Proof by induction: Show that $7|5^{2n}-2^{5n}$ I'm stuck on my inductive step, and can't figure out how to manipulate this algebraically to get the form that I want...
$5^{2k}-2^{5k}=7m,m\in\mathbb{Z}$
\begin{align*}
5^{2k+2}-2^{5k+2}=5^{2k}\cdot 5^{2}-2^{5k}\cdot 2^{2}
\end{align*} 
So i'm not sure what to do now...

$5^{2n}=(5^2)^n=25^n$  Similarly, $2^{5n}=32^n$
$$\text{So, }5^{2n}-2^{5n}=25^n-32^n$$
Now, 
(1)$(a^n-b^n)(a+b)=a^{n+1}-b^{n+1}+ab(a^{n-1}-b^{n-1})$
$\implies a^{n+1}-b^{n+1}=(a^n-b^n)(a+b)-ab(a^{n-1}-b^{n-1})$
If $(a-b)$ divides $(a^{n-1}-b^{n-1}),(a^n-b^n)$
it will divide $ (a^{n+1}-b^{n+1})$
Now, $(a-b)\mid (a-b)$ and $(a-b)\mid (a^2-b^2)$
(2)$(a-b)(a^n+b^n)=a^{n+1}-b^{n+1}-ab(a^{n-1}-b^{n-1})$
$\implies a^{n+1}-b^{n+1}=(a-b)(a^n+b^n)+ab(a^{n-1}-b^{n-1})$
So, $(a-b)\mid (a^{n+1}-b^{n+1})\iff $ it divides $(a^{n-1}-b^{n-1})$
Now, $(a-b)\mid (a-b)$ and $(a-b)\mid (a^2-b^2)$
Here $a=25,b=32$
