Proving that $\gcd(5^{98} + 3, \; 5^{99} + 1) = 14$ 
Prove that $\gcd(5^{98} + 3, \; 5^{99} + 1) = 14$.

I know that for proving the $\gcd(a,b) = c$ you need to prove 


*

*$c|a$ and $c|b$

*$c$ is the greatest number that divides $a$ and $b$ 
Number 2 is what I'm struggling with. Does anybody have any ideas?

First of all, $$5\cdot(5^{98}+3)-1\cdot(5^{99}+1)=14$$
$$\implies (5^{98}+3,5^{99}+1)\mid 14$$
Now both the numbers are even $\implies 2\mid (5^{98}+3,5^{99}+1) $
Using Fermat’s Little theorem $5^6\equiv1\pmod 7$
$\implies 5^{98}=(5^6)^{16}\cdot5^2\equiv 25\pmod 7\equiv-3\implies 7\mid (5^{98}+3)$
and $5^{99}\equiv(5^6)^{16}\cdot5^3\equiv125\pmod 7\equiv-1 \implies 7\mid (5^{99}+1)$
$\implies 7\mid (5^{98}+3,5^{99}+1) $
$\implies (5^{98}+3,5^{99}+1)$ is divisible by lcm$(2,7)=14$
