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 $n \in \mathbb{N}$.



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*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 $n = m+1$, 


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

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

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



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

Use the binomial theorem:
$$5^{3n} + 7^{2n-1}=(4+1)^{3n}+(8-1)^{2n-1}=4a+1^{3n}+8b+(-1)^{2n-1}=4c+1-1=4c$$
