For any integer $n$, $n^2$ is congruent to $0$ or $1$ modulo $3$

Question

I have tried out to prove it by its contrapositive.

Therefore now I am proving that there is no integer whose square is congruent to $2 \pmod 3$.

Consider the case $n^2 \equiv 2 \pmod 3$

$ n^2 = 3k +2$

I am stuck to show $\sqrt{3k+2}$ is not a integer. Can anyone provide some hint to me?

Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. — N. F. Taussig, Mar 06 '20 at 10:44
See what happens when $n=3j$ or $n=3j+1$ or $n=3j+2$. Does one of those three cases have to happen? — robjohn, Mar 06 '20 at 10:46
@robjohn I have tried to substitute k=1,2,3 to familiar the situation more often. I think my problem is that I do not know how to show that form is not a integer... — Chak Wing Mak, Mar 06 '20 at 10:52
If $n=3j$, then $n^2=9j^2=3\left(3j^2\right)$. If $n=3j+1$, then $n^2=9j^2+6j+1=3\left(3j^2+2j\right)+1$. If $n=3j+2$, then $n^2=9j^2+12j+4=3\left(3j^2+4j+1\right)+1$. What other cases can there be? — robjohn, Mar 06 '20 at 10:56
so what it means is that every integer can be written as 3k, 3k+1 only? Sorry, I am confused... — Chak Wing Mak, Mar 06 '20 at 11:02
No. Every integer can be written as $3j$, $3j+1$, or $3j+2$. The squares of these are $3(\dots)$ or $3(\dots)+1$. — robjohn, Mar 06 '20 at 11:04
Got it. However, How can i structure the proof? do I need to prove every integer can be written as 3j, 3j+1, or 3j+2. The squares of these are 3(…) or 3(…)+1 — Chak Wing Mak, Mar 06 '20 at 11:08
We can simply use Euclid's Division Lemma, in this question. — Nikola Alfredi, Mar 06 '20 at 11:11
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score 1 · Answer 1 · answered Mar 06 '20 at 10:43

1

Observe: $[0]_3^2=[0\times 0]_3=[0]_3$ and $[\pm 1]^2_3=[(\pm 1)\times (\pm 1)]_3=[1]_3$; since $\{-1,0,1\}$ is a complete system of residues modulo three, all equivalence classes $[a]_3$ modulo three are accounted for, and none of which square to $[2]_3$.

answered Mar 06 '20 at 10:43

Shaun

47,747

Sorry, I don't understand.......can you explain it ? – Chak Wing Mak Mar 06 '20 at 10:45
2

What's the first thing about this answer that you don't understand, @ChakWingMak? – Shaun Mar 06 '20 at 10:46

score 1 · Answer 2 · answered Mar 06 '20 at 13:59

Fermat's little theorem says that $a^2\cong 1\pmod3$, if $a\not\cong0\pmod 3$.

But this can be done by hand. We can check that $0^2\cong0\pmod 3$, $1^2\cong1\pmod 3$, and $2^2\cong4\cong1 \pmod3$. There are only three cases to check.

For any integer $n$, $n^2$ is congruent to $0$ or $1$ modulo $3$

2 Answers2