Prove that if n is an integer that is not a multiple of 4, then n^2 ≡ 0 mod 4 or n ^2 ≡ 1 mod 4.?

Question

Prove that if n is an integer that is not a multiple of 4, then n^2 ≡ 0 mod 4 or n ^2 ≡ 1 mod 4.?

Answer 1

There are only three cases:



n ≡ 1 mod 4 ... n^2 ≡ 1 mod 4

n ≡ 2 mod 4 ... n^2 = 4 ≡ 0 mod 4

n ≡ 3 mod 4 ... n^2 = 9 ≡ 1 mod 4

Answer 2

consider n = 4k+1, 4k+2, 4k+3



(4k+1)^2 = 16k^2 + 8k + 1 = 4(4k^2+2k) + 1

(4k+2)^2 = 16k^2 + 16k + 4 = 4(4k^2 + 4k + 1)

(4k+3)^2 = 16k^2 + 24k + 9 = (16k^2 + 24k + 8) + 1 = 4(4k^2+6k+2) + 1



in all 3 cases, the remain is either 0 or 1

Answer 3

If n is not a multiple of 4, then we can write

n = 4k + r



where r = 1,2,3



Then

n² = (4k + r)² = 16k² + 8kr + r²



Therefore

n² ≣ r² (mod 4)



Argue from there based on the possible values of r.