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I would like to know how to prove the above statement by contradition. Somebody said that one should prove it by this method but I have no idea what it is.

BIS HD
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Tessa
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  • first thing you should be aware is that $3$ is a prime –  Nov 12 '13 at 11:52
  • That doesn't really help... – Dennis Gulko Nov 12 '13 at 11:53
  • A square is either divisible by three, or it is one larger than a number divisible by $3$. It is never one smaller (i.e. two larger) than such a number. – Arthur Nov 12 '13 at 11:53
  • To prove it by contradiction means that you should see what happens if it's not true that "$3|a$ and $3|b$" given that $3|a^2+b^2$. (i.e., assume one or both of $a$ and $b$ is not divisible by $3$). If you can show that this implies something that you know is false, then your assumption was wrong and it must be the case that $3|a$ and $3|b$. – Casteels Nov 12 '13 at 11:58
  • do i have to prove it case by case? Like proving the first a/3 and then b/3 ? and finally got the results of yes? – Tessa Nov 12 '13 at 13:17

4 Answers4

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For any integer, there are precisely three options for its reminder when divided by $3$, i.e. $0,1,2$. So suppose that $x_i=3k+i$ for $i=0,1,2$. Then $$x_i^2=9k^2+6k+i^2=\begin{cases}3t & i=0\\ 3t+1& i=1,i=2\end{cases}$$ So, for any integer $x$, the reminder of $x^2$ when divided by $3$ is either $0$ or $1$. Now assume, by the way of contradiction, that $a$ or $b$ are not divisible by $3$. Then what can you say about $a^2+b^2$?

Dennis Gulko
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Assume without loss of generality $3\not\mid a$. Then $a^{2}\equiv 1\pmod{3}$. Hence, $a^2+b^2\equiv 1\pmod{3}$ or $a^2+b^2\equiv 2\pmod{3}$. In any case, $3\not\mid (a^2+b^2)$. This proves the contrapositive.

Prism
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  • The contrapositive is $ \left(3\not\mid a \lor 3 \not \mid b \right) \implies 3\not\mid (a^2+b^2)$ , which is equivalent to $ \left(3 \not \mid a \implies 3\not\mid (a^2+b^2)\right) \land \left(3 \not \mid b \implies 3\not\mid (a^2+b^2)\right)$ It seems you have only proven the first statement. Am I wrong in thinking you should next consider the case $ 3 \not \mid b $. – john Sep 30 '19 at 09:43
  • @john: You are right. The situation for the case $3\nmid b$ is exactly the same, because the roles of $a$ and $b$ are symmetric :) That's why I started by saying "without loss of generality $3\nmid a$", which signals that I am considering only one of the cases, while acknowledging that the other case is exactly the same (so we don't bother copying it down). Such use of "without loss of generality" is common. – Prism Sep 30 '19 at 15:17
  • Ah I see. Thanks for the reply! – john Oct 01 '19 at 00:00
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HINT:

If $3\not|a, a\equiv\pm1\pmod 3\implies a^2\equiv1\pmod 3$

and $a^2\equiv0\pmod 3\iff a\equiv0$

lab bhattacharjee
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suppose that a=3k+r , b=3q+r' r,r'=0,1,2

if a=3k , b=3q its ok

if not r=1,2 a=3k+1 or 3k+2

so a^2=9k^2+6k+1=3k'+1 ,or 9k^2+12k +4 =3k'+1

and for b like that

so a^2 mod 3=1 and b^2 mod3 = 1 and it is impossible

then the only condition that's possible is a=3k , b=3q

proof complete

Khosrotash
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