Prime and Perfect Numbers

Home , Mersenne prime, Perfect number, Prime number, Repdigit

Chapter 34

Prime and perfect numbers

34.1 Inﬁnitude of prime numbers

34.1.1 Euclid’s proof If there were only ﬁnitely many primes: 2, 3, 5, 7, ...,P, and no more, consider the number Q = (2 3 5 P ) + 1. · · ··· Clearly it is divisible by any of the primes 2, 3,..., P , it must be itself a prime, or be divisible by some prime not in the list. This contradicts the assumption that all primes are among 2, 3, 5,..., P .

34.1.2 Fermat numbers

2n The Fermat numbers are Fn := 2 + 1. Note that 2n 2n−1 2n−1 Fn 2 = 2 1= 2 + 1 2 1 = Fn 1(Fn 1 2). − − − − − − By induction,

Fn = Fn 1Fn 2 F1 F0 + 2, n 1. − − ··· · ≥ From this, we see that Fn does not contain any factor of F0, F1, ..., Fn 1. Hence, the Fermat numbers are pairwise relatively prime. From this,− it follows that there are inﬁnitely primes. 1

1 It is well known that Fermat’s conjecture of the primality of Fn is wrong. While F0 = 3, F1 = 5, 32 F2 = 17, F3 = 257, and F4 = 65537 are primes, Euler found that F5 = 2 + 1 = 4294967297 = 641 6700417. × 1202 Prime and perfect numbers

34.2 The prime numbers below 20000

The ﬁrst 2262 prime numbers:

10 20 30 40 50 60 70 80 90 100 bbbbbb bb bbb b bb bb b bb b bb bb bb b b b b bb b b b b b b b b b b b b b b b bb b b b b b b b b b b b b b b b b bb b b b b bb b b b b b bb b b b b b b b b b b b b b b b b b b bb b b b b b b b b bb b b b bb b b b b b b b b b b b b bb b b b b b b b b b bb b b b b b b b b b b b b b b b b b b bb b b bb b b b b bb b b b b b b b b b b b b b 100 b b bb b bb b bb b b b b bb b b b b b bb b b b b b b b bb b bb b b b b b bb b b b bb b bb b b b b b b b b bb b b b b b b b bb b b b b b b bb b b b b b b bb b bb b b bb b b b b b b b bb b b b b b bb b b b b b b b bb b b b bb b b b bb b bb b b b 200 b b bb b b b b b b b b b b b b b b b b b bb b b b b bb b b b b bb b b b b b bb b b b b b b b b b b b b bb b b b b b b b b b bb b b bb b b b b b b bb b b b b b b b b bb b b b b b b b b b b b b b b b b b b b bb b b b b b b b b b b b b b b b 300 bb b b b b b b b b b b b b b bb b b b b bb b b b b b b b bb b b b b b b b b b bb b b b b bb b b b b b b b b bb b b bb b b bb b b b b b b b b b b bb b bb b b b b b bb b b bb b b b b b bb b b b b b b b b bb b b b b b bb b b b bb b 400 b bb b b b b b b bb b b b b b b b b b b b b b b bb b b bb b b b b b b bb b b b bb b bb b b b b b b b b b b b b bb b b b b b b b b b bb b b b b b bb b b b b b b bb b b bb b bb b b bb b b b bb b bb b b b b b b b b b b 500 bb b b b b b b b b b b b b b b b bb b b b b b b b b bb b b bb b b bb b b b bb b b b b b bb b b bb b b b b b bb b b b b b b b b b bb b b bb b b b b bb b b b b bb b b b b b b b b b b b b b b b b b b bb b b bb b b b 600 b b bb b b b b b b b b b b b b bb b b b b b b bb b b b bb b b b b b b b b b b b b b b b b b b b bb b b b b b b b b b bb b bb b b bb b b b b b b bb b b b b b b b b bb b b b b b b b bb b b b b b b b b b b b b b b b b b 700 b b bb b b b bb b b bb b b bb b b b b b b bb b b b b b b b b bb b b bb b b bb b b b bb b b b b bb b b b b b b b b b b b b b b b b b b b b b b b b bb b b b bb b b b b b b b b b b b b b bb b b bb b bb b b 800 bb b b b b b b b b bb b b b b b b b b b b b b b b b bb b bb b b bb b b b b b b b bb b b b b b b b b b b b b b b b b b b b b b bb b bb b b b b b b b b bb b b b b b b b b b b b b b b bb b b bb b b b bb b b b b 900 b b b b b b b b b b b b b b b b b b b b bb b b b b b bb b b b b b bb b b b b b b bb b b b b b b b b b b b b b bb b b b b b b bb b b b b bb b b b b b b b b b b b b b bb b b b b b b bb b b b b b b b b b bb b b bb b 1000 b b bb b b b b b b b b b b b bb b b b b b bb b bb b b b b b b b b b b b b b b b b b b b b b b b b b bb b b b b bb b b b b b b bb b b b b b b b bb b b b b b bb b b b bb b b b b b b b b bb b b b b b b 1100 b b b b bb b b b b b b b b b b b b b b b b b b b b b b bb b b b b b b b b bb b b b b b b b b b b b b b b b b b b b b b b b b bb b b bb b bb b b b b b b b b b b bb b b bb b b b b b b b bb b b b b 1200 bb b b b b b b b b b b b b bb b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b bb b bb b b b bb b bb b b b bb b b b b b b b b b b b b b b b b bb b b b b b b b bb b b b b b bb b b bb b b b bb b b b b 1300 b b b b b b b b bb b b bb b b b b b b b b b b bb b b b b b b b b bb b b b b b b b b b b b b b b b b b b b bb b b b b b bb b b b bb b b b b b b bb b b b b b b bb b b bb b b b b b b b b b b b bb b b 1400 b bb b b b b b b b bb b b b b b b b b b b b b b b b b b bb b b b b b b b b bb b b b b b b b bb b b b b b b b b b b b bb b b b b b b b b b b b b b b b bb b b b b bb b b b b b b b b b b b b b b b 1500 b b bb b b b b b bb b bb b b b bb b b b b b bb b b b bb b b b b b b b b b b bb b b b b b b b b b b b b b b b b bb b b b b b b b b bb b b b bb b b b b b b b b b b bb b bb b b b b b b b b b b b b bb b b b b b b 1600 b b b b b b b b b bb b b b b b b b b bb b bb b b bb b b b b b b b bb b b b b b bb b b b b b b bb b b b bb b b b b b b b b b bb b b b b b b b b bb b b b b bb b b bb b b b bb b b b b 1700 bb b b b b bb b b b b b b b bb b b bb b b b b bb b b b b b b b b b b bb b b b b b b b b b bb b b b b b b bb b b b b b b bb b b bb b b bb b b b b b bb b b b b b b b bb b b b b bb b b b b b b b 1800 bb b b b b b b b b b bb b b b b b b b b b bb b b b b b b b b b b b bb b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b bb b b b b b b b b b b b b b bb b b 1900 b b b b b b b b bb b b b b b b b b b b b b bb b b b b b bb b b b bb b b b b b b b b b bb b b b bb b b b b b bb b b b b b b bb b b b b b b b bb b b b b b b b b b b bb b b b b b 34.3 Primes in arithmetic progression 1203

34.3 Primes in arithmetic progression

B. Green and T. Tao (2004) have proved that there are arbitrarily long arithmetic progressions of prime numbers. For k = 0, 1, 2,..., 21, the numbers 376859931192959 + 18549279769020k are all primes.

34.4 The prime number spirals

The ﬁrst 1000 prime numbers arranged in a spiral. = prime of the form 4n + 1; = prime of the form 4n + 3.

34.4.1 The prime number spiral beginning with 17

The numbers on the 45 degree line are n2 + n + 17. Let f(n) = n2 + n + 17. The numbers f(0), f(1),... f(15) are all prime.

n f(n) n f(n) n f(n) n f(n) 0 17 1 19 2 23 3 29 4 37 5 47 6 59 7 73 8 89 9 107 10 127 11 149 12 173 13 199 14 227 15 257 34.4 The prime number spirals 1205

34.4.2 The prime number spiral beginning with 41

The numbers on the 45 degree line are f(n)= n2 + n + 41. f(n)= n2 + n + 41 is prime for 0 n 39. ≤ ≤ n f(n) n f(n) n f(n) n f(n) n f(n) 0 41 1 43 2 47 3 53 4 61 5 71 6 83 7 97 8 113 9 131 10 151 11 173 12 197 13 223 14 251 15 281 16 313 17 347 18 383 19 421 20 461 21 503 22 547 23 593 24 641 25 691 26 743 27 797 28 853 29 911 30 971 31 1033 32 1097 33 1163 34 1231 35 1301 36 1373 37 1447 38 1523 39 1601 1206 Prime and perfect numbers

Prime number spiral beginning with 41: A closer look

b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b 41b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b bb b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b bb b b b b b b b b b b b b b b b bb b b b b b b b b b b b b b b b b b b b b b bb b b b b b b b b b b b b b b b bb b b b b b b bb b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b bb b b b b b b b b b b b b b b b b b b bb b b b b b b b b b bb b b b b b b b b b b bb b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b bb b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b bb b b b b b b b b b b b b b b b b b bb b b b b b b b b b b b bb b b b b b b b b b b b b b b b bb b bb b b b b b b b b b bb b b b b b b b b b b bb b b b b b b b b b bb b b b b b b b b bb b b b b b b b b b b b b b b b b b b b b b b b b bb b b b b b b b b bb b b b b b b b bb b b b b b b b b bb b b b b b bb b b b b b bb b b b b bb b b b b b b b b b b b b b b b b b b b b b b bb b b b b bb b b b b b b bb bb b b b b b b b b b b b bb b b b b bb b b b bb b b b b b b b b b b b bb b b b b b b b b b b b b b b b b b b b b b b b b b b bb b b b b 34.5 Perfect numbers 1207

34.5 Perfect numbers

A number is perfect is the sum of its proper divisors (including 1) is equal to the number itself.

2 k 1 k Theorem 34.1 (Euclid). If 1+2+2 + + 2 − = 2 1 is a prime k 1 k ··· − number, then 2 − (2 1) is a perfect number. − Note: 2k 1 is usually called the k-th Mersenne number and denoted − by Mk. If Mk is prime, then k must be prime. Theorem 34.2 (Euler). Every even perfect number is of the form given by Euclid.

Open problem Does there exist an odd perfect number? Theorem-joke 34.1 (Hendrik Lenstra). Perfect squares do not exist. 2 Proof. Suppose n is a perfect square. Look at the odd divisors of n. They all divide the largest of them, which is itself a square, say d2. This shows that the odd divisors of n come in pairs a, b where a b = d2. Only d is paired to itself. Therefore the number of odd divisors· of n is also odd. In particular, it is not 2n. Hence n is not perfect, a contradiction: perfect squares don’t exist.

2Math. Intelligencer, 13 (1991) 40. 1208 Prime and perfect numbers

34.6 Charles Twigg on the ﬁrst 10 perfect numbers

Let Pn be the n-th perfect number.

k 1 n k Mk Pn = 2 − Mk 123 6 2 3 7 28 3 5 31 496 4 7 127 8128 5 13 8191 33550336 6 17 131071 8589869056 7 19 524287 137438691328 8 31 2147483647 2305843008139952128 9 61 2305843009213693951 26584559915698317446 54692615953842176 10 89 61897001964269013744 19156194260823610729 9562111 47933780843036381309 97321548169216

P1 is the difference of the digits of P2. In P2, the units digit is the • cube of the of tens digit. P and P are the first two perfect numbers prefaced by squares. • 3 4 The first two digits of P3 are consecutive squares. The first and last digits of P4 are like cubes. The sums of the digits of P3 and P4 are the same, namely, the prime 19. P terminates both P and P . 3 • 4 11 14 Three repdigits are imbedded in P . • 5 P contains each of the ten decimal digits except 0 and 5. • 7 P9 is the smallest perfect number to contain each of the nine nonzero • digits at least once. It is zerofree.

P10 is the smallest perfect number to contain each of the ten decimal • digits at least once.

3These contain respectively 65 and 366 digits. 34.7 Mersenne primes 1209

34.7 Mersenne primes

k Primes of the form Mk = 2 1 are called Mersenne prime. The only known Mersenne primes are listed− below.

k Year Discoverer k Year Discoverer 2 Ancient 3 Ancient 5 Ancient 7 Ancient 13 Ancient 17 1588 P.A.Cataldi 19 1588 P.A.Cataldi 31 1750 L.Euler 61 1883 I.M.Pervushin 89 1911 R.E.Powers 107 1913 E.Fauquembergue 127 1876 E.Lucas 521 1952 R.M.Robinson 607 1952 R.M.Robinson 1279 1952 R.M.Robinson 2203 1952 R.M.Robinson 2281 1952 R.M.Robinson 3217 1957 H.Riesel 4253 1961 A.Hurwitz 4423 1961 A.Hurwitz 9689 1963 D.B.Gillies 9941 1963 D.B.Gillies 11213 1963 D.B.Gillies 19937 1971 B.Tuckerman 21701 1978 C.Noll, L.Nickel 23209 1979 C.Noll 44497 1979 H.Nelson, D.Slowinski 86243 1982 D.Slowinski 110503 1988 W.N.Colquitt, L.Welsch 132049 1983 D.Slowinski 216091 1985 D.Slowinski 756839 1992 D.Slowinski,P.Gage 859433 1993 D.Slowinski 1257787 1996 Slowinski and Gage 1398269 1996 Armengaud, Woltman et al. 2976221 1997 Spence, Woltman, et.al. 3021377 1998 Clarksonet.al 6972593 1999 Hajratwala et. al 13466917 2001 Cameron,Woltman, 20996011 2003 MichaelShafer 24036583 2004 Findlay 25964951 2005 Nowak 30402457 2005 Cooper,Booneetal 32582657 2006 Cooper,Booneetal 37156667 9/8/2008 42643801 2009 43112609 8/8/2008

The most recently discovered Mersenne primes M37156667, M42643801, and M43112609 have about 11, 185, 272, 12, 837, 064, million and 12, 978, 189 digits respectively, and are the largest known primes. Chapter 35

Some division problems

35.1 AMM E1

This is Problem E1 of the MONTHLY:

x 7 x x x x x x) xxxxxxxx xxxx x x x x x x xxxx x x x xxxx xxxx

Clearly, the last second digit of the quotient is 0. Let the divisor be the 3-digit number d. Consider the 3-digit number in the seventh line, which is a multiple of d. Its difference from the 4-digit number in the sixth line is a 2-digit number. This must be 9xx. This cannot be the same as the 3-digit number in the fifth line, since the difference between the 3-digit numbers in the fourth and fifth lines is a 3-digit number. Therefore, in the quotient, the digit after 7 is a larger one, which must be smaller than the first and the last digits, since these give 4-digit multiples of d. It follows that the quotient is 97809. Since 8d is a 3-digit number 9xx, the 4-digit number in the third and 1212 Some division problems bottom lines is 9d = 10xx or 11xx. From this 8d must be 99x, and therefore 992 = 8 124. × 97809 124)12128316 1116 9 6 8 8 6 8 1003 9 9 2 1116 1116

35.2 AMM E10

This is Problem E10 of the MONTHLY, by Fitch Cheney. In this case, not even one single digit is given.

xxxxxx x x x) xxxxxxxx x x x xxxx x x x xxxx x x x xxxx xxxx

35.3 AMM E1111

This is said to be the most popular MONTHLY problem. It appeared in the April issue of 1954.

Our good friend and eminent numerologist, Professor Euclide Paracelso Bombasto Umbugio, has been busily engaged test- ing on his desk calculator the 81 109 possible solutions to the problem of reconstructing the following· exact long division 35.4 AMM E971 1213

in which the digits indiscriminately were each replaced by x save in the quotient where they were almost entirely omitted.

xx8xx xxx)xxxxxxxx x x x x x x x x x x x x x x x x x x

Deﬂate the Professor! That is, reduce the possibilities to (18 109)0. ·

Martin Gardner’s remark: Because any number raised to the power of zero is one, the reader’s task is to discover the unique reconstruction of the problem. The 8 is in correct position above the line, making it the third digit of a ﬁve-digit answer. The problem is easier than it looks, yielding readily to a few elementary insights.

35.4 AMM E971

Reconstruct the division problem

) ∗∗∗∗∗2 ∗ ∗ ∗∗∗∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Charles Twigg comments that if the digit 2 is replaced by 9, the answer is also unique. Chapter 36

Transposable integers

36.1 k-right-transposable integers

Let k be a given positive integer. A positive integer X is k-right-transposable if in moving the leftmost digit to the rightmost, the number is multiplied by k. Note that X is a repdigit if and only if k = 1. We shall assume k > 1. Theorem 36.1. The only right-transposable numbers are the repdigits (which are triviallly 1-right transposable, and

(142857)m, (285714)m,

which are 3-right-transposable. Proof. Suppose the number X has n digits, with leftmost digit a. We have n 1 10(X a 10 − )+ a = kX. − · n From this, (10 k)X = a(10 1) = 9a Rn. If k = 3, 10− k can only have− prime divisors· 2, 3, 5. The equation will reduce6 to X−= a repdigit, which is clearly impossible. For k = 3, we have 7X = a(10n 1). If a = 7, then again X is a repdigit. Therefore, we must have 7 dividing− 10n 1. This is possible 106m−1 only if n is a multiple of 6. Therefore X = a 7 − and has ﬁrst digit a. · Now, 106m 1 − = (142857) . 7 m 1216 Transposable integers

It is easy to see that a can only be 1 or 2. Therefore, the only k- transposable numbers are (142857)m and (285714)m with k = 3.

36.2 k-left-transposable integers

Let k be a given positive integer. A positive integer X is k-left-transposable if in moving the rightmost digit to the leftmost, the number is multiplied by k. Note that X is a repdigit if and only if k = 1. We shall assume k > 1. Suppose the number X has n digits, and its rightmost digit is b. We have n 1 X b b 10 − + − = kX. · 10 From this, b 10n + X b = 10kX, and · − (10k 1)X = b(10n 1). − − n n 1 Since X is has n digits, b(10 1) > (10k 1)10 − , and b > n− 10 1(10k 1) 1 − − n − 10 1 = k 10 . This shows that b k. − − ≥ k n 2 19X = b(10n 1) 18m 3 29X = b(10n − 1) 28m 4 39X = b(10n − 1) 12m 5 49X = b(10n − 1) 42m 6 59X = b(10n − 1) 58m 7 69X = b(10n − 1) 22m 8 79X = b(10n − 1) 13m 9 89X = b(10n − 1) 44m − These lead to the following numbers:

X(2) =105263157894736842, X(3) =1034482758620689655172413793, X(4) =102564102564, X(5) =102040816326530612244897959183673469387755, X(6) =10169491525423728813559322033898305084745762711864 40677966, X(7) =1014492753623188405797, X(8) =1012658227848, X(9) =10112359550561797752808988764044943820224719.

b Each of these X(k)k can be replaced by Xk for k = b, . . . , 9. Every k · k-left-transposable number is of the form (X(k))m for X(k) given above and m 1. ≥