For all $n\in \mathbb{N},2^{2n+1}+3^{2n+1}$ is divisible by

For all $n\in \mathbb{N},2^{2n+1}+3^{2n+1}$ is divisible by

Let $P\left(n\right)$ be the statement given by

$P\left(n\right):2^{2n+1}+3^{2n+1}$

here $n=1$

$P\left(1\right):2^{2n+1}+3^{2n+1}$

$=2^{2\times 1+1}+3^{2\times 1+1}$

$=35=5\times 7$

Divisible by $5$

$P\left(1\right)$ is true

Let $P\left(m\right)$ be true

$P\left(m\right)=2^{2m+1}+3^{2m+1}$$=5\lambda$

$2^{2m+1}=5\lambda -3^{2m+1}................\left(i\right)$

$P\left(m\right)$ is true

Let $P\left(m+1\right)$ be true

$P\left(m+1\right):2^{2\left(m+1\right)+1}+3^{2\left(m+1\right)+1}$

$=2^{2m+1+2}+3^{2m+1+2}$

$=2^{2m+1}{2}^2+3^{2m+1}{3}^2$

From equation $\left(i\right)$

$=\left(5\lambda -3^{2m+1}\right)2^2+3^{2m+1}{3}^2$

$=5\lambda .2^2-3^{2m+1}{2}^2+3^{2m+1}{3}^2$

$=5\lambda .2^2-3^{2m+1}\left(2^2-3^2\right)$

$=5\lambda .2^2+5.3^{2m+1}$

$=5\left(\lambda .2^2+3^{2m+1}\right)$

$P\left(m+1\right)$ is true

Hence, it is divisible by $5$