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Research article

On the conjecture of Jeˊsmanowicz

  • Received: 28 December 2022 Revised: 29 March 2023 Accepted: 03 April 2023 Published: 17 April 2023
  • MSC : 11D61, 11D75

  • Let k,l,m1 and m2 be positive integers and let both p and q be odd primes such that pk=2m1am2 and ql=2m1+am2 where a is a positive integer with a±3(mod8). In this paper, using only the elementary methods of factorization, congruence methods and the quadratic reciprocity law, we show that Jeˊsmanowicz' a conjecture holds for the following set of primitive Pythagorean numbers:

    q2lp2k2,pkql,q2l+p2k2.

    We also prove that Jeˊsmanowicz' conjecture holds for non-primitive Pythagorean numbers:

    nq2lp2k2,npkql,nq2l+p2k2,

    for any positive integer n if for a=a1a2 with a11(mod8) not a square and gcd(a1,a2)=1, then there exists a prime divisor P of a2 such that (a1P)=1 and 2|m1,a5(mod8) or 2|m2,a3(mod8).

    Citation: Nan Fan, Jiagui Luo. On the conjecture of Jeˊsmanowicz[J]. AIMS Mathematics, 2023, 8(6): 14232-14252. doi: 10.3934/math.2023728

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  • Let k,l,m1 and m2 be positive integers and let both p and q be odd primes such that pk=2m1am2 and ql=2m1+am2 where a is a positive integer with a±3(mod8). In this paper, using only the elementary methods of factorization, congruence methods and the quadratic reciprocity law, we show that Jeˊsmanowicz' a conjecture holds for the following set of primitive Pythagorean numbers:

    q2lp2k2,pkql,q2l+p2k2.

    We also prove that Jeˊsmanowicz' conjecture holds for non-primitive Pythagorean numbers:

    nq2lp2k2,npkql,nq2l+p2k2,

    for any positive integer n if for a=a1a2 with a11(mod8) not a square and gcd(a1,a2)=1, then there exists a prime divisor P of a2 such that (a1P)=1 and 2|m1,a5(mod8) or 2|m2,a3(mod8).



    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:

    (a,b,c)=(5,12,13),(7,24,25),(9,40,41),(11,60,61).

    Let a,b,c be fixed positive integers. Consider the exponential Diophantine equation

    ax+by=cz. (1.1)

    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 n1,4,5,9,10(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+10(modp).

    It is well known that the numbers

    a=r2s2,b=2rs,c=r2+s2 (1.2)

    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,nN,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 mN. Grytczuk and Grelak [6] proved the conjecture for (r,s)=(2m,1), where mN. Takakuwa and Asaeda [7] generalized the result to (r,s)=(2m,q), where q3(mod4) is a prime if m is odd and a prime divisor p of a satisfies the conditions p1(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

    (an)x+(bn)y=(cn)z (1.3)

    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

    (a,b,c)=(3,4,5),(5,12,13),(7,24,25),(9,40,41),(11,60,61)

    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 n4 be a positive integer and Fn=22n+1. Then, for any positive integer N, the Diophantine equation

    ((Fn2)N)x+(22n1+1N)y=(FnN)z (1.4)

    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

    ((22m1)N)x+(2m+1N)y=((22m+1)N)z, (1.5)

    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

    (15N)x+(8N)y=(17N)z

    is (x,y,z)=(2,2,2) for N1. 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,s0, and proved that (x,y,z)=(2,2,2) is the only solution of the equation

    ((22s+21)N)x+(2s+2N)y=((22s+2+1)N)z. (1.6)

    In 2015, Ma and Wu [16] proved that the only solution of the equation

    ((4n21)N)x+(4nN)y=((4n2+1)N)z (1.7)

    is x=y=z=2 if P(4n21)|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

    (1123222n)x+(3211n)y=(112+3222n)z

    has only the solution x=y=z=2 for any positive integer n. In 2022, Feng and Luo [19] proved that the Diophantine equation

    (q2lp2k2n)x+(pkqln)y=(q2l+p2k2n)z (1.8)

    has only the solution x=y=z=2 for any positive integer n, where p and q are odd primes with pk=2m1am2 and ql=2m1+am2,k,l,m1 and m2 are positive integers and a5(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=2m1am2 and ql=2m1+am2, where a is a positive integer with a±3(mod8). Then the equation

    (q2lp2k2)x+(pkql)y=(q2l+p2k2)z (1.9)

    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 a11(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,a5(mod8) or 2|m2,a3(mod8).

    Then the equation

    (q2lp2k2n)x+(pkqln)y=(q2l+p2k2n)z, (1.10)

    has only the positive integer solution (x,y,z)=(2,2,2) for any n1.

    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 n1.

    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.

    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=2m1am2 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 3l1(mod4). It follows that l is even and

    1=(3la)=(2a)=1,

    which leads to a contradiction.

    If m1 is also even, then we get from the condition

    pk=2m1am2=(2m12+am22)(2m12am22)

    that

    2m12+am22=pk1,2m12am22=pk2,k1>k20.

    So 2m12+1=pk2(pk1k2+1), which would thus give k2=0 and 2m12am22=1. If m1>4, then taking the equation 2m12am22=1 modulo 8 yields am221(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(am1)=v2(m)+2 if v2(m)=h1.

    Proof. We prove the lemma 2.2 by induction on h=v2(m)1. For h=1 we get

    am1=(am1+1)(am11). (2.1)

    If a5(mod8), then we have am1+16(mod8) and am114(mod8), where m=2m1,m1 is odd. Hence (2.1) implies v2(am1)=3=v2(m)+2. If a3(mod8) then we have am1+14(mod8) and am112(mod8). Thus (2.1) also implies v2(am1)=3=v2(m)+2. If the result is shown for some positive integer h=v2(m), then

    a2m1=(am+1)(am1).

    Since am+12(mod8), we get

    v2(a2m1)=v2(am+1)+v2(am1)=v2(m)+3=v2(2m)+2.

    This completes the proof.

    Lemma 2.3. If (x,y,z) is a solution of Eq (1.9) with xyz0(mod2), then (x,y,z)=(2,2,2).

    Proof. It is easy to find that m13 by the condition pk=2m1am2. We may write x=2x1,y=2y1,z=2z1 by the assumption xyz0(mod2). It follows from (1.3) that

    ((22m1+a2m2)z1+am2x12(m1+1)x1)((22m1+a2m2)z1am2x12(m1+1)x1)=p2ky1q2ly1.

    As

    gcd((22m1+a2m2)z1+am2x12(m1+1)x1,(22m1+a2m2)z1am2x12(m1+1)x1)=1,

    then we have

    (22m1+a2m2)z1+am2x12(m1+1)x1=q2ly1

    and

    (22m1+a2m2)z1am2x12(m1+1)x1=p2ky1.

    Taking the difference of the above equations gives

    2(m1+1)x1+1am2x1=((2m1+am2)y1+(2m1am2)y1)((2m1+am2)y1(2m1am2)y1). (2.2)

    If y1 is even, then we have that

    (2m1+am2)y1+(2m1am2)y12(mod8)

    and

    (2m1+am2)y1(2m1am2)y10(mod8).

    It follows from Eq (2.2) that

    (2m1+am2)y1+(2m1am2)y1=2am2x11 (2.3)

    and

    (2m1+am2)y1(2m1am2)y1=2(m1+1)x1am2x12, (2.4)

    where a1a2=a,a1>1. Taking modulo a1 for Eq (2.3) yields to

    2m1y1+10(moda1),

    which leads to a contradiction. Hence y1 is odd; then, we have that

    (2m1+am2)y1+(2m1am2)y10(mod4)

    and

    (2m1+am2)y1(2m1am2)y12(mod4).

    It follows from Eq (2.2) that

    (2m1+am2)y1+(2m1am2)y1=2(m1+1)x1am2x11 (2.5)

    and

    (2m1+am2)y1(2m1am2)y1=2am2x12, (2.6)

    where a1a2=a. Taking modulo a1 for Eq (2.5) yields 2m1y1+10(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

    2(m1+1)x1m11=(y11)/2r=0(y12r)2m1(y12r1)a2rm2.

    Thus y1am2(y11)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

    (288n)x+(175n)y=(337n)z (2.7)

    has only the positive integer solution (x,y,z)=(2,2,2) for any n1.

    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)y1(mod4). It follows that y is even. Taking modulo 32 for Eq (2.7) leads to 34|yz|1(mod32). Then we get by Lemma 2.2 that 5v2(4)+2+v2(|yz|)=4+v2(|yz|). Thus z is even since y is even. We get from Eq (2.7) that

    (337z2+175y2)(337z2175y2)=32x25x.

    If y2 is even, then we have

    337z2+175y22(mod8)

    and

    337z2+175y22(mod3).

    So

    337z2+175y2=2,337z2175y2=32x25x1,

    which is impossible. Hence y2 is odd and

    337z2175y2=232x,337z2+175y2=25x1

    since 337z2175y22(mod8) and 337z2+175y22(mod3). Therefore

    175y2=25x232x.

    Taking modulo 3 yields (1)5x21(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

    (752)y=nzy(337z25x32xnxz). (2.8)

    Since gcd((752)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(zy) and

    52y=337z25x32x7u(xz), (2.9)

    or n=5v for some positive integer v with 2y=v(zy) and

    7y=337z25x32x5v(xz), (2.10)

    or n=7u5v for some positive integers u and v and

    1=337z25x32x7u(xz)5v(xz). (2.11)

    If (2.9) holds, then taking modulo 16 leads to 9y1(mod16). It follows that y is even. If z is even, then we get from Eq (2.9) that

    32x|337z2+5yor32x|337z25y.

    It follows that 32x337z2+5y, which is impossible since

    32x=(7+52)2x/2=(674+1452)x/2>337z2+5y.

    Hence z is odd. Taking modulo 5 for Eq (2.9) gives 2z2x2u(xz)(mod5). So x is also odd since y=u(zy). By taking Eq (2.9) modulo 7, we have 22y1(mod7). It follows that y0(mod3). Taking modulo 9 for Eq (2.9) yields 22z1(mod9). It follows that z0(mod3). Taking modulo 27 for Eq (2.9) leads to 2y13z(mod27). Since

    2y{1(mod27),y0(mod18)10(mod27),y6(mod18)11(mod27),y12(mod18),

    and

    13z{1(mod27),z9(mod18)10(mod27),z3(mod18)11(mod27),z15(mod18),

    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

    225x32x7u(xz)(mod19).

    It follows that

    1=(219)=(119)(219)=1,

    which is a contradiction.

    (ii) (y,z)(6,3)(mod18); then, the congruence modulo 19 of Eq (2.9) leads to

    125x32x7u(xz)(mod19).

    It follows that

    1=(119)=(119)(219)=1,

    which is also a contradiction.

    (iii) (y,z)(12,15)(mod18); then, the congruence modulo 19 of Eq (2.9) leads to

    1425x32x7u(xz)(mod19).

    It follows that

    1=(219)(719)=(119)(219)=1,

    which leads to a contradiction.

    If (2.10) holds, then taking modulo 4 leads to (1)y1(mod4). It follows that y is even. The congruence modulo 5 of Eq (2.10) gives 2y2z(mod5). So z is also even. Then we get from Eq (2.10) that

    32x|337z2+7y2or32x|337z27y2.

    It follows that 32x337z2+7y2, which is impossible since

    32x=(7+52)2x/2=(674+1452)x/2>337z2+7y2.

    If (2.11) holds, then taking modulo 5 for Eq (2.11) would give 12z(mod5). It follows that z is also even. Then we get from Eq (2.11) that

    32x|337z2+1or32x|337z21.

    It follows that 32x337z2+1, which is impossible since

    32x=(7+52)2x/2=(674+1452)x/2>337z2+1.

    The case y>z>x. Then dividing Eq (2.7) by nx yields

    25x32x=nzx(337z175ynyz). (2.12)

    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

    25x=337z175y3s(yz), (2.13)

    or n=2r for some positive integers r and

    32x=337z175y2r(yz), (2.14)

    or n=2r3s for some positive integers r and s and

    1=337z175y2r(yz)3s(yz). (2.15)

    If (2.13) holds, then taking modulo 3 for Eq (2.13) would give (1)x1(mod3). It follows that x is even. Taking modulo 5 for Eq (2.13) leads to 2x2z(mod5). So z is also even. Then we get from Eq (2.13) that

    25y|337z2+25x2or25y|337z225x2.

    It follows that 25y337z2+25x2, which is impossible since

    25y=(24+32)2y/2=(337+2532)y/2>337z2+25x2.

    If (2.14) holds, then taking modulo 5 for Eq (2.14) leads to 32x2z(mod5). It follows that 1=(32x5)=(25)z=(1)z. Thus z is even. Then we get from Eq (2.14) that

    25y|337z2+3xor25y|337z23x.

    It follows that 25y337z2+3x, which is impossible since

    25y=(24+32)2y/2=(337+2532)y/2>337z2+3x.

    Similarly we can prove that (2.15) is impossible.

    This completes the proof.

    Proof. It is easy to find that is enough to prove that xyz0(mod2) by Lemma 2.3. Substituting the conditions pk=2m1am2 and ql=2m1+am2 into Eq (1.9) gives

    am2x2(m1+1)x+(22m1a2m2)y=(22m1+a2m2)z. (3.1)

    Taking modulo 4 for Eq (3.1) gives (1)y1(mod4). It follows that y is even. We now prove that z is also even. Taking modulo 2m1+1 for Eq (3.1) yields

    a2m2u10(mod2m1+1),

    where u=|yz|. It follows that m1+1v2(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

    2m1>am2=a2hr>22h+1.

    It follows that m12h+1+1h+3. If a=3 and r=1, one can easily prove from 2m1=pk+32h that m1h+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

    (1)(m1m2+m1+1)x=(a2m1am2)m2x(22m1am2)(m1+1)x=(22m1+a2m22m1am2)z=1,

    since

    (22m1am2)=1,(a2m1am2)=(2m1a)=(1)m1.

    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

    ((22m1+a2m2)z2+(22m1a2m2)y2)((22m1+a2m2)z2(22m1a2m2)y2)=am2x2(m1+1)x.

    If y2 is even, then similarly we have

    (22m1+a2m2)z2+(22m1a2m2)y2=2am2x1 (3.2)

    and

    (22m1+a2m2)z2(22m1a2m2)y2=2(m1+1)x1am2x2, (3.3)

    where a1a2=a. If 3|a1, then taking modulo 3 for Eq (3.2) yields 20(mod3), which is a contradiction. Thus 3|a2. But this is impossible since

    2(m1+1)x1am2x2>2(m1+1)x>2am2x>2am2x1.

    Therefore y2 is odd; then, we have

    (22m1+a2m2)z2+(22m1a2m2)y2=2(m1+1)x1am2x1

    and

    (22m1+a2m2)z2(22m1a2m2)y2=2am2x2,

    where a1a2=a. Thus taking the difference of the above equations yields

    (2m1+am2)y2(2m1am2)y2=(2(m1+1)x21am2x21+am2x22)(2(m1+1)x21am2x21am2x22).

    So

    (2m1+am2)y2=2(m1+1)x21am2x21+am2x22

    and

    (2m1am2)y/2=2(m1+1)x21am2x21am2x22.

    Adding the two equations one yields

    (2m1+am2)y2+(2m1am2)y2=2(m1+1)x2am2x21.

    We claim that y2=y1=1. On the contrary suppose y1>1. Note that y1 is odd; we get that

    2(m1+1)(x12)2am2x21=(y11)/2r=0(y12r)2m1(y12r1)a2rm2.

    It follows that y1am2(y11)0(mod2), which leads to a contradiction. Therefore y1=1 and 2(m1+1)x2am2x21=2m1+1 yields that x=2.

    This completes the proof.

    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

    (pkql)y=nzy((q2l+p2k2)z(q2lp2k2)xnxz). (4.1)

    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+v1.

    (ⅰ) If u1,v=0, then n=pu. Equation (4.1) becomes

    qly=(q2l+p2k2)z(q2lp2k2)xpu(xz). (4.2)

    Then substituting the conditions pk=2m1am2 and ql=2m1+am2 into Eq (4.2) would give

    2(m1+1)xam2xpu(xz)=(22m1+a2m2)z(2m1+am2)y. (4.3)

    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

    (2\cdot a^{2m_2})^z\equiv (2^{m_1}+a^{m_2})^y \pmod p.

    It follows that

    (-1)^z = \left(\frac{2}{p}\right)^z = \left(\frac{2^{m_1}+a^{m_2}}{p}\right)^y = 1.

    Therefore z is even. Then we get from Eq (4.3) that

    2^{(m_1+1)x}a^{m_2x}p^{u(x-z)} = ((2^{2m_1}+a^{2m_2})^{\frac{z}{2}}+(2^{m_1}+a^{m_2})^{\frac{y}{2}})((2^{2m_1}+a^{2m_2})^{\frac{z}{2}}-(2^{m_1}+a^{m_2})^{\frac{y}{2}}).

    If \frac{y}{2} is odd, then we have

    (2^{2m_1}+a^{2m_2})^{\frac{z}{2}}+(2^{m_1}+a^{m_2})^{\frac{y}{2}}\equiv 4 \pmod 8

    and

    (2^{2m_1}+a^{2m_2})^{\frac{z}{2}}-(2^{m_1}+a^{m_2})^{\frac{y}{2}}\equiv {-2}\pmod 8

    if a\equiv 3\pmod 8 or

    (2^{2m_1}+a^{2m_2})^{\frac{z}{2}}+(2^{m_1}+a^{m_2})^{\frac{y}{2}}\equiv -2 \pmod 8

    and

    (2^{2m_1}+a^{2m_2})^{\frac{z}{2}}-(2^{m_1}+a^{m_2})^{\frac{y}{2}}\equiv {4}\pmod 8

    if a\equiv {-3}\pmod 8 . It follows that

    v_2((2^{2m_1}+a^{2m_2})^z-(2^{m_1}+a^{m_2})^y) = 3.

    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

    {(2^{2m_1}+a^{2m_2})^{\frac{z}{2}}+(2^{m_1}+a^{m_2})^{\frac{y}{2}}}\equiv 2 \pmod 4.

    Thus it follows that

    2^{(m_1+1)x-1}|(2^{2m_1}+a^{2m_2})^{z/2}-(2^{m_1}+a^{m_2})^{\frac{y}{2}};

    however, this is impossible since

    2^{(m_1+1)x-1}\geq 2^{(m_1+1)z} = (4\cdot 2^{2m_1})^{\frac{z}{2}} > (2^{2m_1}+a^{2m_2})^{\frac{z}{2}}-(2^{m_1}+a^{m_2})^{\frac{y}{2}}.

    (ⅱ) If u = 0, v\geq 1 , then n = q^v . Equation (4.1) becomes

    \begin{equation} p^{ky} = \left(\frac{q^{2l}+p^{2k}}{2}\right)^z-\left(\frac{q^{2l}-p^{2k}}{2}\right)^xq^{v(x-z)}. \end{equation} (4.4)

    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

    \begin{equation} 2^{(m_1+1)x}a^{m_2x}q^{v(x-z)} = (2^{2m_1}+a^{2m_2})^z-(2^{m_1}-a^{m_2})^y. \end{equation} (4.5)

    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

    (2\cdot a^{2m_2})^z\equiv (2^{m_1}-a^{m_2})^y \pmod q.

    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

    2^{(m_1+1)x-1}|(2^{2m_1}+a^{2m_2})^{z/2}-(2^{m_1}+a^{m_2})^{y/2},

    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

    \begin{equation} 1 = \left(\frac{q^{2l}+p^{2k}}{2}\right)^z-\left(\frac{q^{2l}-p^{2k}}{2}\right)^xp^{u(x-z)}q^{v(x-z)}. \end{equation} (4.6)

    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

    \begin{equation} 2^{(m_1+1)x}a^{m_2x}p^{u(x-z)}q^{v(x-z)} = (2^{2m_1}+a^{2m_2})^z-1. \end{equation} (4.7)

    Taking modulo q for Eq (4.5) leads to

    (2\cdot a^{2m_2})^z\equiv 1 \pmod q.

    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

    2^{(m_1+1)x-1}|(2^{2m_1}+a^{2m_2})^{z/2}-1,

    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

    \begin{equation} a^{m_2x}2^{(m_1+1)x} = n^{z-x}((2^{2m_1}+a^{2m_2})^z-(2^{2m_1}-a^{2m_2})^yn^{y-z}). \end{equation} (4.8)

    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

    \begin{equation} 2^{(m_1+1)x}a_2^{m_2x} = (2^{2m_1}+a^{2m_2})^z-(2^{2m_1}-a^{2m_2})^ya_1^{s(y-z)}. \end{equation} (4.9)

    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

    2^{(m_1+1)x}a_2^{m_2x}\equiv {(2\cdot a^{2m_2})^z} \pmod {2^{m_1}-a^{m_2}}.

    It follows that

    (-1)^{(m_1+m_1m_2+1)x} = \left(\frac{2}{2^{m_1}-a^{m_2}}\right)^{(m_1+1)x}\left(\frac{a_2}{2^{m_1}-a^{m_2}}\right)^{m_2x} = \left(\frac{2}{2^{m_1}-a^{m_2}}\right)^{z} = (-1)^z,

    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

    (2^{m_1}+a^{m_2})^y|(2^{2m_1}+a^{2m_2})^{z/2}+2^{(m_1+1)x/2}a_2^{m_2x/2}

    or

    (2^{m_1}+a^{m_2})^y|(2^{2m_1}+a^{2m_2})^{z/2}-2^{(m_1+1)x/2}a_2^{m_2x/2}.

    Hence

    (2^{m_1}+a^{m_2})^y\leq (2^{2m_1}+a^{2m_2})^{z/2}+2^{(m_1+1)x/2}a_2^{m_2x/2},

    which is impossible since

    (2^{m_1}+a^{m_2})^y > (2^{2m_1}+a^{2m_2}+2^{m_1+1}\cdot a^{m_2})^{z/2} > (2^{2m_1}+a^{2m_2})^{z/2}+2^{(m_1+1)x/2}a_2^{m_2x/2}.

    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

    2^{(m_1+1)x}a_2^{m_2x}\equiv {(2\cdot a^{2m_2})^z} \pmod {2^{m_1}-a^{m_2}}.

    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

    2^{(m_1+1)x}a_2^{m_2x}\equiv {(2\cdot a^{2m_2})^z} \pmod {2^{m_1}-a^{m_2}}.

    We already prove that is impossible if m_1 is even. If m_1 is odd, then we have

    1 = \left(\frac{2}{2^{m_1}-a^{m_2}}\right)^{(m_1+1)x}\left(\frac{a_2}{2^{m_1}-a^{m_2}}\right)^{m_2x} = \left(\frac{2}{2^{m_1}-a^{m_2}}\right)^{z} = (-1)^z,

    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

    (-1)^{m_2x} = \left(\frac{2}{P}\right)^{(m_1+1)x}\left(\frac{a_2}{P}\right)^{m_2x} = \left(\frac{2}{P}\right)^{z} = 1,

    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

    2^{(m_1+1)x}a_2^{m_2x}\equiv {(2\cdot a^{2m_2})^z} \pmod {2^{m_1}+a^{m_2}}.

    It follows that

    (-1)^{(m_1+1)x} = \left(\frac{2}{2^{m_1}+a^{m_2}}\right)^{(m_1+1)x}\left(\frac{a_2}{2^{m_1}+a^{m_2}}\right)^{m_2x} = \left(\frac{2}{2^{m_1}+a^{m_2}}\right)^{z} = (-1)^z,

    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

    2^{(m_1+1)x}a_2^{m_2x}\equiv {(2\cdot a^{2m_2})^z} \pmod {2^{m_1}+a^{m_2}}.

    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

    2^{(m_1+1)x}a_2^{m_2x}\equiv {(2\cdot a^{2m_2})^z} \pmod {2^{m_1}-a^{m_2}}.

    It follows that

    \begin{equation} (-1)^{(m_1+1)x} = \left(\frac{2}{2^{m_1}-a^{m_2}}\right)^{(m_1+1)x}\left(\frac{a_2}{2^{m_1}-a^{m_2}}\right)^{m_2x} = \left(\frac{2}{2^{m_1}-a^{m_2}}\right)^{z} = (-1)^z, \end{equation} (4.10)

    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

    (-1)^{m_2x} = \left(\frac{2}{2^{m_1}+a^{m_2}}\right)^{(m_1+1)x}\left(\frac{a_2}{2^{m_1}+a^{m_2}}\right)^{m_2x} = \left(\frac{2}{2^{m_1}+a^{m_2}}\right)^{z} = 1,

    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

    2^{(m_1+1)x}a_2^{m_2x}\equiv {(2\cdot a^{2m_2})^z} \pmod {2^{m_1}+a^{m_2}}.

    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

    2^{(m_1+1)x}a_2^{m_2x}\equiv {(2\cdot a^{2m_2})^z} \pmod {2^{m_1}+a^{m_2}}.

    It follows that

    (-1)^{(m_1+m_1m_2+1)x} = \left(\frac{2}{2^{m_1}+a^{m_2}}\right)^{(m_1+1)x}\left(\frac{a_2}{2^{m_1}+a^{m_2}}\right)^{m_2x} = \left(\frac{2}{2^{m_1}+a^{m_2}}\right)^{z} = (-1)^z.

    It follows that x\equiv z\pmod 2 . We already proved that that is impossible.

    If a_1 = a then Eq (4.8) becomes

    \begin{equation} 2^{(m_1+1)x} = (2^{2m_1}+a^{2m_2})^z-(2^{2m_1}-a^{2m_2})^ya^{s(y-z)}. \end{equation} (4.11)

    Taking modulo 2^{m_1}+a^{m_2} for Eq (4.11) leads to

    2^{(m_1+1)x}\equiv {(2\cdot a^{2m_2})^z} \pmod {2^{m_1}+a^{m_2}}.

    It follows that

    (-1)^{(m_1+1)x} = \left(\frac{2}{2^{m_1}+a^{m_2}}\right)^{(m_1+1)x} = \left(\frac{2}{2^{m_1}+a^{m_2}}\right)^{z} = (-1)^z.

    It follows that (m_1+1)x\equiv z\pmod 2 . On the other hand, taking modulo a for Eq (4.11) leads to

    2^{(m_1+1)x}\equiv 2^{2m_1z} \pmod a.

    It follows that

    (-1)^{(m_1+1)x} = \left(\frac{2}{a}\right)^{(m_1+1)x} = \left(\frac{2}{a}\right)^{2m_1z} = 1,

    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

    \begin{equation} a^{m_2x} = (2^{2m_1}+a^{2m_2})^z-(2^{2m_1}-a^{2m_2})^y2^{r(y-z)}. \end{equation} (4.12)

    Thus

    2^{2m_1z}\equiv {2^{2m_1y+r(y-z)}}\pmod a.

    It follows that

    1 = \left(\frac{2}{a}\right)^{2m_1z} = \left(\frac{2}{a}\right)^{2m_1y+r(y-z)} = (-1)^{r(y-z)},

    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

    (-1)^{m_1m_2x} = \left(\frac{a}{2^{m_1}-a^{m_2}}\right)^{m_2x} = \left(\frac{2}{a}\right)^{m_1m_2x} = \left(\frac{2}{2^{m_1}-a^{m_2}}\right)^{z} = (-1)^z,

    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

    \gcd((2^{2m_1}+a^{2m_2})^{\frac{z}{2}}+a^{\frac{m_2x}{2}}, (2^{2m_1}+a^{2m_2})^{\frac{z}{2}}-a^{\frac{m_2x}{2}}) = 2,

    we get

    \begin{equation} (2^{m_1}+a^{m_2})^y|(2^{2m_1}+a^{2m_2})^{\frac{z}{2}}+a^{\frac{m_2x}{2}} \end{equation} (4.13)

    or

    \begin{equation} (2^{m_1}+a^{m_2})^y|(2^{2m_1}+a^{2m_2})^{\frac{z}{2}}-a^{\frac{m_2x}{2}}. \end{equation} (4.14)

    However, the inequalities

    (2^{m_1}+a^{m_2})^y > (2^{2m_1}+a^{2m_2}+2^{m_1+1}a^{m_2})^{\frac{z}{2}} > (2^{2m_1}+a^{2m_2})^{\frac{z}{2}}+a^{\frac{m_2x}{2}}

    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

    (2^{2m_1}+a^{2m_2})^{\frac{z}{2}}+(2^{2m_1}-a^{2m_2})^{\frac{y}{2}}|a^{m_2x},

    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

    \begin{equation} a_2^{m_2x} = (2^{2m_1}+a^{2m_2})^z-(2^{2m_1}-a^{2m_2})^y2^{r(y-z)}a_1^{s(y-z)}. \end{equation} (4.15)

    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.

    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

    \begin{array}{*{20}{c}} {(80, 39, 89), (576,943, 1105), (320,999, 1049), (1344,583, 1465), (1856,183, 1865), }\\ {(11520, 14359, 18409), (168960, 234919, 289369), (46080, 260119, 264169), }\\ {(1757184, 4010263, 4378345)(33409993656, 936621583, 33423116785), }\\ {(8294400, 93679, 8294929), (25029771264, 8063247823, 26296790545). } \end{array}

    Proof. By Theorem 1.1 and Table 1, one can immediately obtain the Corollary 5.1 by a simple calculation.

    Table 1.  The proof of Corollary 5.1.
    p^k=2^{m_1}-a^{m_2} q^l=2^{m_1}+a^{m_2} p^k=2^{m_1}-a^{m_2} q^l=2^{m_1}+a^{m_2}
    3=2^3-5 13=2^3+5 23=2^5-3^2 41=2^5+3^2
    3^3=2^5-5 37=2^5+5 11=2^5-21 53=2^5+21
    3=2^5-29 61=2^5+29 83=2^7-45 173=2^7+45
    347=2^9-165 677=2^9+165 467=2^9-45 557=2^9+45
    23=2^{11}-45 4073=2^{11}+45 1619=2^{11}-429 2477=2^{11}+429
    3623=2^{17}-357^2 258521=2^{17}+357^2 35591=2^{17}-309 226553=2^{17}+309

     | Show Table
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    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.

    Table 2.  Some examples of application of Theorem 1.1.
    p^k=2^{m_1}-a^{m_2} q^l=2^{m_1}+a^{m_2} p^k=2^{m_1}-a^{m_2} q^l=2^{m_1}+a^{m_2}
    59=2^7-69 197=2^7+69 29=2^7-99 227=2^7+99
    107=2^7-21 149=2^7+21 3^3=2^7-101 229=2^7+101
    181=2^8-75 331=2^8+75 3=2^9-509 1021=2^9+509
    3^3=2^9-485 997=2^9+485 71=2^9-21^2 953=2^9+21^2
    251=2^9-261 773=2^9+261 227=2^9-285 797=2^9+285
    61=2^{10}-963 1987=2^{10}+963 1787=2^{11}-261 2309=2^{11}+261
    3709=2^{12}-387 4483=2^{12}+387 7883=2^{13}-309 8501=2^{13}+309
    7643=2^{13}-549 8741=2^{13}+549 32507=2^{15}-261 33209=2^{15}+261
    56923=2^{16}-8613 74149=2^{16}+8613 519527=2^{19}-69^2 529049=2^{19}+69^2
    8060099=2^{23}-69^3 8717117=2^{23}+69^3 33549671=2^{25}-69^2 33559193=2^{25}+69^2

     | Show Table
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    Table 3.  Some examples of application of Theorem 1.1.
    p^k=2^{m_1}-a^{m_2} q^l=2^{m_1}+a^{m_2}
    10887311=2^{25}-69^4 56221553=2^{25}+69^4
    536676431=2^{29}-21^4 537065393=2^{29}+21^4
    536866151=2^{29}-69^2 536875673=2^{29}+69^2
    9007199250656891=2^{53}-21^5 9007199258825093=2^{53}+21^5
    2361183241434822606407=2^{71}-21^2 2361183241434822607289=2^{71}+21^2
    18888953906518175252587=2^{74}-800013^3 18889977956438986456981=2^{74}+800013^3
    9671406556916991286915727=2^{83}-453^4 96714065569170755083830893=2^{83}+453^4
    2475880078570554658666153799=2^{91}-3^{30} 2475880078570966440930343097=2^{91}+3^{30}
    2535301200455545285745922769853=2^{101}-99^9 2535301200457372320240890051651=2^{101}+99^9
    2^{110}-800013^3 1298074214633707419157584487907221=2^{110}+800013^3

     | Show Table
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    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

    \begin{array}{*{20}{c}} {(416n, 87n, 425n), (1728n, 295n, 1753n), (5760n, 2071n, 6121n), (29440n, 3159n, 29609n), }\\ {(31488n, 1255n, 31513n), (47616n, 56887n, 74185n), (59904n, 51847n, 79225n), }\\ {(129536n, 1527n, 129545n), (872448n, 1003207n, 1093945n), (829440n, 884551n, 1212601n), }\\ {(1075200n, 772951n, 1324201n), (1320960n, 632551n, 1464601n), (1382400n, 592951n, 1504201n), }\\ {(2021376n, 74407n, 2022745n), (2058240n, 38551n, 2058601n), (8710976n, 16572007n, 16982425n), }\\ {(4300800n, 16501591n, 17052841n), (5775360n, 16280191n, 17274241n), }\\ {(19562496n, 268079047n, 268791865n), (546963456n, 4277553367n, 4312381225n), }\\ {(1128923136n, 4220783527n, 4369151065n), }\\ {(13435303624704n, 15027021842647n, 171324290231185n), } \end{array}

    for any n\in N .

    Proof. By Theorem 1.2 and Table 4, one can immediately obtain Corollary 5.2 by a simple calculation.

    Table 4.  The proof of Corollary 5.2.
    p^k=2^{m_1}-a^{m_2} q^l=2^{m_1}+a^{m_2} p^k=2^{m_1}-a^{m_2} q^l=2^{m_1}+a^{m_2}
    3=2^4-13 29=2^4+13 5=2^5-3^3 59=2^5+3^3
    19=2^6-45 109=2^6+45 13=2^7-115 3^5=2^7+115
    5=2^7-123 251=2^7+123 163=2^{8}-93 349=2^{8}+93
    139=2^8-117 373=2^8+117 3=2^{8}-253 509=2^{8}+253
    811=2^{10}-213 1237=2^{10}+213 619=2^{10}-405 1429=2^{10}+405
    499=2^{10}-525 1549=2^{10}+525 379=2^{10}-645 1669=2^{10}+645
    349=2^{10}-675 1699=2^{10}+675 37=2^{10}-987 2011=2^{10}+987
    19=2^{10}-1005 2029=2^{10}+1005 3643=2^{12}-453 4549=2^{12}+453
    3571=2^{12}-525 4621=2^{12}+525 3391=2^{12}-705 4801=2^{12}+705
    15787=2^{14}-597 16981=2^{14}+597 61363=2^{16}-4173 69709=2^{16}+4173
    56923=2^{16}-8613 74149=2^{16}+8613 2592691=2^{22}-117 5795917=2^{22}+117

     | Show Table
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    It is easy to see that Je \acute{s} manowicz' a conjecture holds for the following set of primitive Pythagorean numbers:

    \frac{q^{2l}-p^{2k}}{2}, p^kq^l, \frac{q^{2l}+p^{2k}}{2}.

    In addition, Je \acute{s} manowicz' conjecture holds for non-primitive Pythagorean numbers:

    n\frac{q^{2l}-p^{2k}}{2}, np^kq^l, n\frac{q^{2l}+p^{2k}}{2},

    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 .

    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).

    All authors declare no conflict of interest regarding the publication of this paper.



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