Finding the remainder when $5^{55}+3^{55}$ is divided by $16$ 
Find the remainder when $5^{55}+3^{55}$ is divided by $16$.

What I try
$a^{n}+b^{n}$ is divided by $a+b$ when $n\in $ set of odd natural number.
So $5^{55}+3^{55}$ is divided by $5+3=8$
But did not know how to solve original problem
Help me please

$3^4 = 81 = 16\cdot 5 + 1 \equiv 1 \mod(16)$
Then
$3^{52} = (3^{4})^{13} \equiv 1^{13} \mod(16) $
Then
$3^{55} = 3^3 \cdot 3^{52} \equiv 27 \cdot 1 \mod(16)$
And $27 \equiv 11 \mod(16)$
Thus 
$3^{55} \equiv 11 \mod(16)$
Now,
$5^2 \equiv 9 \mod(16)$
So $(5^2)^{26} \equiv (3^{2})^{26} \mod(16)$
Then
$5^{52} \equiv 3^{52} \mod(16)$
And as before, $3^{52} \equiv 1 \mod(16)$
Thus
$5^{52} \equiv 1 \mod(16)$
So
$5^{55} \equiv 5^{3} \mod(16)$
And $5^{3} = 125 = 16 \cdot 7 + 13$
Then $5^{55} \equiv 13 \mod(16)$
And thus $5^{55} + 3^{55}\equiv 13 + 11 \mod(16)$
$24 \equiv 8 \mod(16)$
Finally $5^{55} + 3^{55} \equiv 8 \mod(16)$
So $16 \not| (5^{55} + 3^{55})$
