Primitive Pythagorean triple

[alexmahone]

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.

 

[kaliprasad]

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.

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

$c^2=a^2+b^2\equiv 0+1\equiv 1 \pmod{3}$
you are right that $c^2 = 1 \pmod{3}$ does not mean that $c^2$ is perfect square but in the above you have shown that for a Pythagorean triplet above condition must be true. it is one way and not both ways