Solution-1

Let \(n\) be a positive integer, and let \(m=2^{n}+65\). For the sake of contradiction, suppose that \(m \mid 5^{n}-3^{n}\), so \(5^{n} \equiv 3^{n} \pmod {m}\).

Notice that if \(n\) is even, then \(3 \mid m\), but \(3 \nmid 5^{n}-3^{n}\), contradiction. So, from now on we assume that \(n\) is odd, \(n=2 k+1\). Obviously \(n=1\) is not possible, so \(n \geqslant 3\). Notice that \(m\) is coprime to \(2\), \(3\) and \(5\).

Let \(m_{1}\) be the smallest positive multiple of \(m\) that can be written in the form of either \(\left|5a^{2} - 3b^{2}\right|\) or \(\left|a^{2}-15 b^{2}\right|\) with some integers \(a\) and \(b\).

Note that \(5^{n}-3^{n}=5\left(5^{k}\right)^{2}-3\left(3^{k}\right)^{2}\) is a multiple of \(m\), so the set of such multiples is non-empty, and therefore \(m_{1}\) is well-defined.

Part 1. First we show that \(m_{1} \leqslant 5 m\). Consider the numbers

\[5^{k+1} x+3^{k+1} y, \quad 0 \leqslant x, y \leqslant \sqrt{m}\]

There are \(\lfloor\sqrt{m}\rfloor+1>\sqrt{m}\) choices for \(x\) and \(y\), so there are more than \(m\) possible pairs \((x, y)\). Hence, two of these sums are congruent modulo \(m\) : \(5^{k+1} x_{1}+3^{k+1} y_{1} \equiv 5^{k+1} x_{2}+3^{k+1} y_{2}(\bmod m)\). Now choose \(a=x_{1}-x_{2}\) and \(b=y_{1}-y_{2}\); at least one of \(a, b\) is nonzero, and

\[5^{k+1} a+3^{k+1} b \equiv 0 \quad(\bmod m), \quad|a|,|b| \leqslant \sqrt{m}\]

From

\[0 \equiv\left(5^{k+1} a\right)^{2}-\left(3^{k+1} b\right)^{2}=5^{n+1} a^{2}-3^{n+1} b^{2} \equiv 5 \cdot 3^{n} a^{2}-3^{n+1} b^{2}=3^{n}\left(5 a^{2}-3 b^{2}\right) \pmod {m}\]

we can see that \(\left|5 a^{2}-3 b^{2}\right|\) is a multiple of \(m\). Since at least one of \(a\) and \(b\) is nonzero, \(5 a^{2} \neq 3 b^{2}\). Hence, by the choice of \(a, b\), we have \(0<\left|5 a^{2}-3 b^{2}\right| \leqslant \max \left(5 a^{2}, 3 b^{2}\right) \leqslant 5 m\). That shows that \(m_{1} \leqslant 5 m\). **Part 2.** Next, we show that \(m_{1}\) cannot be divisible by \(2\), \(3\) and \(5\). Since \(m_{1}\) equals either \(\left|5 a^{2}-3 b^{2}\right|\) or \(\left|a^{2}-15 b^{2}\right|\) with some integers \(a, b\), we have six cases to check. In all six cases, we will get a contradiction by presenting another multiple of \(m\), smaller than \(m_{1}\). - If \(5 \mid m_{1}\) and \(m_{1}=\left|5 a^{2}-3 b^{2}\right|\), then \(5 \mid b\) and \(\left|a^{2}-15\left(\frac{b}{5}\right)^{2}\right|=\frac{m_{1}}{5}<m_{1}\). - If \(5 \mid m_{1}\) and \(m_{1}=\left|a^{2}-15 b^{2}\right|\), then \(5 \mid a\) and \(\left|5\left(\frac{a}{5}\right)^{2}-3 b^{2}\right|=\frac{m_{1}}{5}<m_{1}\). - If \(3 \mid m_{1}\) and \(m_{1}=\left|5 a^{2}-3 b^{2}\right|\), then \(3 \mid a\) and \(\left|b^{2}-15\left(\frac{a}{3}\right)^{2}\right|=\frac{m_{1}}{3}<m_{1}\). - If \(3 \mid m_{1}\) and \(m_{1}=\left|a^{2}-15 b^{2}\right|\), then \(3 \mid a\) and \(\left|5 b^{2}-3\left(\frac{a}{3}\right)^{2}\right|=\frac{m_{1}}{3}<m_{1}\). - If \(2 \mid m_{1}\) and \(m_{1}=\left|5 a^{2}-3 b^{2}\right|\), then \(\left|\left(\frac{5 a-3 b}{2}\right)^{2}-15\left(\frac{a-b}{2}\right)^{2}\right|=\frac{m_{1}}{2}<m_{1}\). - If \(2 \mid m_{1}\) and \(m_{1}=\left|a^{2}-15 b^{2}\right|\), then \(\left|5\left(\frac{a-3 b}{2}\right)^{2}-3\left(\frac{a-5 b}{2}\right)^{2}\right|=\frac{m_{1}}{2}. (The last two expressions can be obtained from \((\sqrt{5} a+\sqrt{3} b)(\sqrt{5}-\sqrt{3})=(5 a-3 b)+\sqrt{15}(b-a)\) and \((a+\sqrt{15} b)(\sqrt{5}-\sqrt{3})=\sqrt{5}(a-3 b)+\sqrt{3}(5 b-a)\).) In all six cases, we found that either \(\frac{m_{1}}{2}, \frac{m_{1}}{3}\) or \(\frac{m_{1}}{5}\) is of the form \(\left|5 x^{2}-3 y^{2}\right|\) or \(\left|x^{2}-15 y^{2}\right|\). Since \(m\) is coprime to \(2\), \(3\) and \(5\), the presented number is a multiple of \(m\), but this contradicts the minimality of \(m_{1}\). **Part 3.** The last remaining case is \(m_{1}=m\), so either \(m=\left|5 a^{2}-3 b^{2}\right|\) or \(m=\left|a^{2}-15 b^{2}\right|\). We will get a contradiction by considering the two sides modulo \(3\), \(4\) and \(5\). - \(2^{n}+65=5 a^{2}-3 b^{2}\) is not possible, because \(2^{n}+65 \equiv 1(\bmod 3)\), but \(5 a^{2}-3 b^{2} \not \equiv 1\) \((\bmod 3)\). - \(2^{n}+65=3 b^{2}-5 a^{2}\) is not possible, because \(2^{n}+65 \equiv 1(\bmod 4)\), but \(3 b^{2}-5 a^{2} \not \equiv 1\) \((\bmod 4)\). - \(2^{n}+65=a^{2}-15 b^{2}\) is not possible, because \(2^{n}+65 \equiv \pm 2(\bmod 5)\), but \(a^{2}-15 b^{2} \not \equiv \pm 2\) \((\bmod 5)\). - \(2^{n}+65=15 b^{2}-a^{2}\) is not possible, because \(2^{n}+65 \equiv 1(\bmod 4)\), but \(15 b^{2}-a^{2} \not \equiv 1\) \((\bmod 4)\). We found a contradiction in all cases, that completes the solution. **Comment 1.** Part I is a standard application of Thue's lemma: **Lemma (Thue):** Suppose that \(m>1\) and \(c\) are integers, and \(X, Y\) are positive integers such that \(X \leqslant m<X Y\). Then there exist some integers \(x, y\) with \(|x|<X\) and \(0<y, such that \(x \equiv c y(\bmod m)\). It is a well-known corollary that if \(c, d\) are coprime to \(m\), the congruence \(c x^{2} \equiv d y^{2}(\bmod m)\) has a solution such that \(x, y\) are coprime to \(m\), and \(X, Y\) are positive integers such that \(X Y>m\), then \(c x^{2} \equiv d y^{2}(\bmod m)\) has a solution such that at least one of \(x, y\) is nonzero, \(|x|<X\) and \(|y|. In the solution we applied this corollary with \(c=5, d=3, X=Y=\lfloor\sqrt{m}\rfloor+1\). **Comment 2.** In fact, we proved that a positive integer \(m\) with \(m \equiv 13\) or \(37(\bmod 60)\) cannot divide any nonzero integer of the form \(5 a^{2}-3 b^{2}\) or \(a^{2}-15 b^{2}\) with coprime integers \(a, b\). In other words, if \(m \equiv 13\) or \(37(\bmod 60)\), then 15 is not a quadratic residue modulo \(m\). Using the tools of quadratic reciprocity, the solution can be significantly shortened. Suppose that \(a\) and \(b\) are coprime. For every prime divisor \(p>5\) of \(5 a^{2}-3 b^{2}\) or \(a^{2}-15 b^{2}\), we have

\[\begin{equation*} 1=\left(\frac{15}{p}\right)=\left(\frac{3}{p}\right)\left(\frac{5}{p}\right)=(-1)^{\frac{p-1}{2}}\left(\frac{p}{3}\right)\left(\frac{p}{5}\right) \tag{1} \end{equation*}\]

where \(\left(\frac{a}{p}\right)\) stands for the Legendre symbol. Considering the remainders of \(p\) when divided by \(4\), \(3\) and \(5\), \((1)\) leads to

\[p \equiv \pm 1, \pm 7, \pm 11 \text { or } \pm 17 \bmod 60\]

These remainders form a subgroup of the reduced remainders modulo \(60\). Since \(13\) and \(37\) are not elements in this subgroup, the number \(m=2^{n}+65\) cannot be a product of such primes. Instead of handling the prime divisors of \(m\) separately, we can use Jacobi symbols for further simplification, as shown in the next solution.