proof verification - Prove by induction that $23^{2n} + 31^{2n}+46$ is divisible by 48 for all integers $n ge 0$

The base case, $n = 0$:
$$23^0+31^0+46=48$$ and $48 \bmod 48 = 0$.

Inductive Hypothesis:
Let's assume it is true for $n = k$.
Then $$p(k) = 23^{2k} + 31^{2k}+46$$
$$ \Rightarrow p(k+1) = 23^{2\left(k+1\right)} + 31^{2\left(k+1\right)}+46 = 529\left(23^{2k}\right)+961\left(31^{2k}\right)+46$$

However, I am not sure how to proceed from here in order to get an expression that is divisible by $48$.

Thanks in advance for any hints.

Answer

Note that your $p(k)$ is written wrongly.

\begin{align}
529(23^{2k}+ 31^{2k})+432(31^{2k})+46 &= 529(23^{2k}+ 31^{2k})+9(48)(31^{2k})+46 \\
&= 529(23^{2k}+ 31^{2k}+46)+9(48)(31^{2k})+46(1-529) \\
&= 529(23^{2k}+ 31^{2k}+46)+9(48)(31^{2k})-46(528) \\
&= 529(23^{2k}+ 31^{2k}+46)+9(48)(31^{2k})-46(48)(11) \\
\end{align}