<s>[INST] How to show that for each $n\in\mathbb{N}$, $\gcd(3^{n}+5^{n+1},3^{n+1}+5^{n})=2$ or $ 14$? For each $n\in\mathbb{N}$ show that
$$
\gcd(3^{n}+5^{n+1},3^{n+1}+5^{n})=2\text{ or } 14.
$$
I tried induction but I got stuck.
 [/INST] They are both odd, so their gcd is even. Moreover
\begin{align}
\gcd(3^n+5^{n+1},3^{n+1}+5^n)&=\gcd(3^n+5^{n+1}-5(3^{n+1}+5^n),3^{n+1}+5^n)\\
&=\gcd(3^n (1-3\cdot 5),3^{n+1}+5^n) \\
&=\gcd(3^n \cdot 14,3^{n+1}+5^n) \\
&=\gcd(14,3^{n+1}+5^n) 
\end{align}
divides $14$. So it can be only $2$ or $14$.
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