<s>[INST] Prove that $n^4 \mod 8$ is identically equal to either 0 or 1, $\forall \ n\in \mathbb{N}$. I feel pretty confident about the first half of my solution for this, however I don't like how I used the induction hypothesis on the case for even integers, it feels like it isn't doing anything useful since it is really easy to show that $P(2x+2) \mod 8 \equiv 0$ without using the inductive hypothesis.
This is also the first semester I've been doing proofs, so am I going about this right? 
Proof. Since $1^4 \mod 8 = 1$, $P(1)$ holds. Also, $2^4 \mod 8 = 16 \mod 8 = 0$ so, $P(2)$ holds.
We claim that $P(2x-1)$ holds, then $(2x-1)^4 \mod 8\equiv 1$ for any odd integer $n=2x-1$. So, 
$$(2x-1)^4 = 16x^4-32x^3+24x^2-8x+1 \mod 8\equiv 1,$$
now consider
$$(2x+1)^4 = 16x^4+32x^3+24x^2+8x+1 = (16x^4-32x^3+24x^2-8x+1)+64x^3+16x,$$
then since
$$64x^3+16x \mod 8 = 8(8x^3+2x) \mod 8 \equiv 0,$$
we conclude that
$$(16x^4-32x^3+24x^2-8x+1)+64x^3+16x \mod 8 \equiv 1+0 = 1.$$
Thus, $P(2x+1)$ holds, and all odd integers $n$ hold. 
Next, we claim that $P(2x)$ holds, then $(2x)^4 \mod 8\equiv 0$ for any even integer $n=2x$. So,
$$(2x)^4 \mod 8 = 16x^4 \mod 8 \equiv 0,$$
now consider
$$(2x+2)^4 = 16x^4+64x^3+96x^2+64x+16,$$
then
$$64x^3+96x^2+64x+16 \mod 8 = 8(8x^3+12x^2+8x+2) \mod 8 \equiv 0,$$
we conclude that
$$16x^4+64x^3+96x^2+64x+16 \mod 8 \equiv 0+0 = 0.$$
Thus, $P(2x)$ holds, and all even integers $n$ hold. 
By mathematical induction, if $n=2x-1$ and $n=2x$ hold for the given statement, $n=2x+1$ and $n=2x+2$ also hold. Therefore, the statement holds for all $n \in \mathbb{N}$.
$\mathbb{QED}$
 [/INST] Your idea is okay, but $(2x+2)^4=16 + 64 x + 96 x^2 + 64 x^3 + 16 x^4$, 
unlike what you initially put in the question.

Alternatively, you could argue that if $2|n$ then $8|n^3|n^4$, so $n^4\equiv0\pmod8$,
and if $2\nmid n$ then $2|n-1, n+1, $ and $n^2+1$, so $8|n^4-1$; i.e., $n^4\equiv1\pmod8$.
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