<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] 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 $3^{2n} + 7 = 1+3 = 4 = 0 \pmod 4$.
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