Given natural numbers a and b such that a^2+b^2 is divisible by 21, prove that the same sum of squares is also divisible by 441.

Question

Tried factoring, but I am not getting any ideas.

See https://math.stackexchange.com/questions/122048/1-is-a-quadratic-residue-modulo-p-if-and-only-if-p-equiv-1-pmod4 — lab bhattacharjee, Jul 05 '17 at 08:12
Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title. — Simply Beautiful Art, Dec 14 '17 at 17:53

score 3 · Accepted Answer · edited Jul 05 '17 at 08:55

3

Check out $a^2+b^2$ $\mod 3$ and $\mod 7$.

For any integer $a$ : $a^2\mod3$ is either $0$ or $1$.

for any integer $a$ : $a^2\mod7$ could be $0$, $1$, $2$ or $4$.

So, if $a^2+b^2$ is divisible by $21$, both $a$ and $b$ are.

edited Jul 05 '17 at 08:55

Especially Lime

46,692

answered Jul 05 '17 at 08:24

dEmigOd

3,390

Then, how is it divisible by 441? – Raahil Rai Jul 05 '17 at 08:27
chinese remainder theorem @RaahilRai – Saketh Malyala Jul 05 '17 at 08:28
Oh! Yess. Thanks – Raahil Rai Jul 05 '17 at 08:29
More simply, if $a = 21m, b=21n$, then $a^2 + b^2 = 21^2m^2 + 21^2n^2 = 441(m^2 + n^2)$. – Deepak Jul 05 '17 at 08:29

score 0 · Answer 2 · answered Jul 05 '17 at 08:33

0

HINT: $$a^2+b^2$$ is only divisible by $21$ if $a^2$ and $b^2$ is divisible by $21$ this means that $a$ and $b$ is divisible by $21$ but then must $a^2$ and $b^2$ is divisible by $21^2=441$

answered Jul 05 '17 at 08:33

Dr. Sonnhard Graubner

97,058

Given natural numbers a and b such that a^2+b^2 is divisible by 21, prove that the same sum of squares is also divisible by 441.

2 Answers2