Prove that if a prime $p$ divides $a^2 + b^2$ then it must divide both a and b.

Question

I proved this by taking cases for different remainders and using modular arithmetic, when the prime is $3$.However this method won't give a general proof for all primes.I was eager to get a proof for all primes $p$, but i couldn't get one.I have seen in Pythagorean triples that if both $a$ and $b$ are not divisible by $p$, $c=a^2 + b^2$ is also not, but how do i prove it?

Proving by contradiction may help...maybe induction. Also we can try proving for all numbers. of the form $6n+1$ and $6n-1$ so that it becomes evident for all primes.

The question changed and now is about proving that for all primes equivalent to $3 \bmod 4.$

It's not true for all primes, only for primes $\equiv 3 \pmod{4}$. — Daniel Fischer, Mar 30 '18 at 15:19
Hint. What are the possible values of squares modulo $3$? When might a sum of squares be congruent to $0 \pmod{3}$. — Ethan Bolker, Mar 30 '18 at 15:36
@Saturn: look up the theorem about what numbers can be expressed as a sum of two squares. — Ross Millikan, Mar 30 '18 at 15:36
Thanks for sharing solutions.I will see them surely but after i finish trying to find my own solution. — Saturn, Mar 30 '18 at 15:39
Um, what. So let's say $a = 2$ and $b = 7$ and $a^2 + b^2 = 53$ so $53|2$ and $53|7$???? Since when. And if $\gcd(a,b) = 1$ then $\gcd(a,a^2 + b^2) = 1$ and $\gcd(b, a^2 + b^2) = 1$ so if $p$ divides $a^2 + b^2$ then $p$ divides neither $a$ nor $b$. — fleablood, Mar 30 '18 at 16:04
Maybe you have confused the statement slightly, and this one was the original one? — Dietrich Burde, Mar 30 '18 at 16:04
Possible duplicate of show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$, Let $p$ be a prime so $p\equiv3\pmod4$. If $p|a^2+b^2$, then $p|a,b$ and perhaps If $p\equiv 3\pmod{4}$ and $p\mid x^2+y^2$, prove $p\mid x,y$. — Sil, Mar 30 '18 at 16:09

score 1 · Answer 1 · answered Mar 30 '18 at 16:20

Well it's only true if the prime you are taking about is $\equiv 3 \mod 4$ or $2$ and proof for that is here...

You can consider proving contrapositive i.e. if $p$ does not divide $a$ and $p$ does not divide $b$ then $p$ is not of the form $4m+3$ for integral $m$. Since $p$ is a prime and does not divide $a,b$ so we may assume that $(p,a)=(p,b)=1$.

Thus there exist $a'$ such that $aa' \equiv 1 (\mod p)$ as given $a^2\equiv -b^2 (\mod p)$ on multiplying both side of congruence by $a'a'$ we see that $1\equiv (aa')^2= -b^2(a')^2 (\mod p)$ so if $y =ba' $ then $y^2 \equiv -1(\mod p)$ which implies $p=2 $ or $p\equiv 1 (\mod 4)$.

If you use a \equiv \pmod{b}, it might look better. – Jaideep Khare Apr 12 '18 at 17:29 — Jaideep Khare, Apr 12 '18 at 17:29

Peter · Answer 2 · 2018-03-30T19:58:33.850

0

Suppose $$p\equiv 3\mod 4$$ is a prime and $$a^2+b^2\equiv 0\mod p$$ holds.

$b\equiv 0\mod p$ implies $a\equiv 0\mod p$

If $b\ne 0\mod p$, then multiplying with $(b^{-1})^2$ turns the equation into $$u^2+1\equiv 0\mod p$$ so $$u^2\equiv -1\mod p$$

Hence $-1$ is a quadratic residue mod $p$, which is impossible for a prime $p$ with $p\equiv 3\mod 4$.

edited Mar 30 '18 at 19:58

answered Mar 30 '18 at 15:41

Peter

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Why is it not possible that the quadratic residue mod p be -1 if p is equivalent to 3 mod 4? – Saturn Mar 30 '18 at 17:11
If we have $u^2\equiv -1\mod p$ for an odd prime $p$, obviously the order of $u$ modulo $p$ (the smallest positive integer $m$ with $u^m\equiv 1\mod p$) is $4$. Since we have $u^{p-1}\equiv 1\mod p$ because $p|u$ is obviously impossible because of $p|u^2+1$, $4$ must be a divisor of $p-1$, hence $p$ must have the form $4k+1$ – Peter Mar 30 '18 at 20:02

Prove that if a prime $p$ divides $a^2 + b^2$ then it must divide both a and b.

2 Answers2