Proof by mathematical induction that, for all non-negative integers $n$, $7^{2n+1} + 5^{n+3}$ is divisible by $44$. I have been trying to solve this problem, but I have not been able to figure it out using any simple techniques. Would someone give me the ropes please?
Prove for $n=1$
$7^3 + 5^4 = 968 = 44(22)$ 
Assume $F(k)$ is true and try $F(k+1)+-F(k)$
$F(k) = 7^{2k+1} + 5^{k+3}$
$F(k+1) = 7^{2k+3} + 5^{k+4}$
$49*7^{2k+1} + 5*5^{k+3} +- [7^{2k+1} + 5^{k+3}]$
Neither addition nor subtraction gives us a chance to take $44$ in front of the bracket, the best approach it has come to my mind is to subtract $F(k)$ which gives us:
$4[12*7^{2k+1} + 5^{5+3}]$
And now we need to prove that the inside is divisible by $11$. I have not been able to do so.

Base case : n=1✓
Hypothesis : $44$ divides $7^{2n+1} +5^{n+3}$.
Step $n+1$:
$7^{2(n+1)+1}+ 5^{(n+1)+1}=$
$7^{2n+1}7^2+ 5^{n+1}5=$
$7^{2n+1}(44+5)+5^{n+1}5=$
$(44)7^{2n+1} +5(7^{2n+1}+5^{n+1}).$
First term divisible by $44$, so is the second term by hypothesis.
