International Mathematical Olympiad

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, $n=2k+1$. Obviously $n=1$ is not possible, so $n⩾3$. 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 $|5a^2-3b^2|$ or $|a^2-15b^2|$ with some integers $a$ and $b$. Note that $5^n-3^n=5(5^k{)}^2-3(3^k{)}^2$ is a multiple of $m$, so the set of such multiples is non-empty, and therefore $m_1$ is well-defined.

I. First we show that $m_1⩽5m$. Consider the numbers $$5^{k+1}x+3^{k+1}y,0⩽x,y⩽\sqrt{m}$$ 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 $m$: $5^{k+1}{x}_1+3^{k+1}{y}_1\equiv 5^{k+1}{x}_2+3^{k+1}{y}_2(\mathrm{mod}m)$. Now choose $a=x_1-x_2$ and $b=y_1-y_2$; at least one of $a,b$ is nonzero, and $$5^{k+1}a+3^{k+1}b\equiv 0(\mathrm{mod}m),|a|,|b|⩽\sqrt{m}$$ From $$0\equiv (5^{k+1}a{)}^2-(3^{k+1}b{)}^2=5^{n+1}{a}^2-3^{n+1}{b}^2\equiv 5\cdot 3^n{a}^2-3^{n+1}{b}^2=3^n(5a^2-3b^2)(\mathrm{mod}m)$$ we can see that $|5a^2-3b^2|$ is a multiple of $m$. Since at least one of $a$ and $b$ is nonzero, $5a^2\ne 3b^2$. Hence, by the choice of $a,b$, we have $0<|5a^2-3b^2|⩽\max (5a^2,3b^2)⩽5m$. That shows that $m_1⩽5m$.

II. Next, we show that $m_1$ cannot be divisible by $2$, $3$ and $5$. Since $m_1$ equals either $|5a^2-3b^2|$ or $|a^2-15b^2|$ 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 $m_1=|5a^2-3b^2|$, then $5\mid b$ and $|a^2-15(\frac{b}{5}{)}^2|=\frac{m_1}{5}<m_1$. - If $5\mid m_1$ and $m_1=|a^2-15b^2|$, then $5\mid a$ and $|5(\frac{a}{5}{)}^2-3b^2|=\frac{m_1}{5}<m_1$. - If $3\mid m_1$ and $m_1=|5a^2-3b^2|$, then $3\mid a$ and $|b^2-15(\frac{a}{3}{)}^2|=\frac{m_1}{3}<m_1$. - If $3\mid m_1$ and $m_1=|a^2-15b^2|$, then $3\mid a$ and $|5b^2-3(\frac{a}{3}{)}^2|=\frac{m_1}{3}<m_1$. - If $2\mid m_1$ and $m_1=|5a^2-3b^2|$, then $|{\left(\frac{5a-3b}{2}\right)}^2-15{\left(\frac{a-b}{2}\right)}^2|=\frac{m_1}{2}<m_1$. - If $2\mid m_1$ and $m_1=|a^2-15b^2|$, then $|5{\left(\frac{a-3b}{2}\right)}^2-3{\left(\frac{a-5b}{2}\right)}^2|=\frac{m_1}{2}<m_1$. (The last two expressions can be obtained from $(\sqrt{5}a+\sqrt{3}b)(\sqrt{5}-\sqrt{3})=(5a-3b)+\sqrt{15}(b-a)$ and $(a+\sqrt{15}b)(\sqrt{5}-\sqrt{3})=\sqrt{5}(a-3b)+\sqrt{3}(5b-a)$.) 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 $|5x^2-3y^2|$ or $|x^2-15y^2|$. Since $m$ is coprime to $2$, $3$ and $5$, the presented number is a multiple of $m$, but this contradicts the minimality of $m_1$.

III. The last remaining case is $m_1=m$, so either $m=|5a^2-3b^2|$ or $m=|a^2-15b^2|$. We will get a contradiction by considering the two sides modulo $3$, $4$ and $5$. - $2^n+65=5a^2-3b^2$ is not possible, because $2^n+65\equiv 1(\mathrm{mod}3)$, but $5a^2-3b^2\not\equiv 1(\mathrm{mod}3)$. - $2^n+65=3b^2-5a^2$ is not possible, because $2^n+65\equiv 1(\mathrm{mod}4)$, but $3b^2-5a^2\not\equiv 1(\mathrm{mod}4)$. - $2^n+65=a^2-15b^2$ is not possible, because $2^n+65\equiv \pm 2(\mathrm{mod}5)$, but $a^2-15b^2\not\equiv \pm 2(\mathrm{mod}5)$. - $2^n+65=15b^2-a^2$ is not possible, because $2^n+65\equiv 1(\mathrm{mod}4)$, but $15b^2-a^2\not\equiv 1(\mathrm{mod}4)$. We found a contradiction in all cases, that completes the solution.

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Alternative solution.

Suppose again that $5^n\equiv 3^n(\mathrm{mod}m=2^n+65)$. Like in the first solution, we conclude that $n$ must be odd, and $n⩾3$, so $8\mid 2^n$. Using Jacobi symbols, $$-1=\left(\frac{2^n+65}{5}\right)=\left(\frac{5}{2^n+65}\right)=\left(\frac{5^n}{2^n+65}\right)=\left(\frac{3^n}{2^n+65}\right)=\left(\frac{3}{2^n+65}\right)=\left(\frac{2^n+65}{3}\right)=1,$$ contradiction.