How to show that for each $n\in\mathbb{N}$, $\gcd(3^{n}+5^{n+1},3^{n+1}+5^{n})=2$ or $ 14$?

Question

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.

Answer 1

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 $14$.

Answer 2

HINT

Use that for $k\in \mathbb{Z}$ $\quad\gcd(a,b)=\gcd(a,b-ka)$.

Answer 3

It's $\,\overbrace{(3a\!+\!b,a\!+\!5b)}^{\Large\ \ a,b\ =\ 3^n,5^{n+1}}\!\mid \color{#c00}{14}(a,b)=14\,$ by here, since $\ \begin{align} a&\mapsto\, 3a\!+b\\ b&\mapsto \ a\!+\!5b\end{align}\ $ has det $= 3(5)-1(1) = \color{#c00}{14}$

$a,b$ are odd so the gcd is even, so only the factors $2,14$ of $14$ are possible.

Answer 4

Let $g=\gcd(3^{n}+5^{n+1},3^{n+1}+5^{n})$.

Then $g\mid 3^{n}+5^{n+1}$ and $g\mid 3^{n+1}+5^{n}$ so $$g\mid 3(3^{n}+5^{n+1})-(3^{n+1}+5^{n})= 5^n\cdot 14$$

Since $5\nmid g$ we have $g\in\{1,2,7,14\}$. Since bot expressions are even we can rule out $1$ and $7$.