If $3 \mid a^2+b^2$, then show that $9 \mid ab$.

Question

If $3 \mid a^2+b^2$, then show that $9 \mid ab$.

If $3$ divides $a^2+b^2$ I have that $a^2+b^2 \equiv 0 \pmod{3}$. Looking at the squares it seems that starting from $9$ for every third I’ll get a remainder $0$. Also squares mod $3$ are only $0$ or $1$ so I must have $a^2 \equiv 0 \pmod{3}$ and $b^2 \equiv 0 \pmod{3} $, but still how does this imply the result we’re after?

If $3\mid a^2$, then $3\mid a$; similarly $3\mid b$, so $9\mid ab$ — J. W. Tanner, Nov 01 '21 at 20:47
Short answer, $\forall ~n \in \Bbb{Z^+}$, either $n^2 \equiv 0\pmod{3}$ or $n^2 \equiv 1\pmod{3}$. From the constraints of the problem, you can not have either $a^2 \equiv 1\pmod{3}$ or $b^2 \equiv 1\pmod{3}$. — user2661923, Nov 01 '21 at 21:31

J. W. Tanner · Accepted Answer · 2021-11-01T21:00:14.680

0

$a^2\equiv0\pmod3$ means $3\mid a^2$, which implies $3\mid a$ (by Euclid's lemma) ;

similarly $3\mid b$, so $9\mid ab$.

edited Nov 01 '21 at 21:00

answered Nov 01 '21 at 20:52

J. W. Tanner

60,406

You may add it is Euclid's lemma since OP doesn't seem to know this property. – zwim Nov 01 '21 at 20:58
Ok, @zwim; I did – J. W. Tanner Nov 01 '21 at 21:00
Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Nov 01 '21 at 21:08

If $3 \mid a^2+b^2$, then show that $9 \mid ab$.

1 Answers1