3
$\begingroup$

Prove that for any primitive Pythagorean triple (a, b, c), exactly one of a and b must be a multiple of 3, and c cannot be a multiple of 3.

My attempt:

Let a and b be relatively prime positive integers.

If $a\equiv \pm1 \pmod{3}$ and $b\equiv \pm1 \pmod{3}$,

$c^2=a^2+b^2\equiv 1+1\equiv 2 \pmod{3}$

This is impossible as the only quadratic residues modulo 3 are 0 and 1.

So far, so good.

If one of a, b is $\equiv 0 \pmod{3}$ and the other is $\equiv \pm1 \pmod{3}$,

$c^2=a^2+b^2\equiv 0+1\equiv 1 \pmod{3}$

This is the part I don't understand. Just because $c^2\equiv 1\pmod{3}$ doesn't mean that $c^2$ must be a perfect square. For example, $a=12$ and $b=13$ satisfy the above conditions but $c^2=a^2+b^2=313$, which isn't a perfect square.

$\endgroup$
5
  • $\begingroup$ If $a,b\equiv 0$ (mod $3$) then you are done! $\endgroup$ – Sam Weatherhog Nov 10 '15 at 22:30
  • $\begingroup$ $c^2$ must be a perfect square for $(a,b,c)$ to be a Pythagorean triple, primitive or not. Pythagorean triples are specified to be triples of integers, after all, so $(12,13,\sqrt{313})$ is not valid. $\endgroup$ – Arthur Nov 10 '15 at 22:33
  • $\begingroup$ I would like to note this problem has to do entirely with the congruence of a,b, and c in mod 3, and the what any x value in Z is in mod 3. The answers given show this great. Similarly, a, b, or c must be a multiple of 5 in a PPT. $\endgroup$ – Nucl3ic Jan 18 '16 at 2:00
  • $\begingroup$ In the second part all you need to prove is that $c\not \equiv 0\pmod 3$. You have shown that $c^2\equiv 1\pmod 3$ so you conclude that $3$ does not divide $c$. $\endgroup$ – DanielWainfleet Sep 27 '17 at 19:14
  • $\begingroup$ See also: Proof: primitive pythagorean triple, a or b has to be divisible by 3. $\endgroup$ – Martin Sleziak Jul 10 '20 at 13:05
2
$\begingroup$

Your first step was a good start: It proves that at least one of $a$ and $b$ must be a multiple of $3$. We will have to eliminate the case of both $a$ and $b$ being multiples of $3$, but that is easy: If $a=3p$ and $b=3q$ then $$c^2 = a^2+b^2 = (3p)^2 + (3q)^2 = 9(p^2+q^2) = 9m$$ so $c$ is also a multiple of $3$, in which case the triplet is not primitive, as $3$ divides all three of them.

So exactly one of $a$ and $b$ must be a multiple of $3$.

Your second step has shown that since exactly one of $a$ and $b$ is a multiple of $3$, $c \equiv 1 \pmod{3}$. You don't care that not every number of the form $3k+1$ is a square; all you care about is that the number that happens to be $c^2$ must be expressible as $3k+1$. And this completes the proof that $c$ is not a multiple of $3$.

$\endgroup$
0
$\begingroup$

Your proof is complete. You are asked to show that one of $a,b,c$ is divisible by $3$. In the first part you show that $a$ and $b$ can't both be non-divisible by $3$. In the second part, you assume that one of $a,b$ is divisible by $3$ and show that $c^2\equiv 1$ (mod $3$) which implies that $c$ is not divisible by $3$ and hence exactly one of $a,b,c$ is divisible by $3$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.