<s>[INST] Proof by induction that $3^{2n} + 7$ is divisible by $4$ Demonstrate by induction: $3^{2n} + 7 = 4k$ is true, for any $n\in \mathbb N$. I need to demonstrate this using the induction principle.
So far I have:
$n = 1$
$$3^{2\cdot 1} + 7 = 4\cdot k $$
$$9 + 7 = 4k$$
$$16 = 4k$$
$$k = 4$$
So it checks for $n=1$.
$n = h$
$$3^{2\cdot h} + 7 = 4\cdot k$$
$n = h +1$
$$3^{2\cdot (h + 1)} + 7 = 4\cdot k'$$
(I use $k'$ to note that it's not the same $k$ as in $n = h$)
And I don't know how to continue.
 [/INST] Just to be contrary,
here is a non-induction proof.
$\begin{array}\\
3^{2n}+7
&=3^{2n}-1+8\\
&= (3^n+1)(3^n-1)+8
\end{array}
$
Since $3^n$ is odd
(for $n \ge 1$),
$3^n-1$ and
$3^n+1$
are consecutive even numbers,
so one is exactly divisible by $2$
and the other is divisible by
(at least) $4$.
Therefore,
their product is divisible
by at least 8,
so
$3^{2n}+7$
is divisible by 8.
</s>