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?

Are you allowed to use Fermat's Little theorem?
$5^{\phi(14)} = 5^6 \equiv 1 \mod 14$
$5^{98} \equiv 5^{96}5^2 \equiv 25 \equiv -3 \mod 14$ so $14|5^{98} + 3$
and $5^{99} \equiv -3*5 \equiv -15 \equiv -1 \mod 14$ so $14|5^{99} + 1$
So $14$ is a common divisor.
Hmm, but how to show it is the greatest common divisor?
Well, we know  $14|\gcd(5^{98} + 3, 5^{99} + 1)$
And we can see $5 \not \mid 5^{98} + 3$ nor $5^{99}+1$ 
so $\gcd(5^{98} + 3, 5^{99} + 1) = \gcd(5^{98} + 3, [5^{99} + 1] - 5[5^{98} + 3])$
$= \gcd(5^{98} + 3, [5^{99} + 1] - [5^{99} + 15]) = \gcd(5^{98} + 3, - 14)$
$\gcd(5^{98} + 3, 14)$
So $\gcd(5^{98} + 3, 5^{99} + 1)|14$ and $14|\gcd(5^{98} + 3, 5^{99} + 1)$.
So $14 = \gcd(5^{98} + 3, 5^{99} + 1)$.
