<s>[INST] Prove that the square of any given odd number gives a remainder of 1 when divided by 4 I'm supposed to prove this by induction, and I get that if I assume $n$ is odd such that $n^2$ gives a remainder of 1, then $n^2=4m+1$ for a number $m$ where $m \in \mathbb{Z}$,  but I honestly can't get any further than that. Could someone complete the proof for me so that I can get a proper idea of how I can do this? 
 [/INST] Here is a proof that doesn't require induction:
We want to show that $n^2 - 1 \equiv 0 \mod{4}$ for $n$ odd.
Assume n is odd. Note that $n^2 = (n-1)^2 + 2(n-1) + 1$, or, visually for n =5:

Since $n$ is odd, $n-1$ is even and can be written as $2k$ for some $k$. So we have that:
$$\begin{align}
n^2 - 1 &= (n-1)^2 + 2(n-1) \\
        &= (2k)^2 + 2(2k) \\
        &= 4k^2 + 4k \\
        &= 4(k^2 + k)
\end{align}$$
So we have shown that $n^2 -1$ is a multiple of 4, so by definition $n^2 - 1 \equiv 0 \mod{4}$ or $n^2 \equiv 1 \mod{4}$ 
$\blacksquare$
Unfortunately, some professors require a certain type of proof even when many are valid. We can try to turn this reasoning into an induction proof. First, we prove the claim for the base case:
Let $n = 1$. Then $1^2 = 1 \equiv 1 \mod{4}$, so the claim holds for the base case. Now, we move to prove the inductive step. Assume the claim holds for some odd n. We want to show that it holds for $n+2$ (the next odd n). See the figure below to visualize this. 

What we are saying is that $\color{red} n^2 \color{black} \equiv 1 \mod{4}$, so we want anything we add in to get to $(n+2)^2$ to be equivalent to 0 mod 4. That is, we need to check that
$$
2\color{blue}n  + 2\color{orange}{(n+1)}\color{black} + \color{purple} 2 \color{black} \equiv 0 \mod{4}
$$ 
Well, we see this must be the case because
$$\begin{align}
2n + 2(n+1) + 2 &= 2n + 2n + 2 + 2 \\
        &= 4n + 4 \\
        &= 4(n+1) \\
\end{align}$$
is a multiple of four, so $2n + 2(n+1) + 2 \equiv 0 \mod{4}$. So putting this together we get that:
$$
(n+2)^2 = n^2 + 2n + 2(n+1) + 2 \equiv 1 \mod{4}
$$
as desired. $\blacksquare$.
The other proofs here are perfectly valid, I was hoping only to give a little perspective.
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