TWO THEOREMS ON THE STRUCTURE OF PYTHAGOREAN TRIPLES AND SOME DIOPHANTINE CONSEQUENCES By Konstantine ‘Hermes’ Zelator
1. INTRODUCTION
The major theorem of this paper (Theorem 1) deals with the structure of primitive pythagorean triangles. In Theorem 1, it is proven that there are no primitive 2 2 pythagorean triangles of the form ( 1 2 ,, zysxs ), i.e., triangles with one leg equal to s1 times a perfect square and the other equal to s2 times an integer square, if the integers s1 and s2 satisfy the following conditions:
a) s1 and s2 are both positive odd squarefree integers.
b) = 121 n 1 2 pppnppss n ≡≡≡≡≥ 8mod1...,5,2,... , where 1 ,..., pp n are primes.
c) If s1≡ ,8mod1 (and then of course s2≡ )8mod5 , then s1 is a quadratic
nonresidue of every divisor d of s2 , with d > 1.
(If s2 ≡ 8mod1 , then the same condition is assumed on s2 with respect to the
divisors d of 1 ds > .)1,
One can easily find many examples of integers s1 , s2 satisfying the above conditions. Below, we offer a few.
1. s1 , s2 primes with ≡ ss 21 ≡ ;8mod5,1 s1 a quadratic nonresidue ofs2 .
2. = , pppps 21211 ≡≡ ,,8mod1 pp 21 primes, s2 also a prime, p1 a quadratic
nonresidue of s2 , while p1 a quadratic residue of s2 .
3. s1 ≡ ,8mod1 s1 a prime, = ,, pppps 21212 prime with
1 pp 2 ≡≡ ;8mod5,1 s1 a quadratic nonresidue of both p1 and p2 .
4. = 11 n npps ≥ ,3;... n an odd integer, s2 a prime, s2 ≡ 8mod5 ; 1 ,..., pp n
primes with 1 ≡≡ pp n ≡ ;8mod1... an even number of integers among the
primes 1 ,..., pp n being quadratic residues of s2 , while the remaining
primes (odd in number) being quadratic nonresudues of s2 .
Theorem 2 is somewhat more technical in nature. It is shown that if n is a
positive odd squarefree integer such that whenever 21 =nss , then 2 2 24 ss 21 ≡≡ 8mod5,1 (or vice versa), the diophantine equation ( ) =+ zynx , with (ynx )= 1, is solvable in Z+, if and only if the diophantine equation 2 42 42 + 1 2 += 3 ydxdzd is solvable in Z , for some positive divisors ,, ddd 321 of
n with 321 , 2 <= ndnddd and 3 < nd . It is fairly easy to see that n, satisfying the hypothesis of Theorem 2, is either the product of an odd number of primes p ≡ ,8mod5 or otherwise the product of an odd of primes p ≡ ,8mod5 and any number of primes q ≡ 8mod1 .
Theorem 3 is simple in its statement and proof. It states that if qp ≡≡ 4mod1 and q primes, p a quadratic nonresidue of q, then the diophantine equation 2 2 ()()2 2 =+ zqypx 2 with ( qypx ) =1, , has no solution in Z+. ~ 2 ~
By combining Theorems 2 and 3, Theorem 4 is established. It states that if qp ≡≡ ,8mod5,1 p a quadratic nonresidue of q , the diophantine equation 2 ()2 =+ zypqx 24 , with ()ypqx = ,1, has a solution in Z+, if and only if the equation += yxpqz 442 , with ()yx = ,1, has a solution in Z+.
2. THEOREM 1
Let , ss 21 be two odd squarefree positive integers greater than 1, with (ss 21 )= 1, and
such that ss 21 is the product of a prime congruent to 5 mod 8 and primes congruent
to 1 mod 8,i.e., = 121 n ≥ pnppss 1 ≡ 2 ≡ ≡ pp n ≡ 8mod1...,8mod5,2,... .In addition,
assume that if s1 ≡ 8mod1 (same condition on s2 , if s2 ≡ 8mod1 ), then s1 is a quadratic nonresidue of every divisor d of s2 , including s2 itself, with d > 1.
42 242 Under the above assumptions, the diophantine equation 1 2 =+ zysxs has no + solution in the set of positive integers Z , with ( 21 ysxs )=1, .
Remark
One can easily find examples of , ss 21 satisfying the hypothesis of the theorem. For
instance, , ss 21 being primes with ≡ ss 21 ≡ 8mod5,1 and s1 a quadratic nonresidue of
s2 . Also, = pppps 21211 ≡≡ 8mod1, , where , pp 21 are primes, s2 also a prime,
s2 ≡ 8mod5 such that p1 being a quadratic nonresidue of s2 , while p1 a quadratic
residue of ss 12 ≡ 8mod1; , s1 a prime, while = ,, pppps 21212 primes,
1 pp 2 ≡≡ 8mod5,1 , and with s1 a quadratic nonresidue of both p1 and p2 .
Proof.
Let us assume that (,, zyx ) is a solution with ( 21 ysxs ) = 1, in positive integers x , y, z to the equation.
2 2 2 2 2 ()()1 2 =+ zysxs (1)
In virtue of (21 ysxs )= 1, , we see that equation (1) describes a primitive Pythagorean triangle.
Assume first x to be odd any y even. From (1), it follows that
222 2 22 1 −= nmxs , 2 = 2mnys , += nmz (2)
For positive integers , nm with ()nm = 1, and + nm ≡ 2mod1
The first equation in (2) implies, since nm = 1),( ,
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2 2 2 2 11 − ndmd 12 11 + ndmd 12 = nkmx , = kn , = km , (3) 11 2 2
for positive integers 2111 ,,,, kddnm with 11 = ,1),( = sddnm 121 and k = 1or 2. (Refer
to References[1]or [2].) Note that s1 is squarefree, so must be d1 and d 2 .Now, x is odd and so, from the first equation in (3), we see that k = 1 and ≡ nm 11 ≡ .2mod1
On the other hand, the second equation in (2) implies the following two possibilities or sub cases:
2 2 = 1 ,2 = δδ 2 NnMm (4)
2 2 δ1 == 2, δ 2 NnMm (5)
Where δ ,δ 21 are positive integers with δ δ = s221 , (they are squarefree since s2 is), and , NM are positive integers with ( NM = 1, ), since nm = .1),(
Also, since, as we have shown above, k=1, the last two equations in (3) take the form
2 2 2 2 2 11 −= 12 2, 11 += ndmdmndmdn 12 (6)
First assume (4) to be the case.Then of course, since m and n have different parities, m must be even and n odd, and so N must also be odd.
Combining equations (4) and (6), we obtain
2 2 2 2 2 2 2δ 2 11 −= mdmdN 12 4, δ1 11 += ndmdM 12 . (7)
By virtue of nmN 11 ≡≡≡ 2mod1 (7) yields
2 2δ −≡ dd 212 4,8mod δ1 +≡ ddM 21 .8mod (8)
According to the hypothesis of Theorem 1, either s1 is a prime congruent to 8mod5 or
a product of a prime congruent to8mod5 and primes congruent to 8mod1 and s2 is a product of primes congruent to 8mod1 , or vice versa. Suppose, under the hypothesis
of Theorem 1, that s2 ≡ ,8mod5 while s1 ≡ 8mod1 ; then from = sdd 121 and δ δ = s221 , we conclude dd 21 ≡≡ 8mod1 and δ1 ≡ δ 2 ≡5,1 or δ1 ≡ δ1 ≡ 8mod1,5 . Then, however, the first congruence in (8) implies
δ 2 ≡ 8mod02 , a contradiction since δ 2 is odd.
Now, if s1 ≡ 5and s2 ≡ 8mod1 , we obtain, from = dds 211 and s = δ δ 212 , under the hypothesis of the theorem,
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1 dd 2 ≡≡ 5,1 or 1 dd 2 ≡≡ 1,5 and δ ≡ δ 21 ≡ 8mod1
The first equation in (8) implies δ 2 ≡ 8mod42 , a contradiction since δ 2 is odd.
Next, assume equation (5) to be the case; then M is odd. By combining equations (5) and (6), we arrive at
2 2 2 2 2 2 4δ 2 ( 11 −±= 12 ) 2, δ1 11 += ndmdMndmdN 12 . (9)
Repeating the reasoning we applied when we considered (8), we see that if s1 ≡ 5 and
s2 ≡ ,8mod1 then 1 dd 2 ≡≡ 5,1 or 1 ≡ dd 2 ≡ 8mod1,5 while δ ≡ δ 21 ≡ .8mod1
2 The second of (9) yields 1M 22 δδ +≡≡ dd 211 8mod ; but δ1 ≡ ,22 while
dd 21 ≡+ ,8mod6 whence a contradiction. Finally, assume s1 ≡ 1and s2 ≡ .8mod5
According to Legendre’s theorem, since nNm 11 ≠ 0, the first equation in (9) implies that ⋅ dd 21 is a quadratic residue of δ 2 ; but = sdd 121 ≡ 8mod1 and δ 2 divides
s2 .Therefore, if δ 2 is a divisor of s2 , with δ 2 > 1, by the hypothesis of the theorem, it
would follow that s1 is a quadratic nonresidue of δ 2 , whence a contradiction.
Now, if δ 2 = 1 then δ δ ⇒= δ = ss 21221 .
Then, however, the second equation of (9) and Legendre’s theorem imply that
−=− sdd 121 is a quadratic residue of δ = s21 ; but -1 is a quadratic residue of
s2 (since s2 is a product of primes congruent to 1 mod 4). Hence s1 is a quadratic
residue of s2 , contrary to the hypothesis of the theorem.
To conclude the proof of Theorem 1, let us go back to (1) and consider the case where
x is even and y odd. Unless we use the special assumption that if s1 ≡ 8mod1 , then
s1 is a quadratic nonresidue of every divisor if s2 (except 1), it is clear that the case x = even and y = odd is identical in treatment with the case x = odd and y =even which we have already treated.
Now, if s1 ≡ ,8mod1 then s2 ≡ 8mod5 ; then the treatment of the case x = even, y =
odd, s1 ≡ 1and s2 ≡ 8mod5 , is obviously identical with the treatment of the case x =
odd, y = even, s1 ≡ 5and s2 ≡ ,8mod1 which has already been done.
3. THEOREM 2
Let n be a squarefree positive integer such that if s1 , s2 are any divisors of n with
21 = nss , then ss 21 ≡≡ 8mod5,1 or vice versa. Then the diophantine equation =+ zyxn 2442 with(ynx )= 1, , has a solution in the set of positive integers Z+, if and 2 42 42 only if, the diophantine equation 1 2 += 3 ydxdzd has a solution, ~ 5 ~
with (32 ydxd )= 1, , in positive integers x, y, z, for some positive divisors ,, ddd 321 of n , such that ddd 321 = n , with 2 < nd and 3 < nd .
Remark.
It follows from the hypothesis of the theorem that n ≡ 8mod5 (take 1 = ns and s2 = )1 ; one easily finds examples of numbers n that satisfy the hypothesis.
1. n = prime, n ≡ .8mod5
2. = ppn 21 , p1 , p2 primes with 1 ≡ pp 1 ≡ 8mod1,5 .
3. = 321 i ≥ 1 ≡ ,5;3... ≡ 32 ≡ ≡ ppppnppppn i ≡ 8mod1... . 4. n is the product of an odd number of primes congruent to 5 mod 8 and any number of primes congruent to 1 mod 8. 5. In fact, n is either the product of an odd number of primes p ≡ 8mod5 , or otherwise the product of an odd number of primes p ≡ 8mod5 and any number of primes q ≡ 8mod1 .
Proof. Let x , y ,z be positive integers, with( ynx ) = 1, , satisfying the equation
2 2 ()()2 2 =+ zynx 2 (10)
We distinguish between the cases (x = even, y = odd) and (x = odd, y = even).
Case 1; x = odd , y = even
Equation (10), together with the condition ( ynx ) = 1, , imply
−= trnx 222 , 2 = 2rty , += trx 22 (11)
For positive integers r, t with (r, t) = 1 and + tr ≡ .2mod1
The second equation (11) implies
r = R 2 , = 2Tt 2
or ()TR = 1, r = 2R 2 , = Tt 2
The latter possibility is ruled out, for if it holds, the first equation in (11) gives
2 4 −≡ TRnx 44 8mod (12)
But T is odd (since 2 rTt ≡= mod0, 2 and + tr ≡ 2mod1 ) , hence (12) implies 2 4 ≡−≡ orRnx 8mod7314 ; however, x is odd and so 2 ≡≡ ornnx mod73 8, contrary to the fact that n ≡ 8mod5 (refer to the remark underneath Theorem 2).
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Now, suppose that 2 == 2, TtRr 2 . The first equation in (11) implies, from References [2] or [1], (and since ( tr ) = 1, ),
2 2 2 2 11 −⋅ TsRsk 12 ( + TsRsk ) ⋅= TRkx , t = , r = 11 12 (13) 11 2 2
for integers ,TR 11 with()TR 11 = 1, and integers , ss 21 with 21 = nss and also k = 1 or 2.
Now, since x is odd it follows from the first equation of (13), k=1 and
TR 11 ≡≡ 2mod1 . Since r = R 2 , the third equation in (13) implies
2 2 2 2 11 += TsRsR 12 , 2 2 2 2 11 12 +≡+≡ ssTsRsR 21 8mod .
However, R is odd (recall we are in the subcase 2 == 2, TtRr 2 ), and so the last congruence gives ss 21 ≡+ 8mod2 ; however, we have 21 = nss and so, according to the hypothesis of the theorem, we must have s1 ≡ 1, s2 ≡ 8mod5 or vice versa. At any rate, we obtain + ss 21 ≡ 8mod6 , contradicting the congruence
ss 21 ≡+ 8mod2 obtained above.
2 Note that since − ss 21 ≡ 8mod4 and = 2Tt and therefore the second equation in (13) cannot be rendered impossible via a congruence modulo 8.
Case 2: x = even , y = odd.
Equation (1) gives
2 = ,2 222 , +=−= trztryrtnx 22 (14) for positive integers r, t with ()tr = 1, and + tr = 2mod1
The second equation of (14) shows, since y is odd, that r is odd and t even (consider it mod 4). Thus, the first equation in (14) implies (in virture of ( tr ) = )1,
2 2 = 1 = 2, 2TdtRdr (15) with 21 ddndd 21 >= )0,( .
On the other hand, from −= try 222 and ( tr ) = 1, , we obtain in 2 2 1 +≡ TRr 1 , = 2 TRt 11 (16)
for positive integers ,TR 11 with ( TR 11 ) = 1, and + TR 11 ≡ 2mod1 . The second equations of (16) and (15) give ~ 7 ~
2 2 = 22 TRTd 11 , ( TR 11 ) = 1, (17)
Therefore 2 2 = RdR 231 , = TdT 241 (18) and with = ()ddddd 43243 > 0,, .
By using the first of (15), (16) and equations (18), we obtain
2 2 4 2 4 1 3 2 += 4 TdRdRd 2 (19)
with (TdRd 2423 )= 1, (since ( R1 ,T1 ) =1). Note that 431 = nddd , since 21 = ndd and
= ddd 243 . We claim that 3 < nd and 4 < nd in (19). For if, say 3 = nd , then from
431 = nddd , it follows that = dd 41 = 1(remember ,, ddd 321 > 0 ). Also, since
= ddd 432 , we must have 2 = nd .
From 3 = , dnd 4 = 1and equation (18), we obtain
2 2 = TT 21 , = nRR 21 (20)
From 2 = nd and equation (17), we obtain
2 2 2 = 2 TRT 2 (21)
From the second equation of (16), we arrive at
2 2 = 2 2 TnRt 2 (22)
Combining (21) and (22), we obtain
= 2nTt 2 (23)
222 On the other hand, 1 ,1 2 == ndd (15) and (16) imply = TRx and so by (21) we obtain
22 2 2 = 2 TRRx 2 (24)
2 2 Equation (24) shows that ≥ Ts 2 . We claim that the equal sign cannot hold, for if 2 2 2 2 = Tx 2 , then (24) implies RR 2 == 1. However, (19) is then rendered impossible 2 4 for it gives, on account of 1 = 3 = ,,1 dndd 4 = 1, 1 += Tn 2 , which is of course impossible.
22 Hence (24) shows that it mush be 0 2 << xT , and so we are led to an indefinite descent with respect to the initial equation (1). The same argument is made if we assume 4 = nd in (19). ~ 8 ~
Finally, to construct a solution to (1), from a solution to (19) is an easy matter. Given + a solution TRR 22 ),,( in Z to (19), with 431 = nddd , 3 < nd , 4 < nd , and + (TdRd 2423 )= 1, , one can construct a solution ( ,, zyx ) in Z to (19), by simply tracing back through equations (19) to (14)
4. THEOREM 3
Let p , q be primes with qp ≡≡ 4mod1 . Assume that p is a quadratic nonresidue of q (and thus q is a nonresidue of p, by the reciprocity law). Then the diophantine 2 2 equation ()2 ( 2 ) =+ zqypx 2 , with( qypx ) = 1, has no solution in the set of positive integers Z+.
Proof. Assume x ,, zy in Z+, to satisfy
2 2 ()()2 2 =+ zqypx 2 (25)
With ()qypx = 1, Then equation (25) describes a primitive pythagorean triangle, and so by assuming x to be odd and y even (without any loss of generality) , we obtain
−= nmpx 222 , 2 = 2mnqy , += nmz 22 (26) for positive integers m , n with ( nm ) = 1, and + nm ≡ 2mod1 . Since x is odd and p ≡ 4mod1 , we have px2 = mod1 4 , and so a congruence modulo 4 in the first of (26) show that m ≡ 2mod1 and n ≡ 2mod0 .
Then the second equation of (26) yields two possibilities, since ( nm ) = 1, , either
2 2 = qmm 1 , = 2nn 1 or ( nm ) = 1, (27) 2 2 = mm 1 , = 2qnn 1
If the first possibility of (27) holds, then the first equation in (26) implies
2 −≡ 2 mod qnpx (28)
But, -1 is a quadratic residue of q (since q ≡ 4mod1 ), and so by virtue of ≠ mod0 qn (because ≡ mod0 qm and ( nm ) = 1, , it follows from(27)
that p is a quadratic residue of q, contradicting the hypothesis of the theorem.
If the second possibility of (27) is the case, we then obtain from the first of (26), ≡ 22 mod qmpx , and a similar contradiction, as previously, is obtained.
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5. THEOREM 4
Let p, q be primes with p ≡ 1and q ≡ 8mod5 . Also, suppose that p is a quadratic nonresidue of q, (and so q is a quadratic nonresidue of p).
2 Under the above assumption, the diophantine equation ( 2 ) =+ zypqx 24 , with ypqx = 1),( , has a solution in positive integers x ,, zy if and only if the diophanitne equation += yxpqz 442 has solution, with ( yx ) = 1, , in positive integers x ,, zy .
Proof.
Clearly, the hupothesis of this theorem satisfies the hypothesis of Theorem 2, with = pqn .
2 Hence, according to Theorem 2, ( 2 ) =+ zypqx 24 has a solution, with ypqx = 1),( , in Z+, if and only if the equation
2 42 42 1 2 += 3 ydxdzd (29)
+ has a solution in Z , for some ,, ddd 321 with ( 32 ydxd ) = 1, , 321 = nddd , 2 < nd and 3 < nd ,()ddd 321 > 0,, .
But = pqn , = dddpq 321 , thus if d1 = 1, then on account of , 32 < pqdd , it must be
2 = pd , 3 = qd or vice versa of course.
However, this would imply, by equation (29), an equation += yqxpz 42422 with ()qypx = 1, , Which is, by Theorem 3, impossible in Z+.
Hence we see that we cannot have d1 = 1 in (29). Now, if 1 = pd , then 2 = qd and d 3 = 1or d 2 = 1and 3 = qd . So (29) would yield
+= yxqpz 4422 , ( yqx ) = 1, or (30)
+= yqxpz 4242 , ( qyx ) = 1,
However, under the conditions ()yqx = 1, , ( qyx ) = 1, , the equations in (30) imply that p is a quadratic residue of q, contradicting the hypothesis of the theorem.
A similar argument is left for the case of 1 = qd . Consequently, we see that (29) may + have a solution Z , only for 1 = pqd and = dd 32 = 1.
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2 Thus, the equation ()2 =+ zypqx 24 , with ( ypqx ) = 1, , is equivalent in Z+ , with the equation += yxpqz 442 , ()yx = 1, .
REFERENCES
[1] L.E Dickson History of the Theory of Numbers, Vol. II, pages 421, 422 AMS Chelsea Publishing, ISBN 0-8218-1935-6; 1992
[2] Konstantine Zelator, “The Diophantine equation =+ zkyx 222 and integral 1 triangles with a cosine value of " 2 , Mathematics and Computer Education, Volume 40, No. 3 (Fall 2006), pp 191-197
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