If $n$ is an integer then $n^2$ is the same as $0$ or $1\pmod 4$?

Question

I have been stuck on this problem for awhile.

How would i go about solving it, an explanation would be helpful as well.

Show that if $n$ is an integer then $n^2 \equiv 0$ or $1 \pmod 4$?

Answer 1

Let $k\in\mathbb Z$.

If $n=2k$, then $n^2=4k^2$.

If $n=2k-1$, then $n^2=4k^2-4k+1$.

Answer 2

HINT: An integer $n$ is either even or odd. If $n$ is even, then $n=2m$ for some integer $m$. If $n$ is odd, then $n=2m+1$ for some integer $m$. In each case express $n^2$ in terms of $m$.

Answer 3

An important and fundamental idea:

There are only four numbers mod $4$. You can try them all.

Answer 4

Did you try calculating? There are only four cases mod 4:

$n \equiv 0 \Rightarrow n^2 \equiv 0^2 \equiv 0$

$n \equiv 1 \Rightarrow n^2 \equiv 1^2 \equiv 1$

$n \equiv 2 \Rightarrow n^2 \equiv 2^2 \equiv \ ...$

$n \equiv 3 \Rightarrow \ ...$

Answer 5

Suppose $n$ is even. Then $n=2k$ for some integer $k$.

Then $n^2 = (2k)^2 = 4k^2 \equiv 0$ (mod 4).

You can do something similar with $n$ odd

Answer 6

To prove $n^2\not\equiv 2 \mod 4$, we can see that $n$ must be even, and therefore $4|n^2$.

To prove $n^2\not\equiv 3 \mod 4$, if it is true we have $n^2-1=4k+2\to (n+1)(n-1)=4k+2$, but as $n$ is odd both $n+1$ and $n-1$ are even and so $4|(n+1)(n-1)$.

We can establish the presence of $n^2\equiv 0\mod4$ and $n ^2\equiv 1\mod4$ by examining $n=0$ and $n=1$.