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How can I prove that square of every integer is of the form either 3k+1 or 3k. but not 3k+2. This was my approach: I considered first, the integer n to be even and then; n= 2m and if n is odd then n=2m+1 but this step takes me to nowhere. I get usually struck in these kind of questions. and this time I seriously need help . So, please post answer in detail and also mention how and why you arrived at that particular step. Thanks in advance.

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I did not see you take the square of the integers you made. –  Raskolnikov Jul 24 '12 at 7:01
Take $n=3k$ or $n=3k\pm 1$ –  Kns Jul 24 '12 at 7:05
Nearly the same: math.stackexchange.com/questions/172535/… –  Zander Jul 24 '12 at 14:31
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2 Answers 2

HINT: Instead of looking at $n = 2m$ or $n = 2m+1$, look at $n = 3m$ or $n = 3m \pm 1$.

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Really a simple problem but I failed to find the hint , that is in the problem itself. Thanks... understood the approach. –  shrey Jul 24 '12 at 7:11
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So you mean, that the quadratic residue $x^2 \neq 2 \bmod 3$: $$ (3m\pm 1)^2=9m^2\pm6m+1=3m'+1 =1 \bmod 3 \neq 2 \bmod 3 $$ $(3m)^2$ is obvious...

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