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.
 A: Be careful not to assume what you're trying to prove. After the induction hypothesis, i.e.,
$$3^{2h} + 7 = 4k \quad \textrm{for some } k,\tag1$$
you can't jump to
$$3^{2(h+1)} + 7 = 4k' \quad \textrm{for some } k',\tag2$$
because that's what you must prove. Instead, start with just the left-hand side of $(2)$, and manipulate it so that you can use what you know from the hypothesis in $(1)$:
$$
\begin{align}
3^{2(h+1)} + 7 &= 3^{2h+2} + 7\\
 &= 3^2 3^{2h} + 7\\
 &= 9\cdot 3^{2h} + 7\\
 &= 8\cdot 3^{2h} + \color{maroon}{3^{2h} + 7}\\
\end{align}$$
Can you take it from here?
A: You have shown for n=1 which is the first step so 
Now assume or state that it is true for n=h
$$3^{2h}+7=4k$$
You want to show that it is true for n=h+1
So start with  $3^{2h+2}+7$
 $$3^{2h+2}+7= 9(3^{2h})+7=9(4k-7)+7$$
$$3^{2h+2}+7=36k-56=4(9k-14)=4m$$ where m is another natural number thus proving our assumption
A: 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.
A: Observe that $3^{2n} = 9^n = (9^n-1^n) + 1 = 8\cdot(...)+1=0+1 =1\pmod 4$, and $7 = 3 \pmod 4$, hence $3^{2n} + 7 = 1+3 = 4 = 0 \pmod 4$.
A: I'll give a proof with congruences: as $3^2\equiv 1\mod8$,
$$3^{2^n}+7=(3^2)^n+7\equiv 1^n+7\equiv 0\mod 8.$$ 
A fortiori, $\;3^{2^n}+7\equiv 0\mod 4$.
A: $3^{2n} + 7 \equiv (-1)^{2n} + 7 \equiv 1^n+7 \equiv 0 \pmod 4$
