elementary number theory - 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 $n \in \mathbb{N}$.

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


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


3. 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.

Answer

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.)