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.
 A: 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}
A: Note that $529 \equiv 961 \equiv 1 \mod 48$.
\begin{align}
p(k+ 1) &\equiv 23^{2(k+1)} + 31^{2(k+1)} + 46 &\mod 48\\
p(k+ 1) &\equiv 529 \cdot 23^{2k} + 961 \cdot 31^{2k} + 46 &\dots\\
p(k+ 1) &\equiv 23^{2k} +31^{2k} + 46 &\dots\\
p(k+ 1) &\equiv p(k) &\dots
\end{align}
Now use $p(k+1)\equiv p(k) \mod 48$ as your inductive step to show that $p(k) \equiv 0 \mod 48$ for all $k$.
A: If $48$ divides $23^{2k} + 31^{2k}+46$ then there exists an integer $m$ such that $48m= 23^{2k} + 31^{2k}+46$, solving for say $31^{2k}$ (you could have done $23^{2k}$) we get $31^{2k}=48m-23^{2k}-46$ and use that when showing the statement is true at $k+1$. All that is left to do is show that $p(k+1)=48r$ for some integer $r$ which will be in terms of $m$
A: Note that $$ 529\left(23^{2k}\right)+961\left(31^{2k}\right)+46$$
$$=(11\times 48 +1)\left(23^{2k}\right)+(20\times 48 +1)\left(31^{2k}\right)+46$$
$$\equiv  \left(23^{2k}\right)+\left(31^{2k}\right)+46  \text { mod (48)} $$
Now use the induction hypothesis and you will be done. 
A: Method$\#1:$
$$p(k+1)-23^2p(k)=31^{2k}(31^2-23^2)=31^{2k}\cdot9\cdot48$$
Or $$p(k+1)-31^2p(k)=-23^{2k}(31^2-23^2)=\cdots$$
In either case,$$48|p(k+1)\iff48|p(k)$$  as $(23,48)=(31,48)=1$
Method$\#2:$
W/O induction, $23^2=(24-1)^2\equiv1\pmod{48}\implies23^{2n}\equiv(1)^n$
and $31^2=961\equiv1\pmod{48}\implies31^{2n}\equiv(1)^n$
A: Here is another take:
$$
23^{2n} + 31^{2n}= (23^2)^n + (31^2)^n \equiv 1^n + 1^n = 2 \bmod 48
$$
