Let $a,b\in\mathbb Z$ such that $\gcd(a,b)=5$. Find $\gcd(25ab^2, a^4+b^4)=d$.
This is all I managed to think of: $a=5k_1$ and $b=5k_2$. Therefore, $25ab^2=3125k_1k_2^2$ and $a^4+b^4=625(k_1^4+k_2^4)$. So $d|625(k_1^4+k_2^4+5k_1k_2^2)$. Does this mean that $d|625$?
Help?
Actually, we see that $625$ divides both number, so that $625|d$. Now the question is, can $d$ be greater than $625$? We have that $\gcd(k_1,k_2)=1$. Any prime $p\neq5$ that divides $3125k_1k_2^2$ divides one of $k_1,k_2$ but not the other, so it doesn't divide $k_1^4+k_2^4$. The question remains whether it's possible that $5\mid(k_1^4+k_2^4)$.
Consider $k^4\pmod{10}$, that is, the last digit of $k^4$. A little arithmetic shows that the only possibilities are $0,1,5,6$. The only way to get a number ending in $0$ or $5$ is to add two chosen from $\{0,5\}$, so that both $k_1,k_2$ would have to be divisible by $5$, contradiction.
Therefore, the gcd is $625$.