(given to Jean-Denis Zafar). Let $m$ and $n$ be two integers that are coprime. Calculate the GCD of $5^{m}+7^{m}$ and $5^{n}+7^{n}$.
. We denote $S_{k}=5^{k}+7^{k}$. We will show that the GCD of $S_{n}$ and $S_{m}$ is 12 if $m$ and $n$ are odd, and 2 otherwise. We denote $\delta$ as the GCD we wish to calculate. For $m=n=1$, we directly have $\delta=12$. We assume $m>n$ and set $m=n+a$; $a$ and $n$ are coprime. Let $U_{a}=7^{a}-5^{a}$. Then $S_{m}-5^{a} S_{n}=7^{n} U_{a}$. Now, we have $\operatorname{GCD}(\delta, 7)=\operatorname{GCD}(\delta, 5)=1$, hence $\delta=\operatorname{GCD}\left(U_{a}, S_{n}\right)$. If $a$ is odd, we set $S_{n}=\delta \ell$ and $U_{a}=\delta k$ with $k$ and $\ell$ coprime. Then $7^{a n}=\left(\delta \ell-5^{n}\right)^{a}$ can be written as $\delta L-5^{a n}$ using the binomial formula. Similarly, $7^{a n}=\delta K+5^{a n}$ for some integer $K$. Thus $\delta(K-L)=2 \times 5^{a n}$. According to Gauss's theorem, $\delta$ is a divisor of 2. It is 2 because the two numbers are even. If $a$ is even, we set $a=2 b$ so that $\operatorname{GCD}(b, n)=1$ and $n$ is odd. Then $\delta$ divides $\left(7^{n}-5^{n}\right)\left(7^{n}+5^{n}\right)=7^{2 n}-5^{2 n}$ as well as $U_{a}=7^{2 b}-5^{2 b}$. It therefore divides their GCD, which is $7^{2}-5^{2}=24$. Furthermore, since $m$ and $n$ are odd, we have $7^{m} \equiv-1(\bmod 8)$ and $5^{m} \equiv 5(\bmod 8)$, and thus $S_{m} \equiv 4$ (mod 8). Similarly, $S_{n} \equiv 4(\bmod 8)$. Consequently, $S_{n}$ and $S_{m}$ are divisible by 4 but not by 8. Similarly, we show that they are divisible by 3. The value of $\delta$ is therefore 12 as announced.