1 Introduction

In 1956, Sierpiński[13] showed that the equation \(3^x + 4^y = 5^z\) has only the positive integer solution \((x, y, z) = (2, 2, 2)\). Jeśmanowicz [8] conjectured that if a, b, c are Pythagorean numbers, i.e., positive integers satisfying \(a^2 + b^2 = c^2\), then

$$\begin{aligned} a^x + b^y = c^z \end{aligned}$$

has only the positive integer solution \((x, y, z) = (2, 2, 2)\). As an analogue of Jeśmanowicz’ conjecture, the author [15] proposed the following:

Conjecture 1

If a, b, c are positive integers satisfying \(a^2 + b^2=c^2\) with \(\gcd (a, b, c)=1\) and b odd, then the equation

$$\begin{aligned} x^2 + b^m=c^n \end{aligned}$$
(1.1)

has only one positive integer solution \((x, m, n)=(a, 2, 2)\).

The author[15] proved that if b and c are primes such that (i) \(b^2 + 1=2c\) and (ii) \(d=1\) or even if \(b \equiv 1\pmod 4\), then Conjecture 1 is true, where d is the order of a prime divisor of (c) in the ideal class group of \({\mathbb {Q}}(\sqrt{-b} )\). In [3, 9, 10] and [19], it was shown that if \(b\not \equiv 1\pmod 8\), and b or c is a prime power, then Conjecture 1 is true. It has been verified that Conjecture 1 holds for many other Pythagorean numbers. But Conjecture 1 is still unsolved. (See Cao[2], Terai[16, 17] and [18] for another analogue of Jeśmanowicz’ conjecture.)

Related to Conjecture 1, in [5] and [6], Cenberci and Senay proposed the following:

Conjecture 2

If a, b, c are positive integers satisfying \(a^2 + b^2=c^4\) with \(\gcd (a, b, c)=1\) and b odd, then equation (1.1) has only one positive integer solution \((x, m, n)=(a, 2, 4)\).

As another analogue of Conjecture 1, we also propose the following:

Conjecture 3

If a, b, c are positive integers satisfying \(a^2 + b^4=c^2\) with \(\gcd (a, b, c)=1\) and b odd, then equation (1.1) has only one positive integer solution \((x, m, n)=(a, 4, 2)\).

By Magma[1], we verified that Conjecture 3 is true in the range \(c \le 10^5\) and \(\max \{m,n\} \le 20\).

In this paper, when b has at most two distinct primes, we show that Conjecture 3 is true under some conditions. The proof is based on elementary methods and results concerning the equation \(x^2+1=2y^n\) due to Ljunggren - Störmer and the equation \(x^2 +y^4=z^n\) due to Ellenberg. We also verify that when \(b=3p, 5p\) with p odd prime, Conjecture 3 is true under a certain condition.

2 Lemmas

As well-known, primitive Pythagorean numbers a, b, c with b odd can be parametrized as follows:

$$\begin{aligned} a=2u_0v_0,~~~ b=u_0^2-v_0^2,~~~c=u_0^2+v_0^2, \end{aligned}$$

where \(u_0,v_0\) are positive integers such that \(\gcd (u_0,v_0)=1\), \(u_0 \not \equiv v_0 \pmod 2\) and \(u_0>v_0\).

In view of the above prametrization, we obtain all positive integers a, b, c satisfying \(a^2+b^4=c^2\) with b odd. In fact, since \(b^2=u_0^2-v_0^2\), we have

$$\begin{aligned} u^2=u_0+v_0,~~~v^2=u_0-v_0 \end{aligned}$$

with \(b=uv\). Then

$$\begin{aligned} u_0=\frac{~u^2+v^2~}{2},~~~v_0=\frac{~u^2-v^2~}{2}. \end{aligned}$$

Thus we have shown the following:

Lemma 1

All positive integer solutions of the equation \(a^2+b^4=c^2\) with \(\gcd (a, b,c)=1\) and b odd are given by

$$\begin{aligned} a=\frac{~u^4-v^4~}{2},~~ b=uv,~~ c=\frac{~u^4+v^4~}{2},~ \end{aligned}$$
(2.1)

where u, v are positive integers such that \(\gcd (u,v)=1\), \(u \equiv v \equiv 1\pmod 2\) and \(u>v\).

We also need the following lemmas to prove Theorems 1,2,3.

Lemma 2

(Störmer[14]) The Diophantine equation

$$\begin{aligned} x^2 + 1=2y^n \end{aligned}$$

has no solutions in integers \(x>\)1, \(y \ge 1\) and n odd \(\ge 3\).

Lemma 3

(Ljunggren[11]) The Diophantine equation

$$\begin{aligned} x^2 + 1=2y^4 \end{aligned}$$

has only the positive integer solution \((x, y)=(1, 1), (239, 13)\).

Lemma 4

(Ellenberg[7]) Let n be a positive integer with \(n \ge 4\). Then the equation

$$\begin{aligned} x^2 +y^4=z^n \end{aligned}$$

has no solutions in nonzero pairwise coprime integers x, y, z.

The following lemma is immediate from [4] and [12].

Lemma 5

  1. (1)

    The Diophantine equation

    $$\begin{aligned} x^2+9^m=3281^n \end{aligned}$$

    has only the positive integer solution \((x, m, n)=(3280,4,2)\).

  2. (2)

    The Diophantine equation

    $$\begin{aligned} x^2+15^m=25313^n \end{aligned}$$

    has only the positive integer solution \((x, m, n)=(25312,4,2)\).

  3. (3)

    The Diophantine equation

    $$\begin{aligned} x^2+25^m=195313^n \end{aligned}$$

    has only the positive integer solution \((x, m, n)=(195312 ,4,2)\).

  4. (4)

    The Diophantine equation

    $$\begin{aligned} x^2+33^m=7361^n \end{aligned}$$

    has only the positive integer solution \((x, m, n)=(7280,4,2)\).

3 The cases \(b=p^{\alpha },~p^{\alpha }q^{\beta }\) with p, q odd primes.

In this section, when b has at most two distinct primes, we show that Conjecture 3 is true under some conditions.

Theorem 1

Let a, b, c be as in (2.1). Suppose that \(b \equiv \pm 3\pmod 8\) and that b satisfies at least one of the following :

  1. (i)

    \(b=u\) with \(u=p^{\alpha }\) and \(v=1\).

  2. (ii)

    \(b=uv\) with \(u=p^{\alpha }\) and \(v=q^{\beta }.\)

  3. (iii)

    \(b=u\) with \(u=p^{\alpha }q^{\beta }\) and \(v=1\),

where p, q are distinct odd primes and \(\alpha ,\beta \) are positive integers.

Then Conjecture 3 is true.

Proof

Let a, b, c be as in Theorem 1. Let (x, m, n) be a positive integer solution of equation (1.1).

Since \(b \equiv \pm 3\pmod 8\) and \(c \equiv 1\pmod 4\), we have

$$\begin{aligned} \left( \frac{b}{c}\right) = \left( \frac{c}{b}\right) = \left( \frac{2}{b}\right) =-1. \end{aligned}$$

Hence we see that m and n are even.

Put \(n=2N\). From (1.1), we have

$$\begin{aligned} b^m=(c^N + x)(c^N - x). \end{aligned}$$
  1. (i)

     \(b=u\) with \(u=p^{\alpha }\) and \(v=1\). Then we have

    $$\begin{aligned} b^m + 1=2c^N. \end{aligned}$$
    (3.1)

    Note that \(b^4 + 1=2c\), since \(v=1\). If N=1 or 2, then we easily see that equation (3.1) has only the solution \((m, N)=(4, 1)\). If \(N \ge 3\), then it follows from Lemmas 2, 3 that equation (3.1) has no solutions.

  2. (ii)

     \(b=uv\) with \(u=p^{\alpha }\) and \(v=q^{\beta }.\) Then we have

    $$\begin{aligned} b^m + 1=2c^N, \end{aligned}$$
    (3.2)

    or

    $$\begin{aligned} u^m + v^m=2c^N. \end{aligned}$$
    (3.3)

    First consider equation (3.2). If \(N \ge 3\), then it follows from Lemmas 2, 3 that equation (3.2) has no solutions. If N=1 or 2, then equation (3.2) can be written as

    $$\begin{aligned} (uv)^m+1=u^4+v^4 ~~\text{ or }~~(u^4+v^4)^2/2. \end{aligned}$$

    It is easy to show that the above equation has no solutions except for the equation

    $$\begin{aligned} (uv)^6+1=(u^4+v^4)^2/2. \end{aligned}$$
    (3.4)

    In view of a property of an elliptic curve, we see that equation (3.4) has no solutions. Indeed, by putting \(X=2u^2v^2\) and \(Y=2(u^4+v^4)\), (3.4) can be reduced to the ellitptic curve

    $$\begin{aligned} E:~Y^2=X^3+8 \end{aligned}$$

    with rank \(E(\mathbb {Q})=1\) and all integer points on E are \((X,Y)=(-2, 0), ~(1, \pm 3),~(2,\pm 4),\) \((46, \pm 312).\) Next consider equation (3.3). Since m is even, the proof is divided into two cases (a) \(m \equiv 0\pmod 4\) and (b) \(m \equiv 2\pmod 4\). Case (a) \(m \equiv 0\pmod 4\). If \(n \ge 4\), then it follows from Lemma 4 that equation (1.1) has no solutions. If \(n=2\), i.e., \(N=1\), then the relation \(c=(u^4+v^4)/2\) yields \(m=4\) from (3.3), and hence \((x,m,n)=(a,4,2).\) Case (b) \(m \equiv 2\pmod 4\), say \(m=2l\) with l odd. Then equation (3.3) becomes

    $$\begin{aligned} \left( \frac{~u^2+v^2~}{2} \right) \left( \frac{~u^{2l}+v^{2l}~}{u^2+v^2}\right) = \left( \frac{~u^4+v^4~}{2} \right) ^N \end{aligned}$$

    Let r be an odd prime of \((u^2+v^2)/2\). Then \(u^4+v^4 \equiv 0\pmod r\). This implies that \(u \equiv 0\pmod r\) and \(v \equiv 0\pmod r\), which contradicts \(\gcd (u,v)=1\).

  3. (iii)

    \(b=u\) with \(u=p^{\alpha }q^{\beta }\) and \(v=1\). Then we have

    $$\begin{aligned} b^m + 1=2c^N, \end{aligned}$$
    (3.5)

    or

    $$\begin{aligned} u_1^m + v_1^m=2\left( \frac{~u_1^4v_1^4+1~}{2}\right) ^N \end{aligned}$$
    (3.6)

    with \(u_1=p^{\alpha }\) and \(v_1=q^{\beta }\). First consider equation (3.5). Note that \(b^4 + 1=2c\), since \(v=1\). As above, equation (3.5) has only the solution \((m, N)=(4, 1)\). Next consider equation (3.6). In view of \(b=u_1v_1\) and \(b \equiv \pm 3\pmod 8\), we may suppose that \(u_1 \equiv \pm 3\pmod 8\) and \(v_1 \equiv \pm 1\pmod 8\) . Since m is even, the proof is divided into two cases (a) \(m \equiv 0\pmod 4\) and (b) \(m \equiv 2\pmod 4\). Case (a) \(m \equiv 0\pmod 4\). If \(n \ge 4\), then it follows from Lemma 4 that equation (1.1) has no solutions. If \(n=2\), i.e., \(N=1\), then (3.6) has no solutions. Case (b) \(m \equiv 2\pmod 4\), say \(m=2l\) with l odd. Then equation (3.6) becomes

    $$\begin{aligned} \left( \frac{~u_1^2+v_1^2}{2}\right) \left( \frac{~u_1^{2l}+v_1^{2l}~}{u_1^2+v_1^2}\right) = \left( \frac{~u_1^4v_1^4+1~}{2} \right) ^N \end{aligned}$$
    (3.7)

    The right hand side of (3.7) is divisible by \((u_1^2+v_1^2)/2 \equiv 5\pmod 8\). This contradicts the fact that an odd prime factor r of \(A^4+B^4\) satisfies \(r \equiv 1\pmod 8\). This completes the proof of Theorem 1.

\(\square \)

4 The cases \(b=3p, 5p\) with p odd prime.

In this section, when \(b=3p,5p\) with p odd prime, we consider eqaution (1.1). We do not assume that \(b \equiv \pm 3\pmod 8\), i.e., \(p \equiv \pm 1\pmod 8\).

Theorem 2

  1. (i)

    Let p be an odd prime with \(p>3.\) Put \(u=p\) and \(v=3\) in (2.1). If \((3p+1)/2\) has a prime factor other than 17 and 193, then Conjecture 3 is true. In particular, if p is an odd prime with \(3< p<10^7\), then Conjecture 3 is true.

  2. (ii)

    Let p be an odd prime. Put \(u=3p\) and \(v=1\) in (2.1). If \((p+3)/2\) and \((p^2+3^2)/2\) have a prime factor other than 17 and 193, respectively, then Conjecture 3 is true. In particular, if p is an odd prime satisfying \(3\le p<10^7\) with \(p\not =31,383,167039\), then Conjecture 3 is true.

Proof

(i) \(u=p\) and \(v=3\) in (2.1). Then \(a=(p^4-3^4)/2,~b=3p,~c=(p^4+3^4)/2.\) Suppse that our assumptions are all satisfied. Let (x, m, n) be a positive integer solution of equation (1.1).

From (1.1), we have

$$\begin{aligned} \left( \frac{c}{3}\right) = \left( \frac{4c}{3}\right) = \left( \frac{2}{3}\right) \left( \frac{2c}{3}\right) =-1. \end{aligned}$$

Hence we see that n are even, say \(n=2N\). (We do not know if m is even or not.) Thus

$$\begin{aligned} (3p)^m=(c^N + x)(c^N - x). \end{aligned}$$

Then we have

$$\begin{aligned} (3p)^m + 1=2c^N \end{aligned}$$
(4.1)

or

$$\begin{aligned} p^m + 3^m=2c^N. \end{aligned}$$
(4.2)

First consider equation (4.1). If m is odd, then equation (4.1) becomes

$$\begin{aligned} \left( \frac{~3p+1~}{2} \right) \left( \frac{~(3p)^m+1~}{3p+1}\right) = \left( \frac{~p^4+3^4~}{2} \right) ^N. \end{aligned}$$

Note that \((3p+1)/2\) is odd, since \((p^4+3^4)/2\) is odd. Let r be an odd prime factor of \((3p+1)/2\). Then \((p^4+3^4)/2 \equiv 0\pmod r\). This implies that \(3^8+1 \equiv 0\pmod r\), so \(r=17,193\), which contradicts our assumption. In the same way as in the proof of Theorem 1 (i), if m is even, we easily see that equation (4.1) has no solutions.

Next consider equation (4.2). If m is odd, then equation (4.2) becomes

$$\begin{aligned} \left( \frac{~p+3~}{2} \right) \left( \frac{~p^m+3^m~}{p+3}\right) = \left( \frac{~p^4+3^4~}{2} \right) ^N. \end{aligned}$$

Let r be an odd prime factor of \((p+3)/2\). Then \(p^4+3^4 \equiv 0\pmod r\). This implies that \(p \equiv 0\pmod r\) and \(3 \equiv 0\pmod r\), which contradicts \(\gcd (p,3)=1\). If \(m \equiv 0\pmod 4\), then it follows from Lemma 4 that equation (1.1) has only the solution \((x,m,n)=(a,4,2)\). We may suppose that \(m \equiv 2\pmod 4\), say \(m=2l\) with l odd. Then equation (4.2) becomes

$$\begin{aligned} \left( \frac{~p^2+3^2}{2}\right) \left( \frac{~p^{2l}+3^{2l}~}{p^2+3^2}\right) = \left( \frac{~p^4+3^4~}{2} \right) ^N. \end{aligned}$$

Let r be an odd prime factor of \((p^2+3^2)/2\). Then \(p^4+3^4 \equiv 0\pmod r\). This implies that \(p \equiv 0\pmod r\) and \(3 \equiv 0\pmod r\), which contradicts \(\gcd (p,3)=1\).

In particular, by Magma[1], we verified that \((3p+1)/2\) has a prime other than 17 and 193 in the range \(3<p<10^7\) with \(p\not =11\). When \(p=11\), Conjecture 3 is true from Lemma 5,(4).

(ii) \(u=3p\) and \(v=1\) in (2.1). Then \(a=((3p)^4-1)/2,~b=3p,~c=((3p)^4+1)/2.\) Suppse that our assumptions are all satisfied. Let (x, m, n) be a positive integer solution of equation (1.1).

When \(p = 3\), Conjecture 3 is true from Lemma 5,(1). We may suppose that \(p>3\). As in (i), equation (1.1) is reduced to solving the following:

$$\begin{aligned} (3p)^m + 1=2\left( \frac{~(3p)^4+1~}{2} \right) ^N \end{aligned}$$
(4.3)

or

$$\begin{aligned} p^m + 3^m=2\left( \frac{~(3p)^4+1~}{2} \right) ^N. \end{aligned}$$
(4.4)

First consider equation (4.3). If m is odd, then an odd prime factor r of \((3p+1)/2\) divides \(((3p)^4+1)/2\). This implies that \(2 \equiv 0\pmod r\), which is impossible. If m is even, then it follows from Lemmas 1,2 that equation (4.3) has only the solution \((m, N)=(4, 1)\).

Next consider equation (4.4). If m is odd, then equation (4.4) becomes

$$\begin{aligned} \left( \frac{~p+3~}{2} \right) \left( \frac{~p^m+3^m~}{p+3}\right) = \left( \frac{~(3p)^4+1~}{2} \right) ^N. \end{aligned}$$

Let r be an odd prime factor of \((p+3)/2\). Then \((3p)^4+1 \equiv 0\pmod r\). This implies that \(3^8+1 \equiv 0\pmod r\), so \(r=17,193\), which contradicts our assumption. If \(m \equiv 0\pmod 4\), then it follows from Lemma 4 that equation (1.1) has no solutions. We may suppose that \(m \equiv 2\pmod 4\), say \(m=2l\) with l odd. Then equation (4.4) becomes

$$\begin{aligned} \left( \frac{~p^2+3^2}{2}\right) \left( \frac{~p^{2l}+3^{2l}~}{p^2+3^2}\right) = \left( \frac{~(3p)^4+1~}{2} \right) ^N. \end{aligned}$$

Let r be an odd prime factor of \((p^2+3^2)/2\). Then \((3p)^4+1 \equiv 0\pmod r\). This implies that \(3^8+1 \equiv 0\pmod r\), so \(r=17,193\), which contradicts our assumption.

In particular, by Magma[1], we verified that \((p+3)/2\) has a prime other than 17 and 193 in the range \(3<p<10^7\) with \(p\not =31,383,167039\), and that \((p^2+3^2)/2\) has a prime other than 17 and 193 in the range \(3<p<10^7\) with \(p\not =5\). When \(p=5\), Conjecture 3 is true from Lemma 5,(2). This completes the proof of Theorem 2. \(\square \)

Similarly, when \(b=5p\), we can show the following:

Theorem 3

  1. (i)

    Let p be an odd prime with \(p>5.\) Put \(u=p\) and \(v=5\) in (2.1). If \((5p+1)/2\) has a prime factor other than 17 and 11489, then Conjecture 3 is true. In particular, if p is a prime with \(5< p<10^7\), then Conjecture 3 is true.

  2. (ii)

    Let p be an odd prime. Put \(u=5p\) and \(v=1\) in (2.1). If \((p+5)/2\) and \((p^2+5^2)/2\) have a prime factor other than 17 and 11489, respectively, then Conjecture 3 is true. In particular, if p is a prime satisfying \(3 \le p<10^7\) with \(p\not =29,22973\), then Conjecture 3 is true.

Proof

(i) \(u=p\) and \(v=5\) in (2.1). Then \(a=(p^4-5^4)/2,~b=5p,~c=(p^4+5^4)/2.\) Suppse that our assumptions are all satisfied. Let (x, m, n) be a positive integer solution of equation (1.1).

From (1.1), we have

$$\begin{aligned} \left( \frac{c}{5}\right) = \left( \frac{4c}{5}\right) = \left( \frac{2}{5}\right) \left( \frac{2c}{5}\right) =-1. \end{aligned}$$

Hence we see that n are even, say \(n=2N\). (We do not know if m is even or not.) Thus

$$\begin{aligned} (5p)^m=(c^N + x)(c^N - x). \end{aligned}$$

Then we have

$$\begin{aligned} (5p)^m + 1=2c^N \end{aligned}$$
(4.5)

or

$$\begin{aligned} p^m + 5^m=2c^N. \end{aligned}$$
(4.6)

First consider equation (4.5). If m is odd, then equation (4.5) becomes

$$\begin{aligned} \left( \frac{~5p+1~}{2} \right) \left( \frac{~(5p)^m+1~}{5p+1}\right) = \left( \frac{~p^4+5^4~}{2} \right) ^N. \end{aligned}$$

Note that \((5p+1)/2\) is odd, since \((p^4+5^4)/2\) is odd. Let r be an odd prime factor of \((5p+1)/2\). Then \((p^4+5^4)/2 \equiv 0\pmod r\). This implies that \(5^8+1 \equiv 0\pmod r\), so \(r=17,11489\), which contradicts our assumption. In the same way as in the proof of Theorem 1 (i), if m is even, we easily see that equation (4.5) has no solutions.

Next consider equation (4.6). If m is odd, then equation (4.6) becomes

$$\begin{aligned} \left( \frac{~p+5~}{2} \right) \left( \frac{~p^m+5^m~}{p+5}\right) = \left( \frac{~p^4+5^4~}{2} \right) ^N. \end{aligned}$$

Let r be an odd prime factor of \((p+5)/2\). Then \(p^4+5^4 \equiv 0\pmod r\). This implies that \(p \equiv 0\pmod r\) and \(5 \equiv 0\pmod r\), which contradicts \(\gcd (p,5)=1\). If \(m \equiv 0\pmod 4\), then it follows from Lemma 4 that equation (1.1) has only the solution \((x,m,n)=(a,4,2)\). We may suppose that \(m \equiv 2\pmod 4\), say \(m=2l\) with l odd. Then equation (4.6) becomes

$$\begin{aligned} \left( \frac{~p^2+5^2}{2}\right) \left( \frac{~p^{2l}+5^{2l}~}{p^2+5^2}\right) = \left( \frac{~p^4+5^4~}{2} \right) ^N. \end{aligned}$$

Let r be an odd prime factor of \((p^2+5^2)/2\). Then \(p^4+5^4 \equiv 0\pmod r\). This implies that \(p \equiv 0\pmod r\) and \(5 \equiv 0\pmod r\), which contradicts \(\gcd (p,5)=1\).

In particular, by Magma[1], we verified that \((5p+1)/2\) has a prime factor other than 17 and 11489 in the range \(3<p<10^7\).

(ii) \(u=5p\) and \(v=1\) in (2.1). Then \(a=((5p)^4-1)/2,~b=5p,~c=((5p)^4+1)/2.\) Suppse that our assumptions are all satisfied. Let (x, m, n) be a positive integer solution of equation (1.1).

When \(p = 5\), Conjecture 3 is true from Lemma 5,(3). We may suppose that \(p \not =5\). As in (i), equation (1.1) is reduced to solving the following:

$$\begin{aligned} (5p)^m + 1=2\left( \frac{~(5p)^4+1~}{2} \right) ^N \end{aligned}$$
(4.7)

or

$$\begin{aligned} p^m + 5^m=2\left( \frac{~(5p)^4+1~}{2} \right) ^N. \end{aligned}$$
(4.8)

First consider equation (4.7). If m is odd, then an odd prime factor r of \((5p+1)/2\) divides \(((5p)^4+1)/2\). This implies that \(2 \equiv 0\pmod r\), which is impossible. If m is even, then it follows from Lemmas 1,2 that equation (4.7) has only the solution \((m, N)=(4, 1)\).

Next consider equation (4.8). If m is odd, then equation (4.8) becomes

$$\begin{aligned} \left( \frac{~p+5~}{2} \right) \left( \frac{~p^m+5^m~}{p+5}\right) = \left( \frac{~(5p)^4+1~}{2} \right) ^N. \end{aligned}$$

Let r be an odd prime factor of \((p+5)/2\). Then \((5p)^4+1 \equiv 0\pmod r\). This implies that \(5^8+1 \equiv 0\pmod r\), so \(r=17,11489\), which contradicts our assumption. If \(m \equiv 0\pmod 4\), then it follows from Lemma 4 that equation (1.1) has no solutions. We may suppose that \(m \equiv 2\pmod 4\), say \(m=2l\) with l odd. Then equation (4.8) becomes

$$\begin{aligned} \left( \frac{~p^2+5^2}{2}\right) \left( \frac{~p^{2l}+5^{2l}~}{p^2+5^2}\right) = \left( \frac{~(5p)^4+1~}{2} \right) ^N. \end{aligned}$$

Let r be an odd prime factor of \((p^2+5^2)/2\). Then \((5p)^4+1 \equiv 0\pmod r\). This implies that \(5^8+1 \equiv 0\pmod r\), so \(r=17,11489\), which contradicts our assumption.

In particular, by Magma[1], we verified that \((p+5)/2\) has a prime factor other than 17 and 11489 in the range \(3 \le p<10^7\) with \(p\not =29,22973\), and that \((p^2+5^2)/2\) has a prime factor other than 17 and 11489 in the range \(3\le p<10^7\) with \(p\not =3\). When \(p=3\), Conjecture 3 is true from Lemma 5,(2). This completes the proof of Theorem 3. \(\square \)