prove that $n^2 \bmod 4 = 0$ or $1$ for all integers

Question

(Use the fact that every integer is either even or odd to prove that $n^2 \bmod 4 = 0$ or $1$ for all integers)

Let $n \in \mathbb{Z}$, then $n$ is either even or odd.

Case 1: ($n$ is odd): By definition of odd $n = 2j + 1$ with $j \in \mathbb{Z}$, then $(2j + 1)^2 = (4j^2 + 4j + 1)$

By definition of mod $4j^2 + 4j + 1 = 4q + r$

$\forall q, r \in \mathbb{Z}$ and $0 \le r \le q$

$4(j^2 + j) + 1 = 4q + r$

$4(j^2 + j) - 4q = -1 + r$

$4(j^2 + j - q) = -1 + r$

Let $s \in \mathbb{Z}$ with $s = j^2 + j - q$, then

$4s = -1 + r$

So by definition of divisibility $4$ divides $n^2$ with a remainder of $1$

I have a feeling I got this wrong and I just bulled my way to get an answer. But I wasn't given any marks here:

case 2: ($n$ is even):

I got this section wrong.

Can someone help solve me this proof that was on my test?

I'm confused: did you get case 2 wrong because you left it blank or did you write something there that you neglected to copy over to here? — Robert Soupe, May 12 '15 at 01:53
Thoight of the same thing! There's a interesting anlogy when you think of positions of a point in a n x n matrix rotating 360 degrees! — MathematicsBeginner, Aug 23 '22 at 18:46

score 3 · Accepted Answer · answered May 12 '15 at 01:47

3

If $n$ is even $\implies n=2k\implies n^2=(2k)^2=4k^2\equiv 0 \pmod 4$.

If $n$ is odd $\implies n=2k+1\implies n^2=4k^2+4k+1=4(k^2+k)+1\equiv 1 \pmod 4$

answered May 12 '15 at 01:47

Teoc

On the android app this looks like "If $n$ is even [Math processing error]..." – Matt Samuel May 12 '15 at 01:52
We weren't taught congruence in this exam until the next one, so there must be a different way to solve it? sorry i didn't specify later, but ill take it as an answer – hit May 12 '15 at 01:53
At the second to last steps for both cases, you could say that when this is divided by 4, it would result in a remainder of 0 or 1. – Teoc May 12 '15 at 01:54
ok, thanks for the help – hit May 12 '15 at 01:54

score 1 · Answer 2 · answered May 12 '15 at 01:45

1

If $n$ is even then $n=2m$, and $(2m)^2=4m^2\equiv 0\pmod{4}$.

answered May 12 '15 at 01:45

Ramified_Minds

we weren't taught congruence at that point yet – hit May 12 '15 at 01:47
What do you mean? We say $k$ is congruent to $0$ modulo $4$ if $\exists : p$ s.t. $k=4p+0=4p$. Thus, by definition, $4m^2$ is congruent to $0$ modulo $4$.. – Ramified_Minds May 12 '15 at 01:50

2 Answers2