Perfect square is $0$ or $1$ modulo $4$ .

Question

Prove that for every integer $n$ either

Couldn't you just write out the 4 cases, of 0,1,2, and 3 mod 4 and get the desired answers? — JB King, Jan 14 '16 at 17:41
Just calculate $1^2, 2^2, 3^2, 0^2$ and look for the remainder... — , Jan 14 '16 at 17:41
Possible duplicate of The square of an integer is congruent to 0 or 1 mod 4 — , Jan 14 '16 at 22:29

score 3 · Answer 1 · answered Jan 14 '16 at 17:43

3

Hint:

If $n = 2k$, $$n^2 = 4k^2$$ If $n = 2k+1$, $$n^2 = (2k+1)^2 = 4k^2+4k+1 = 4(k^2+k)+1$$

answered Jan 14 '16 at 17:43

fosho

score 3 · Answer 2 · answered Jan 14 '16 at 17:45

3

$n$ can be even or odd.

When $n=2k$ we get

$(2k)^2=4k^2 \equiv 0 \pmod 4$.

When $n=2k+1$ we get

$n^2 = (2k+1)^2 = 4k^2 + 4k + 1 \equiv 1 \pmod 4$.

answered Jan 14 '16 at 17:45

Maffred

2 Answers2