<s>[INST] Express $ \operatorname{gcd}\left(5^{m}+7^{m}, 5^{n}+7^{n}\right) $ in terms of $m$ and $n$ 
Let $m$ be a positive integer with $\operatorname{gcd}(m, n)=1 .$ Express
  $
\operatorname{gcd}\left(5^{m}+7^{m}, 5^{n}+7^{n}\right)
$
  in terms of $m$ and $n$

My work -
let $d=\operatorname{gcd}(5^m +7^m,5^n +7^n)$ then 
$5^{2m} \equiv 7^{2m}$ mod(d)
$5^{2n} \equiv 7^{2n}$ mod(d)
and obviously $gcd(5,d)=gcd(7,d)=1$ so,
$5^{gcd(2m,2n)} \equiv 7^{gcd(2m,2n)}$ (mod d)
$5^2         \equiv 7^2$          (mod d)
$d= 1,2,3,4,6,8,12,24$
now i find values of d how to express this in terms of $m$ and $n$  ???
 [/INST] When $m$ and $n$ are both odd and coprime and $a,b$ are coprime, $gcd(a^m + b^m, a^n + b^n) = a + b$.
As stated by @Geoffrey in the comments, $a+b \mid a^k + b^k$ for odd $k$:
$$a^m+b^m = (a + b)\underbrace{(a^{m-1} - a^{m-2}b + a^{m-3}b^{2} - \dots + b^{m-1})}_{m\space terms}$$
$$a^n+b^n = (a + b)\underbrace{(a^{n-1} - a^{n-2}b + a^{n-3}b^{2} - \dots + b^{n-1})}_{n\space terms}$$
The rightmost polynomials in both equations have no common factor since $m$ and $n$ have no common factor.
If $m,n$ are both even, then they cannot be coprime. If one of $m, n$ is even, then...?
</s>