Induction: Prove that $5^{3n} + 7^{2n-1}$ is divisible by $4$ 
Prove that $5^{3n} + 7^{2n-1}$ is divisible by $4$ for all $n \in \mathbb{N}$.



*

*For $n=1$, 
$\Rightarrow 5^3 + 7^1 \Rightarrow 132 \mid 4$ (which is divisible by $4$)

*Let us assume given equation holds true for $n = m$, 
$\Rightarrow 5^{3m} + 7^{2m-1} | 4$

*Now for $n = m+1$, 


*

*$5^{3m+3} + 7^{2m+2-1}$

*$5^{3m} \cdot 5^3 + 7^{2m-1} \cdot 7^2$

*$5^{3m} \cdot 125 + 7^{2m-1} \cdot 49$



How do I go ahead from here? I am kind of stuck.

Let $M_{n}=5^{3n}+7^{2n-1}$, $n\in\mathbb{N}$. Now, as you well say, take $n=1$ then $M_{1}=5^{3}+7=132$ which is divisible by 4. 
So, let us assume the equation holds true for $k$, i.e., $M_{k}=5^{3k}+7^{2k-1}$, and we try to prove that holds true for $k+1$. Therefore,
$$M_{k+1}-M_{k}=(5^{3k+3}+7^{2k+1})-(5^{3k}+7^{2k-1})=5^{3k}(5^{3}-1)+7^{2k-1}(7^{2}-1)$$
$$=(7^{2}-1)(5^{3k}+7^{2k-1})+(5^{3}-7^{2})5^{3k}$$
$$\therefore M_{k+1}=M_{k}+48(5^{3k}+7^{2k-1})+76(5^{3k})$$
Where, each term is divisible for 4, and we're done.
