a perfect cube. So the possibility n =1 is also ruled out. So n cannot odd. Case II If n is even. Then n + 1 is prime to n, n + 2 and n + 3. Consequently n + 1 is relatively prime to n()() n+2 n + 3 . Since the product of relatively prime numbers n + 1 and n()() n+2 n + 3 is a perfect cube, each of them must be a perfect cube. n3 < n()()() n +2 n + 3 < n + 3 3 ∴ n()() n+2 n + 3 = either ()n + 1 3 or ()n + 2 3since n()() n+2 n + 3 and n + 1 are relatively prime ∴ First possibility ruled out. Also n()()() n+2 n + 3 = n + 2 3 ⇒n +4 = 0 which is out of question. Consequently n cannot be even. Thus, we find that the product of 4 consecutive integers cannot be a perfect cube. − Example 18 Find all primes p for which the quotient (2p 1 − 1 )| p is a square. Solution Supposem() m+1 = 7 n 2,m andn are integers sincem andm + 1are relatively prime. ∴ m and m + 1 must be the numbers 7p2, q 2 (in some order) p andq are relatively prime and pq= n; Since the product of 2 consecutive integer is even. ∴ m() m + 1 is even, which means that one of the numbers m, m + 1 must be even. Suppose m= q 2 (so that m+1 = 7 p2 ). Since every square number is of one of the forms 4k, 4 k + 1. Consequently m + 1 must be of one of the forms 4k + 1, 4k + 2. However this is not possible for if p is even, then 7p2 is of the form 4k. If p is odd, then 7p2 is of the form 4k + 3. ∴ m+1 ≠ 7 p2. So m= 7 p2 and m+1 = q 2 Concept of Finding Number of Positive Integral Solutions for the Equation of the Form x2+ y 2 = k We know that, (n )22 ≡ 0 mod 4 and ()2n + 12 ≡ 1 mod 4 Now, if (a) x and y are both even then, x2+ y 2 ≡ 0 mod 4 (b) x and y are both odd then, x2+ y 2 ≡ 2 mod 4 Theory of Numbers 37 (c) one is even and other is odd then, x2+ y 2 ≡ 1 mod 4 {∴x2+ y 2 ≡ 0, 1, 2 mod 4 andx2+ y 2 ≡/ 3 and 4} the above discussion implies, if x2+ y 2 = k and if k is of the form of (4m + 3 ), then x2+ y 2 = k does not have any integral solution. e.g., suppose, we are asked to find integral solution for equation x2+ y 2 = 19, then it will not have any integral solution because 19 is of the form ()4m + 3 . Now, if we have x4+ y 4 = k, then we know that (n )24 ≡ 0 mod 16 and ()2n + 14 ≡ 1 mod 16 Again, if (a) x and y are both even then, x4+ y 4 ≡ 0 mod 16 (b) x and y are both odd then, x4+ y 4 ≡ 2 mod 16 (c) one is even and other is odd, then x4+ y 4 ≡ 1 mod 16 ∴ x4+ y 4 ≡ 0,, 1 2 mod 16 and x4+ y 4 ≡/ i mod 16, where i = (,,,..,)3 4 5 15 So, the above discussion implies, if x4+ y 4 = k and k is of the form ()16m+ i , wherei = 3,,,, 4 5K 15, then x4+ y 4 = k will not have any integral solution. e.g., suppose we are asked to find integral solutions for equation. x4+ y 4 = 16003, then it will not give any integral solution because 16003 is of the form ()16m + 3 . The concept can be extended to more than two variables expression, suppose the equation is 4 +4 +4 +4 + +4 = x1 x2 x3 x4 ... x14 1599 now, we know that (n )24 ≡ 0 mod 16 and ()2n + 14 ≡ 1 mod 16 14 ∴ Σ 4 ≡ K xi 0,, 1 2 14 mod 16 i = 1 but our RHS is 1599 ≡ 15 mod 16. ∴ No integral solution can be obtained for the above equation. Reason [If all the variables are considered to be odd, then maximum remainder which can come out is “14” and if any of the variable is an even number then remainder will be less than 14.] Now, let us consider another discussion. If we have a2+ b 2 + c 2 = a 2 b 2 we know, (n )22 ≡ 0 mod 4 and ()2n + 12 ≡ 1 mod 4 38 Indian National Mathematics Olympiad Case I If a, b and c all are odd then, a2+ b 2 + c 2 ≡ 3 mod 4 whereas a2 b 2 ≡ 1 mod 4. It will never give an equality, so the given equation has no integral solution. Case II If two numbers are odd and one is even, then a2+ b 2 + c 2 ≡ 2 mod 4 whereas a2 b 2 ≡ 0, 1 mod 4 Again we get no integral solution. Case III If two even and one odd. a2+ b 2 + c 2 ≡ 1 mod 4 whereas a2 b 2 ≡ 0 mod 4 Again, no integral solution Case IV If all are even then a2+ b 2 + c 2 ≡ 0 mod 4 whereas a2 b 2 ≡ 0 mod 4 Now, there is a possibility to have a solution and the only possible solution is (0, 0, 0) which is the only trivial solution. Now, let us come again to the discussion of x2+ y 2 = k we have seen, ifk=4 m + 3 there is no integral solution now, if k is even then it will be either of the form k= 4 m or k=4 m + 2 considering k= 4 m If m can be expressed as i2+ j 2 , where i and j are non-negative integers such that (i) i≠ j, then there will be four integral solutions and 8 ordered pairs. (ii) if i= j or m can be written as i 2+ 0 2, then there will be two integral solutions and 4 ordered pairs. Let us consider some examples. e.g., x2+ y 2 = 20 here, 20 is of the form 4(5) and 5 can be expressed as 5= 12 + 2 2 which gives i = 1 and j = 2 so, it implies there are 4 integral solutions, which will be of the form ± 2i and ± 2j also we have 8 ordered pairs. Therefore in this case we have (2, 4) as one of the solutions and other solutions are (2, –4), (–2, –4), and (–2, 4) also (4, 2), (4, –2), (–4, –2) and (–4, 2) keep this thing in mind there are only 4 integers used. e.g., x2+ y 2 = 8 Here, 8= 4() 2 and 2= 12 + 1 2 which gives i= j . So, it implies there are 2 integral solutions which will be of the form 2i and 2j also we have 4 ordered pairs therefore in this case we have our solutions are (2, 2), (–2, –2), (2, –2) and (–2, 2) keep this thing in mind there are only 2 integers used. Theory of Numbers 39 e.g., x2+ y 2 = 4 Here, 4= 4() 1 and 1= 12 + 0 2 which gives i = 1 and j = 0 So, it implies there are 2 integral solutions, which are of the form 2i and 2j also we have 4 ordered pairs. Therefore in this case the solutions are (,2 0 ),(− 2 , 0 ),(, 0 2 ) and (0, –2). e.g., x2+ y 2 = 24 Here, 24= 4 × 6 and 6 can't be represented as i2+ j 2 so it will not have any integral solution. e.g., x2+ y 2 = 12 Here, 12= 4 × 3 and 3 again can't be expressed as i2+ j 2 so it will also not have any integral solution. Now, considering k=4 m + 2 and if m can be written as ()i2+ j 2 + i + j and if i≠ j, then these will be 4 integers and 8 ordered pairs of solutions. And if i= j, then there will be 2 integers and 4 ordered pairs of solutions. e.g., x2+ y 2 = 10 Here, 10= 4() 2 + 2 and 2=()()()() 12 + 0 2 + 1 + 0 Therefore there will be four integral solutions which will be given as ±()2i + 1 and ±()2j + 1 and in this case the solutions are ± 3 and ± 1 which will give eight ordered pairs as (3, 1), (3, –1), (–3, 1) and (–3, –1) also (1, 3), (1, –3), (–1, 3) and (–1, –3). e.g., x2+ y 2 = 2 Here, 2= 4() 0 + 2 and 0=()()()() 02 + 0 2 + 0 + 0 Therefore there are only two integral solutions which will again be given as ±()2i + 1 and ±()2j + 1 and in this case the solutions are ± 1 and ± 1, which will give four ordered pairs as (1, 1), (1, –1), (–1, 1) and (−1, −1). e.g., x2+ y 2 = 18 Here, 18= 4() 4 + 2 and 4=()()()() 12 + 1 2 + 1 + 1 So i = 1 and j = 1 ⇒ i= j Therefore it will have 2 integral solutions which will be given as ±()2i + 1 and ±()2j + 1 and in this case the solutions are ± 3 and ± 3 which gives four ordered pairs as (,),(,3 3 3− 3 ),( − 3 ,) 3 and (,)−3 − 3 e.g., x2+ y 2 = 14 Here, 14= 4() 3 + 2 and 3 can't be expressed as ()i2+ j 2 + i + j as it is always an even number and an even number can't be equal to an odd number. So it implies if right hand side is ()4m + 2 and m is an odd number. So the equation will never produce any integral solution. Now, we will extend this concept for an odd number in right hand side of the equation. 40 Indian National Mathematics Olympiad x2+ y 2 = k i.e., k=4 m + 1 or k=4 m + 3 As it has been already discussed ()k=4 m + 3 will not produce any integral solutions. So, considering k=4 m + 1, only. If m= i2 + j 2 + j, then there will be an odd integer and an even integer, if i ≠ 0 and j ≠ 0 or i ≠ 0 and j = 0 , then there are four integers and 8 ordered pairs which will satisfy the equation. So, one of the integral solution is ± 2i and other is ±()2j + 1 . Now, ifi = 0, j ≠ 0, then there are two integers and four ordered pairs which will satisfy the equations. So, one of the integral solution is 0 and other is ±()2j + 1 . e.g., x2+ y 2 = 21 Here, 21= 4 × 5 + 1 But 5 cannot be written as i2+ j 2 + j, so it will not give any integral solution. Exceptional case : If x2+ y 2 = k and k is an odd and a perfect square, then perform the following test always. Take square root of k, which will come out to be as k, now subtract “1” from it we get ()k − 1 always double it, so it becomes 2(k − 1 ), now add “1” to it which becomes 2()k − 1 . If this value is a perfect square say, it is a 2 , then the equation will always have 6 integers and 12 ordered pairs as its solutions and the integers will be ±a, ± ( k − ),1 ± k and 0. Always keep this thing in mind k is an integer. And if the test fails, then equation will be solved by the method discussed earlier. e.g., x2+ y 2 = 169 here 169= 4(), 42 + 1 which is of the form ()4m + 1 and also it is an odd perfect square so we will have to perform the mentioned test. e.g., 169= 13 13− 1 = 12 12× 2 = 24 24+ 1 = 25 and 25= 52 ∴ we will have 4 integers in which ±5,, ± 12 will form 8 ordered pairs (5, 12), (5, –12), (–5, 12), (–5, –12), (12, 5), (12, –5), (–12, 5), (–12, –5) also there will be three ± 13, 0 which will form four pairs (13, 0),(–13, 0), (0, 13), (0, –13) e.g., x2+ y 2 = 49 here 49= 4 × 12 + 1, which is of the form ()4m + 1 and also an odd perfect, so we will again perform the mentioned test. 49= 7 7− 1 = 6 6× 2 = 12 12+ 1 = 13 but 13 is not a perfect square therefore the solution will be checked by the earlier method. 12=()()() 02 + 3 2 + 3 Here, i = 0 and j = 3 so the solutions will be 0 and ± 7. Also the ordered pairs will be (0, 7), (0, –7), (7, 0) and (–7, 0) Theory of Numbers 41 Concept Solving of the equation of the form xy= n. If we are asked to find the number of positive integral solution for xy= n, we first write n is the form α α α 1 2 3 p1.. p 2 p 3 . The number of positive integral solution is same as the number of divisors of n which is α+ α + α + equal to (11 )( 2 1 )( 3 1 )… Let us consider example xy = 8 = 23 Number of divisors of 8 are 3+ 1 = 4. So there are 4 integral solution and 4 ordered pair namely (1, 8) (8, 1) (2, 4) (4, 2) Now let us consider another example xy=72( x + y ) we can write it as (x−72 )( y − 72 ) = 722. Let x−72 = X, y − 72 = 722 ∴ XY =722 = 2 6. 3 4 Number of solutions are 35. A Concept The area of a ∆ formed by pythogorean triplet with integer sides is always divisible by 3. 1 Area of ∆ ABC= BC × AB 2 k2 + 1 1 =2x() x 2 − 1 2 k2 – 1 =()x3 − x Let p() x= x3 − x BC 2k For x > 1 we use induction p(),2= 8 − 2 = 6 p( )2 is true Let p() x be true for n= m p() m= m3 − m = 3 c ∴ p()()() m+1 = m + 13 − m + 1 =()()m +13 − m + 1 = m 3 + 3 m 2 + 2 m =3c + m + 3 m2 + 2 m =3()c + m + m 2 ∴ P() m + 1 is true. ∆ Concept The radius of the circumcircle of a formed by pythogorean A triplet cannot be integer. The hypotenuse of the ∆ ABC is the diameter of the circle. Let us consider the pythogorean triplet x 2 − 1, 2x, x 2 + 1 Here x 2 + 1 is hypotenuse. Since x is an even number its square is also even, therefore an even number plus one is an odd number. ∴ 2 + x 1 is an odd number B C x 2 + 1 ∴ Radius of circumcircle is 2 Concept For any natural number x for x= 0,,,..., 1 2 n − − 2n+ 1,() 2 x 22 x 2 x − 1 , 2x() 22 n 2 x + 1 − AC2=[2x ( 2 2 n 2 x + 1 )] 2 42 Indian National Mathematics Olympiad − − + =22x() 2 4 n 4 x + 2 2 n 2 x 1 + 1 C − + − + + =()24n 4 x 2 x + 2 2 n 2 x 1 2 x + 2 2 x − + AC2=()2 4n 2 x + 2 2 n 1 + 2 2 x x 2n–2x x − + 2 (2 + 1) 2 + 1 AB2+ BC 2 =[()]()2 2x 2 2 n 2 x − 1 2 + 2 n 1 2 − − + =22x() 2 4 n 4 x − 2 ⋅ 2 2 n 2 x + 1 + 2 2 n 2 − =24n 2 x + 2 2 x + 2 2 n ⋅ 2 2 − 2 2 n ⋅ 2 − A x 2n–2x B =24n 2 x + 2 2 x + 4 ⋅ 2 2 n − 22. 2 n 2 (2 – 1) − =24n 2 x + 2 2 x + 22. 2 n − + =24n 2 x + 2 2 x + 2 2 n 1 Example 1 Prove that there are no natural numbers, which are solutions of15x2− 7 y 2 = 9. Solution 15x2− 9 = 7 y 2 3() 5x2− 3 = 7 y 2 ⇒7y 2 is a multiple of 3. ⇒ y is a multiple of 3. Let y= 3 z 3() 5x2− 3 = 7 × 9 z 2 5x2− 3 = 21 z 2 5x2= 21 z 2 + 3 5x 2 is a multiple of 3. ⇒ x is a multiple of 3 Let x= 3 u 15u2= 1 + 7 z 2 15u2− 6 z 2 = 1 + z 2 1 + z 2 is a multiple of 3. But for any z between 0 to 9,1 + z 2 is not a multiple of 3. For any z, the given equation has no integral solution. Aliter Since RHS is odd, x and y must be opposite i.e., one even and one odd. As 3|15 and 3|9 ∴ 3 must divide 7y 2. = 2 −2 = Let y3 y1 so 5x 21 y1 3 Again, since 3 divide 21 so 3 must divide 5x 2. = Let x3 x1 , we get 2 −2 = 15x1 7 y1 1 ⇒ 2 =2 + 15x1 7 y1 1 Theory of Numbers 43 2 Last digit of perfect square y1 may be one of these values 0, 1, 4, 9, 6, 5. 2 + Hence, last digit of 7y1 1 will be 1, 8, 9, 4, 3, 6 respectively. 2 But 15x1 ends in 0 or 5. ∴ 2 =2 + 15x1 7 y1 1 has no solutions. Example 2 Show that x2 +1 = 3 y has no solutions in integers. Solution Since LHS cannot be a multiple of 3 for any x between 0 to 9. RHS is always a multiple of 3. ∴ x2 +1 = 3 y has no integral solutions. Example 3 Show that 21x2− 10 y 2 = 9 has no solution. Solution 21x2− 9 = 10 y 2 ⇒ 3() 7x2− 3 = 10 y 2 ⇒ 10y 2 is a multiple of 3. ⇒ y is a multiple of 3. = Let y3 y1 2 − = × 2 3() 7x 3 10 9 y1 2 − = 2 7x 3 30 y1 2 = +2 ⇒ 2 = + 2 7x 3 30 y1 7x 3() 1 10 y1 ⇒ 7x 2 is a multiple of 3 So, x is multiple of 3. = Let x3 x1 ×2 = + 2 7 9x1 3() 1 10 y1 2 = + 2 21x1 1 10 y1 2 −2 = + 2 21x1 9 y1 1 y1 2 −2 = + 2 3() 7x1 3 y1 1 y1 ⇒ + 2 1 y1 is a multiple of 3. + 2 But1 y1 is not a multiple of 3. ∴ The given equation has no integral solution. Note Every integer m can be written in the form x2+ y 2 − 5 z 2. If m= 2 n, then =2n =()()() n − 22 + 2 n − 1 2 − 5 n − 1 2 If m=2 n + 1 =()() n + 12 + 2 n 2 − 5 n 2 Verification, 7= 2()()()() 3 + 1 = 3 + 12 + 2 × 3 2 − 5 3 2 =16 + 36 − 45 =52 − 45 = 7 Similarly, every integer can be written in the form of x2+ y 2 + z 2 − 5 u 2 44 Indian National Mathematics Olympiad 1 Example 4 Prove 2n C is an integer. n + 1 n Solution If a and b are integers and a− b = c, then c is also an integer. = 2n = 2n Let a Cn ; b Cn − 1 2n 2n 2n! 2n! 2n! n CC−− = − =1 − n n 1 n!! n (n+1 )!( n − 1 )! n!! n n +1 2n! 1 = ⇒ it is an integer. n!! n n + 1 kn kn (kn )! kn! CC− − = − n n 1 (kn− n )! n ! (kn− n +1 )!( n − 1 )! (kn )! (kn− n )! n ! = 1 − (kn− n )! n ! (kn− n +1 )!( n − 1 )! (kn )! n = 1 − (kn− n )! n ! kn− n + 1 kn− n +1 − n kn−2 n + 1 = knC = knC . n kn− n + 1 n kn− n + 1 It is an integer. Example 5 If xy=22 ⋅ 3 4 ⋅ 5 7() x + y , find the number of integral solution. Solution Let N =22 ⋅ 3 4 ⋅ 5 7 xy= N() x + y xy= Nx + Ny xy− Nx − Ny = 0 ()()x− N y − N = N 2 ()()..x− N y − N = 24 3 8 5 14 Number of integral solution =()()()4 + 1 8 + 1 14 + 1 =5 × 9 × 15 = 675 Example 6 Find all positive integers x, y satisfying 1+ 1 = 1 x y 20 Solution Suppose x, y are two positive integers such that 1 1 1 + = …(i) x y 20 1 1 1x − 20 then = − = y20 x 20 ⋅ x + − ∴ 1= x 20 4 5 x y 20x Implying that 5x is rational. Theory of Numbers 45 Now x∈ N ⇒ 5x ∈N. Hence 5x is the square of an integer which is divisible by 5. ∴ 5x= () 5 a 2 for some a∈ N i.e., x= 5 a 2 similarly y= 5 b2 for some b∈ N. Now, Eq. (i) becomes 1 1 1 + = ⇒ 2(a+ b ) = ab a b 2 ()()a−2 b − 2 = 4 ⇒ (,)(,),(,),(,)a b ∈{}3 6 4 4 6 3 . ∴ Solution set is {(45, 180), (80, 80), (180, 45)}. Example 7 Find the number of solutions in positive integers of the equation 3x+ 5 y = 1008. Solution Let x, y∈ N such that 3x+ 5 y = 1008 then 3 5| y ⇒ 3|y ⇒ y= 3 k for some k∈ N Now, 3x+ 15 k = 1008 ⇒ x+5 k = 336 ⇒ 5k ≤ 335 ⇒ k ≤ 67 Thus, any solution pair is given by (,)(,)x y=336 − 5 k 3 k where1≤k ≤ 67. ∴ Number of solutions is 67. Example 8 Prove that there do not exist positive integrs x,, y z satisfying 2xz= y 2 and x+ z = 997. Solution 2|y 2 ⇒4|2xz ⇒2|x or z ⇒2|x and z [Q 2 |(x+ z )] = = = Let x2 x1, y 2 y 1 and z2 z1 = 2 + = Then 2x1 z 1 y 1 and x1 z 1 997 = = Again y1 and one of x1,, z 1 say x1, are even writing x12 x 2 and y12 y 2 = 2 + = We have x2 z 1 y 2 and 2x2 z 1 997 Since, 997 is a prime, x 2 and z1 are relatively prime. ∴ Each is a square since their product is square. + + ≡ ≡ Since any square is of the form 8n, 8n 1 or 8n 4, 2x 2 0 or 2(mod 8) and z1 1 Q (mod 8) ( z1 is odd). + ≡ Hence 2x2 z 1 1 or 3 (mod 8). A contradiction as 997≡ 5 (mod 8). Example 9 Show that the equation 3x10− y 10 = 1991 has no integral solution. Solution Suppose the existence of x, y∈ Z such that 3x10− y 10 = 1991. Note that 11|1991. ∴ Neither x nor y is divisible by 11 for otherwise 11 would divide both . ⇒ 1110| 3x 10− y 10 = 11 10 |1991 an impossibility. Hence x and y are prime to 11. ∴ x10≡ y 10 ≡1 (mod 11) ⇒ 1991= 3x10 − y 10 ≡ 3 − 1 = 2 a contradiction. 46 Indian National Mathematics Olympiad Example 10 Find all integral solutions of x4+ y 4 + z 4 − t 4 =1991 Solution Let n be any integer ; when it is oddn4−1 =()()() n − 1 n + 1 n 2 + 1 is divisible by 16 as n−1,, n + 1 n 2 + 1are all even andn −1,n + 1being consecutive even integers one of them is divisible by 4 . When n is even n 4 ≡ 0 (mod 16). Thus, n 4 ≡ 0 or 1, Now for any x,,,, y z t∈ Z x4+ y 4 + z 4 − t 4 ≡ α where α ∈{} −1,,,, 0 1 2 3 , since1991≡ 7 x4+ y 4 + z 4 − t 4 ≠1991 Example 11 For n∈ N, let s(n) denote the number of ordered pairs (,)x y of positive integers for 1 1 1 which + = . Determine the set of positive integers n for which s() n = 5 . x y n 1 1 1 Solution + = ⇒ x, y> n x y n ∴ x= n + a and n+ b, a, b∈ N. 1 1 1 Now, + = n+ a n+ b n ⇒ ()()()n+ b + n + a n = n + a n + b ⇒ n2 = ab ∴ s() n is the number of divisors of n 2 α α = 1K m α≥K ≥ α Let n p1 pm be prime factorization of n where 1 m. = +αK + α Then s()()() n 1 21 1 2 m ∴ s() n = 5 ⇒ +α = 1 21 5 and m =1 ∴ = 2 n p1 Required set is {p2 : p is prime}. 1 1 1 Example 12 If + = ;a,, b c are positive integers with no common factors. Prove that()a+ b is a a b c square. a+ b 1 Solution By the hypothesis = ab c i.e., ()a+ b c = ab Let p be any prime which divides ();a+ b then p divides one of a, b and therefore both. Since gcd(,,); a b c =1 p does not divide c. ∴ for any k∈ N, pk | a ⇔ pk | b Hence the maximum power of p which divides a+ b is the square of the maximum power of p which divides a. ∴ a+ b is a square. Theory of Numbers 47 3n − 5 Example 13 Find all integers n such that is also an integer. n + 1 3n − 5 8 Solution Since, =3 − is an integer. n+ 1 n + 1 8 ∈{} ±1,,, ± 2 ± 4 ± 8 n + 1 8 and 3 − is a square. n + 1 8 Consequently = −1 or 2 ; n + 1 i.e., n = − 9 or 3 Example 14 Find the number of pairs of integers (,)a b such that a3+ a 2 b + ab 2 + b 3 +1 = 2002. Solution a3+ a 2 b + ab 2 + b 3 =()()a + b a2 + b 2 = 2001 =3 × 23 × 29 a+ b is therefore one of the three numbers 3,,. 23or 29 If a+ b = 3, then a=1, b = 2 (or a=2, b = 1) so that a2+ b 2 = 5. But in this case a2+ b 2 =23 × 29 ∴a+ b is not 3. If a+ b = 23, then a2+ b 2 = 87 so that both a and b will be less than 10 and a+ b < 20, a contradiction. If a+ b = 29, then a2+ b 2 = 69 so that both a and b will be less than 10 and a+ b < 20, a contradication. Thus, the number of pairs (,)a b satisfying the given condition is zero.