1.
Introduction
In 1955/1956, Sierpinˊski [1] showed that the equation 3x+4y=5z has x=y=z=2 as its only solution in positive integers. In the same year, Jeˊsmanowicz [2] proved that Sierpinˊski's result holds also for the following Pythagorean numbers:
Let a,b,c be fixed positive integers. Consider the exponential Diophantine equation
Jeˊsmanowicz [2] proposed the following problem:
Conjecture 1.1. Assume that a2+b2=c2. Then Eq (1.1) has no positive integer solution (x,y,z) other than x=y=z=2.
The pioneering works related to Conjecture 1.1 were obtained by Ke [3]. Ke proved the conjecture for the Pythagorean number 2n+1,2n(n+1),2n(n+1)+1 if n≡1,4,5,9,10(mod12) or n is odd and there exist a prime p and a positive integer s such that 2n+1=ps or n is the sum of two squares and there exists a prime p that is congruent to 3 modulo 4 such that 2n+1≡0(modp).
It is well known that the numbers
form all primitive Pythagorean numbers, where gcd(r,s)=1,r>s and r and s have opposite parity.
Jˊozefiak [4] confirmed the conjecture for (r,s)=(2mpn,1), where m,n∈N,N={1,2,⋯} denotes the set of positive integers and p is a prime number. Dem'janenko [5] proved the conjecture for (r,s)=(m,1), where m∈N. Grytczuk and Grelak [6] proved the conjecture for (r,s)=(2m,1), where m∈N. Takakuwa and Asaeda [7] generalized the result to (r,s)=(2m,q), where q≡3(mod4) is a prime if m is odd and a prime divisor p of a satisfies the conditions p≡1(mod4) and (qp)=−1. Most of the existing works on Conjecture 1.1 concern the coprimality case, that is, gcd(a,b)=1. Indeed, all of the above mentioned results treat the coprimality case, and such a case is essential in the study of Eq (1.1). Several authors studied the more general equation
under several conditions with n>1 and a2+b2=c2 with gcd(a,b)=1.
Deng and Cohen [8] proved that the only solution of (1.3) is x=y=z=2 if a is a prime power and n is a positive integer such that P(b)|n or P(n)⧸|b, where P(n) is the product of all distinct prime divisors of n. They proved also that the only solution of (1.3) is x=y=z=2 for each of the Pythagorean triples
and for any positive integer n. Following Deng and Cohen's work, Le [9] gave the following more general result in 1999: If (x,y,z) is a solution of (1.3) with (x,y,z)≠(2,2,2), then one of the following conditions is satisfied:
(1) max{x,y}>min{x,y}>z,P(n)|c and P(n)<P(c);
(2) x>y>z and P(n)|b;
(3) y>z>x and P(k)|a.
Sixteen years later in 2015, Yang and Fu [10] simplified the conditions given in the above result by removing all conditions on P(n). Meanwhile between 1999 and 2015, many mathematicians considered several specific cases of Eq (1.3). In 2013, Yang and Tang [11] proved the following: Let n≥4 be a positive integer and Fn=22n+1. Then, for any positive integer N, the Diophantine equation
has no solution other than (x,y,z)=(2,2,2).
In 2014, Tang and Weng [12] generalized the above result and proved that the unique solution of (1.4), for any positive integers n and N, is (x,y,z)=(2,2,2). The same year, Xinwen Zhang and Wenpeng Zhang [13] proved that the only solution of the equation
for any positive integers m and N, is (x,y,z)=(2,2,2). Finally, another special case for m=2 of Eq (1.4) was recently studied by Yang and Tang [14]. They proved that the only solution of
is (x,y,z)=(2,2,2) for N≥1. In 2014, Deng [15] considered another special case of Eq (1.4) by putting m=s+1 and accepting some divisibility conditions such as P(a)|N or P(N)⧸|a,s≥0, and proved that (x,y,z)=(2,2,2) is the only solution of the equation
In 2015, Ma and Wu [16] proved that the only solution of the equation
is x=y=z=2 if P(4n2−1)|N. Very recently, Miyazaki [17] nicely proved the Jeˊsmanowicz conjecture when a or b is a power of 2 by extending the result of Tang and Weng [12]. In 2017, Soydan, Demirci, Cangul and Togbe [18] proved that the Diophantine equation
has only the solution x=y=z=2 for any positive integer n. In 2022, Feng and Luo [19] proved that the Diophantine equation
has only the solution x=y=z=2 for any positive integer n, where p and q are odd primes with pk=2m1−am2 and ql=2m1+am2,k,l,m1 and m2 are positive integers and a≡5(mod8) is a prime.
After these works, Conjecture 1.1 has been proved to be true for various particular cases. For recent results, we only refer to the papers of Deng, Yuan and Luo [20], Hu and Le [21], Miyazaki [17,22], Miyazaki, Yuan and Wu [23], Terai [24], Yuan and Han [25] and the references given there.
In this paper, we will prove that the result of [19] holds when a is a positive integer with a≡±3(mod8). We have following:
Theorem 1.1. Let k,l,m1,m2 be positive integers and let p and q be odd primes such that pk=2m1−am2 and ql=2m1+am2, where a is a positive integer with a≡±3(mod8). Then the equation
has only the positive integer solution (x,y,z)=(2,2,2).
Theorem 1.2. Let the assumptions of k,l,m1,m2,p,q be as in Theorem 1.1. Suppose that the following conditions hold:
(i) If a=a1a2 with a1≡1(mod8) not a square and gcd(a1,a2)=1, then there exists a prime divisor P of a2 such that (a1P)=−1,
(ii) 2|m1,a≡5(mod8) or 2⧸|m2,a≡3(mod8).
Then the equation
has only the positive integer solution (x,y,z)=(2,2,2) for any n≥1.
Corollary 1.1. Let the assumptions of k,l,m1,m2,p,q be as in Theorem 1.1. If a is a product of a square and a prime that is congruent 3 modulo 8 and m2 is odd, or if a is a product of a square and a prime that is congruent 5 modulo 8 and m1 is even, then Eq (1.10) has only the positive integer solution (x,y,z)=(2,2,2) for any n≥1.
We organize this paper as follows. In Section 2, we present some lemmas which are needed in the proofs of our main results. Consequently, in Sections 3 to 4, we give the proofs of Theorem 1.1 to 1.2 and Corollary 1.1, respectively. In Section 5, we give some applications of Theorem 1.1 and Theorem 1.2.
2.
Lemmas
In this section, we present some lemmas that will be used in the proof of results.
Lemma 2.1. Let k,l,m1,m2 be positive integers and let p and q be odd primes such that pk=2m1−am2 and ql=2m1+am2, where a is a positive integer with a≡±3(mod8). Then m1 is odd or m2 is odd except for (p,q,k,l,m1,m2,a)=(7,5,1,2,4,2,3). Moreover if m1 is odd and m2 is even, then 3|a.
Proof. It is easy to find that is enough to prove that 2⧸|m1 and 3|a except for (p,q,k,l,m1,m2,a)=(7,5,1,2,4,2,3) when m2 is even.
If m1 is odd, we claim that 3|a. On the contrary suppose that 3⧸|a; then, we get from ql=2m1+am2 that 3|ql, and so q=3 since q is prime. Taking modulo 4 for the equation 3l=2m1+am2 would give 3l≡1(mod4). It follows that l is even and
which leads to a contradiction.
If m1 is also even, then we get from the condition
that
So 2m12+1=pk2(pk1−k2+1), which would thus give k2=0 and 2m12−am22=1. If m1>4, then taking the equation 2m12−am22=1 modulo 8 yields am22≡−1(mod8), which leads to a contradiction since a≡±3(mod8). Hence m1=4,a=3,m2=2, which implies that p=7,k=1,q=5,l=2. This completes the proof. □
Lemma 2.2. Let v2(n) denote nonnegative integer t such that 2t|n and 2t+1⧸|n. Let a be an odd integer with a≡±5(mod8). Then we have that v2(am−1)=v2(m)+2 if v2(m)=h≥1.
Proof. We prove the lemma 2.2 by induction on h=v2(m)≥1. For h=1 we get
If a≡5(mod8), then we have am1+1≡6(mod8) and am1−1≡4(mod8), where m=2m1,m1 is odd. Hence (2.1) implies v2(am−1)=3=v2(m)+2. If a≡3(mod8) then we have am1+1≡4(mod8) and am1−1≡2(mod8). Thus (2.1) also implies v2(am−1)=3=v2(m)+2. If the result is shown for some positive integer h=v2(m), then
Since am+1≡2(mod8), we get
This completes the proof. □
Lemma 2.3. If (x,y,z) is a solution of Eq (1.9) with x≡y≡z≡0(mod2), then (x,y,z)=(2,2,2).
Proof. It is easy to find that m1≥3 by the condition pk=2m1−am2. We may write x=2x1,y=2y1,z=2z1 by the assumption x≡y≡z≡0(mod2). It follows from (1.3) that
As
then we have
and
Taking the difference of the above equations gives
If y1 is even, then we have that
and
It follows from Eq (2.2) that
and
where a1a2=a,a1>1. Taking modulo a1 for Eq (2.3) yields to
which leads to a contradiction. Hence y1 is odd; then, we have that
and
It follows from Eq (2.2) that
and
where a1a2=a. Taking modulo a1 for Eq (2.5) yields 2m1y1+1≡0(moda1). So a1=1,a2=a. We claim that y1=1. On the contrary suppose y1>1. Note that y1 is odd; Eq (2.5) would give that
Thus y1am2(y1−1)≡0(mod2), which is a contradiction. Therefore y1=1 and 2(m1+1)x1=2m1+1 yield that x1=1. Substituting x=y=2 into Eq (1.9) gives z=2.
This completes the proof. □
In this section, we present two useful results necessary for the proof of our main results.
Lemma 2.4. ([9]) If (x,y,z) is a solution of (1.3) with (x,y,z)≠(2,2,2), then one of the following conditions is satisfied
(i) max{x,y}>min{x,y}>z; (ii) x>z>y; (iii) y>z>x.
Lemma 2.5. ([15], [17]) Assume that n>1. Then (1.3) has no solution (x,y,z) with max{x,y}>min{x,y}>z.
Lemma 2.6. The equation
has only the positive integer solution (x,y,z)=(2,2,2) for any n≥1.
Proof. We first consider the case n=1. Assume that (x,y,z) is a positive integer solution. Taking modulo 4 for Eq (2.7) leads to (−1)y≡1(mod4). It follows that y is even. Taking modulo 32 for Eq (2.7) leads to 34|y−z|≡1(mod32). Then we get by Lemma 2.2 that 5≤v2(4)+2+v2(|y−z|)=4+v2(|y−z|). Thus z is even since y is even. We get from Eq (2.7) that
If y2 is even, then we have
and
So
which is impossible. Hence y2 is odd and
since 337z2−175y2≡2(mod8) and 337z2+175y2≡2(mod3). Therefore
Taking modulo 3 yields (−1)5x−2≡1(mod3). It follows that x is even. Thus we get by Lemma 2.3 that x=y=z=2.
We now consider the case n>1. Assume that (x,y,z) is a positive integer solution with (x,y,z)≠(2,2,2). Then we have by Lemmas 2.4 and 2.5 that x>z>y or y>z>x. We shall discuss separately two cases.
The case x>z>y. Then dividing Eq (2.7) by ny yields
Since gcd((7⋅52)y,337z)=1, we can observe that the two factors on the right-hand side are co-prime. Hence Eq (2.8) yields n=7u for some positive integer u with y=u(z−y) and
or n=5v for some positive integer v with 2y=v(z−y) and
or n=7u⋅5v for some positive integers u and v and
If (2.9) holds, then taking modulo 16 leads to 9y≡1(mod16). It follows that y is even. If z is even, then we get from Eq (2.9) that
It follows that 32x≤337z2+5y, which is impossible since
Hence z is odd. Taking modulo 5 for Eq (2.9) gives 2z≡−2x⋅2u(x−z)(mod5). So x is also odd since y=u(z−y). By taking Eq (2.9) modulo 7, we have 22y≡1(mod7). It follows that y≡0(mod3). Taking modulo 9 for Eq (2.9) yields 22z≡1(mod9). It follows that z≡0(mod3). Taking modulo 27 for Eq (2.9) leads to 2y≡13z(mod27). Since
and
we have (y,z)≡(0,9)(mod18) or (y,z)≡(6,3)(mod18) or (y,z)≡(12,15)(mod18).
(i) (y,z)≡(0,9)(mod18); then, the congruence modulo 19 of Eq (2.9) leads to
It follows that
which is a contradiction.
(ii) (y,z)≡(6,3)(mod18); then, the congruence modulo 19 of Eq (2.9) leads to
It follows that
which is also a contradiction.
(iii) (y,z)≡(12,15)(mod18); then, the congruence modulo 19 of Eq (2.9) leads to
It follows that
which leads to a contradiction.
If (2.10) holds, then taking modulo 4 leads to (−1)y≡1(mod4). It follows that y is even. The congruence modulo 5 of Eq (2.10) gives 2y≡2z(mod5). So z is also even. Then we get from Eq (2.10) that
It follows that 32x≤337z2+7y2, which is impossible since
If (2.11) holds, then taking modulo 5 for Eq (2.11) would give 1≡2z(mod5). It follows that z is also even. Then we get from Eq (2.11) that
It follows that 32x≤337z2+1, which is impossible since
The case y>z>x. Then dividing Eq (2.7) by nx yields
It is easy to see that the two factors on the right-hand side are co-prime. Thus, Eq (2.12) yields n=3s for some positive integer s and
or n=2r for some positive integers r and
or n=2r3s for some positive integers r and s and
If (2.13) holds, then taking modulo 3 for Eq (2.13) would give (−1)x≡1(mod3). It follows that x is even. Taking modulo 5 for Eq (2.13) leads to 2x≡2z(mod5). So z is also even. Then we get from Eq (2.13) that
It follows that 25y≤337z2+25x2, which is impossible since
If (2.14) holds, then taking modulo 5 for Eq (2.14) leads to 32x≡2z(mod5). It follows that 1=(32x5)=(25)z=(−1)z. Thus z is even. Then we get from Eq (2.14) that
It follows that 25y≤337z2+3x, which is impossible since
Similarly we can prove that (2.15) is impossible.
This completes the proof. □
3.
Proof of Theorem 1.1
Proof. It is easy to find that is enough to prove that x≡y≡z≡0(mod2) by Lemma 2.3. Substituting the conditions pk=2m1−am2 and ql=2m1+am2 into Eq (1.9) gives
Taking modulo 4 for Eq (3.1) gives (−1)y≡1(mod4). It follows that y is even. We now prove that z is also even. Taking modulo 2m1+1 for Eq (3.1) yields
where u=|y−z|. It follows that m1+1≤v2(2m2)+v2(u)+2=v2(m2)+3+v2(u) by Lemma 2.2. Let v2(m2)=h and m2=2hr,2⧸|r. If a>3 or r>1, then we have
It follows that m1≥2h+1+1≥h+3. If a=3 and r=1, one can easily prove from 2m1=pk+32h that m1≥h+3. Hence v2(u)≥1. It follows that z is also even. Finally we prove that x is even. If m2 is odd, we get from Eq (3.1) that
since
It follows that x is even since m2 is odd. If m2 is even, then we have by Lemmas 2.1 and 2.6 that m1 is odd and 3|a. We get from Eq (3.1) that
If y2 is even, then similarly we have
and
where a1a2=a. If 3|a1, then taking modulo 3 for Eq (3.2) yields 2≡0(mod3), which is a contradiction. Thus 3|a2. But this is impossible since
Therefore y2 is odd; then, we have
and
where a1a2=a. Thus taking the difference of the above equations yields
So
and
Adding the two equations one yields
We claim that y2=y1=1. On the contrary suppose y1>1. Note that y1 is odd; we get that
It follows that y1⋅am2(y1−1)≡0(mod2), which leads to a contradiction. Therefore y1=1 and 2(m1+1)x2⋅am2x21=2m1+1 yields that x=2.
This completes the proof. □
4.
Proof of Theorem 1.2 and Corollary 1.1
Proof. Assume that (x,y,z) is a positive integer solution with (x,y,z)≠(2,2,2). Then we have by Lemmas 2.4, 2.5 and Theorem 1.1 that n>1 and either x>z>y or y>z>x. By the assumptions and Lemma 2.1, we have that m2 is odd except for (p,q,k,l,m1,m2,a)=(7,5,1,2,4,2,3). We know that the case (p,q,k,l,m1,m2,a)=(7,5,1,2,4,2,3) is impossible by Lemma 2.6. We shall discuss separately two cases.
Consider the case x>z>y. Then dividing both sides of Eq (1.10) by ny yields
If gcd(pq,n)=1, Eq (4.1) and n>1 imply that y=z<x. We deduce a contradiction to the fact that y<z. Therefore, we suppose gcd(pq,n)>1. We write n=puqv, where u+v≥1.
(ⅰ) If u≥1,v=0, then n=pu. Equation (4.1) becomes
Then substituting the conditions pk=2m1−am2 and ql=2m1+am2 into Eq (4.2) would give
Taking modulo 8 for Eq (4.3) gives a^{m_2y}\equiv 1 \pmod 8 . So y is even since m_2 is odd. By taking equation p^k = 2^{m_1}-a^{m_2} modulo 8 , we have p\equiv -a\equiv {\mp 3} \pmod 8 . Taking modulo p for Eq (4.3) leads to
It follows that
Therefore z is even. Then we get from Eq (4.3) that
If \frac{y}{2} is odd, then we have
and
if a\equiv 3\pmod 8 or
and
if a\equiv {-3}\pmod 8 . It follows that
But the left hand side of Eq (4.3) is divided by 2^4 , which leads to a contradiction. Hence \frac{y}{2} is even and
Thus it follows that
however, this is impossible since
(ⅱ) If u = 0, v\geq 1 , then n = q^v . Equation (4.1) becomes
Then substituting the conditions p^k = 2^{m_1}-a^{m_2} and q^l = 2^{m_1}+a^{m_2} into Eq (4.4) would give
Taking modulo 8 for Eq (4.5) gives (-a)^{m_2y}\equiv 1 \pmod 8 . So y is even since m_2 is odd. By taking the equation q^k = 2^{m_1}+a^{m_2} modulo 8 leads to q\equiv a\equiv {\pm 3} \pmod 8 . Taking modulo q for Eq (4.5) leads to
It follows that (-1)^z = \left(\frac{2}{q}\right)^z = \left(\frac{2^{m_1}-a^{m_2}}{q}\right)^y = 1 . Therefore z is even. Then similarly we get from Eq (4.5) that
which is impossible by the above result that has been proved (see discussion of Eq (4.2)).
(ⅲ) If u\geq 1, v\geq 1 , then n = p^uq^v . Equation (4.1) becomes
Then substituting the conditions p^k = 2^{m_1}-a^{m_2} and q^l = 2^{m_1}+a^{m_2} into Eq (4.6) would give
Taking modulo q for Eq (4.5) leads to
It follows that (-1)^z = \left(\frac{2}{q}\right)^z = \left(\frac{1}{q}\right) = 1 . Therefore z is even. Then similarly we get from Eq (4.7) that
which is impossible by the above result that has been proved (see discussion of Eq (4.2)). This completes the proof of the first case.
Consider the case y > z > x . Then dividing both sides of Eq (1.10) by n^x yields
If \gcd(2a, n) = 1 , Eq (4.8) and n > 1 imply that x = z < y . We deduce a contradiction to the fact that x < z . Therefore, we suppose \gcd(2a, n) > 1 . We write n = 2^ra_1^s , where r+s\geq 1, a_1 > 1 is a divisor of a .
(ⅰ) If r = 0, s\geq 1 , then n = a_1^s and m_2x = s(z-x) . If a_1 < a , then Eq (4.8) becomes
Since a\equiv \pm 3 \pmod 8 , we have to consider the eight cases.
Case 1: (a_1, a_2)\equiv {(1, 3)}\pmod 8. Taking modulo 2^{m_1}-a^{m_2} for Eq (4.9) leads to
It follows that
which leads to x\equiv z\pmod 2 since m_2 is odd. So we get from m_2x = s(z-x) that x\equiv z\equiv 0\pmod 2 . Then we get from Eq (4.9) either
or
Hence
which is impossible since
Case 2: (a_1, a_2)\equiv {(1, 5)}\pmod 8. Taking modulo 2^{m_1}-a^{m_2} for Eq (4.9) leads to
We know that is impossible by the result proved in Case 1.
Case 3: (a_1, a_2)\equiv {(3, 1)}\pmod 8. Then taking modulo 2^{m_1}-a^{m_2} for Eq (4.9) leads to
We already prove that is impossible if m_1 is even. If m_1 is odd, then we have
which leads to z is even. Similarly we know that a_2 is not a square. So by the assumption, there is an odd prime divisor P of a_1 such that
which leads to x being even. We know that is impossible by the result proved in Case 1.
Case 4: (a_1, a_2)\equiv {(3, 7)}\pmod 8. Taking modulo 2^{m_1}+a^{m_2} for Eq (4.9) leads to
It follows that
which leads to x\equiv z\pmod 2 . We know that is impossible by the result proved in Case 1.
Case 5: (a_1, a_2)\equiv {(5, 1)}\pmod 8. Then taking modulo 2^{m_1}+a^{m_2} for Eq (4.9) leads to
We already proved that that is impossible.
Case 6: (a_1, a_2)\equiv {(5, 7)}\pmod 8. Taking modulo 2^{m_1}-a^{m_2} for Eq (4.9) leads to
It follows that
which leads to x\equiv z\pmod 2 if m_1 is even. We have already proven that that is impossible. If m_1 is odd, then we get from Eq (4.10) that z is even. On the other hand, taking modulo 2^{m_1}+a^{m_2} for Eq (4.9) leads to
which leads to x being even. We know that is impossible by the result proved in Case 1.
Case 7: (a_1, a_2)\equiv {(7, 3)}\pmod 8. Taking modulo 2^{m_1}+a^{m_2} for Eq (4.9) leads to
We already proved that that is impossible.
Case 8: (a_1, a_2)\equiv {(7, 5)}\pmod 8. Taking modulo 2^{m_1}+a^{m_2} for Eq (4.9) leads to
It follows that
It follows that x\equiv z\pmod 2 . We already proved that that is impossible.
If a_1 = a then Eq (4.8) becomes
Taking modulo 2^{m_1}+a^{m_2} for Eq (4.11) leads to
It follows that
It follows that (m_1+1)x\equiv z\pmod 2 . On the other hand, taking modulo a for Eq (4.11) leads to
It follows that
which leads to (m_1+1)x\equiv z\equiv 0 \pmod 2 . We know that is impossible by the above result.
(ⅱ) If r\geq 1, s = 0 , then n = 2^r and (m_1+1)x = r(z-x) . Equation (4.8) becomes
Thus
It follows that
which yields that r(y-z) is even. Taking modulo 2^{m_1}-a^{m_2} for Eq (4.12) leads to a^{m_2x}\equiv {(2\cdot a^{2m_2})^z} \pmod {2^{m_1}-a^{m_2}}. It follows that
which yields m_1m_2x\equiv z\pmod 2 . If r(y-z) > 2 , we consider Eq (4.12) modulo 8 ; we have a^{m_2x}\equiv 1\pmod 8 . This means that m_2x is even. As
we get
or
However, the inequalities
contradict (4.13) and (4.14). Hence r(y-z) = 2 ; considering Eq (4.12) modulo 8 , we obtain a^{m_2x}\equiv 5\pmod 8 . This means that m_2x\equiv 0\pmod 2 or a\equiv 5\pmod 8 . We know that the case m_2x\equiv 0\pmod 2 is impossible by the proof of the case r(y-z) > 2 . Hence a\equiv 5\pmod 8 . We have that m_1 is even by the assumption. Thus we get from (m_1+1)x = r(z-x) and r(y-z) = 2 that r = 1, y = z+2 . Note that m_1m_2x\equiv z\pmod 2 . We get that both y and z are even. Therefore we get from Eq (4.12) that
which is impossible.
(ⅲ) If r\geq 1, s\geq 1 , then n = 2^ra_1^s and (m_1+1)x = r(z-x), m_2x = s(z-x) . Equation (4.8) becomes
Similarly we can prove that Eq (4.15) is impossible. This completes the proof of the second case. This completes the proof. □
By the proof of Theorem 1.2, one can immediately obtain Corollary 1.1.
5.
Applications
Corollary 5.1. Equation (1.1) has only the positive integer solution (x, y, z) = (2, 2, 2) if (a, b, c) is one of the following primitive Pythagorean numbers
Proof. By Theorem 1.1 and Table 1, one can immediately obtain the Corollary 5.1 by a simple calculation. □
Remark 5.1. There are many prime numbers p, q and positive integers k, l, m_1, m_2, a satisfying the conditions of Theorem 1.1. One can see Tables 2 and 3.
Corollary 5.2. Equation (1.1) has only the positive integer solution (x, y, z) = (2, 2, 2) if (a, b, c) is one of the following Pythagorean numbers
for any n\in N .
Proof. By Theorem 1.2 and Table 4, one can immediately obtain Corollary 5.2 by a simple calculation. □
6.
Conclusions
It is easy to see that Je \acute{s} manowicz' a conjecture holds for the following set of primitive Pythagorean numbers:
In addition, Je \acute{s} manowicz' conjecture holds for non-primitive Pythagorean numbers:
for any positive integer n if for a = a_1a_2 with a_1\equiv 1 \pmod 8 not a square and \gcd(a_1, a_2) = 1 , then there exists a prime divisor P of a_2 such that \left(\frac{a_1}{P}\right) = -1 and 2|m_1, a\equiv 5 \pmod 8 or 2\not|m_2, a\equiv 3\pmod 8 .
Acknowledgments
This research was supported by the Major Project of Education Department in Sichuan (No. 16ZA0173), NSF of China (No. 11871058) and Nation project cultivation project of China West Normal University (No. 22KA018).
Conflict of interest
All authors declare no conflict of interest regarding the publication of this paper.