<s>[INST] 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.
 [/INST] To complete your argument try to use the induction hypothesis. For example, write 
$$5^{3m} \cdot 125 + 7^{2m-1} \cdot 49 = (5^{3m} + 7^{2m-1})  + 5^{3m} \cdot 124 + 7^{2m-1} \cdot 48.$$ 
Now $4$ divides $5^{3m} + 7^{2m-1}$ by induction hypothesis, and $4$ divides $5^{3m} 124$ as $4$ divides $124$, and $4$ divides $7^{2m-1} \cdot 48$. 
Whence the sum of the three is also divisible by $4$ and you are done. 
(This is basically the argument in the comments, but avoiding the modulo concept.) 
</s>