Pythagorean triple

Quote:

Originally Posted by curiousmuch

This problem has got me stumped.

Let (x,y,z) be a primitive Pythagorean triple. Prove that one of the integers in the triple is divisible by 3, one is divisible by 4, and one is divisible by 5.

3,4,5 do not have to divide different elements of the triple.

I mean we know that by definition a primitive Pythagorean triple has to satisfy x^2+y^2=z^2. We also know that one leg is odd, the other is even, and that the hypotenuse is odd. Also, gcd(x,y,z)=1.

I tried isolating one variable and then trying to show it is divisible by 3,4,5, but am running into trouble.

Help would be awesome. Thanks.

Hi curiousmuch.

Let

$x=m^2-n^2$

$y=2mn$

$z=m^2+n^2$

$m$ and $n$ cannot be both odd, otherwise $x,y,z$ would all be even. Hence one of $m$ and $n$ must be even. Hence $y$ is always divisible by 4.

If one of $m$ and $n$ is divisible by 3, then $y$ would also be divisible by 3. Suppose neither $m$ nor $n$ is a multiple of 3. Then 3 does not divide $m^2$ or $n^2$ either. Now, either $m^2\equiv n^2\pmod3$ or $m^2\not\equiv n^2\pmod3.$ In the former case, $x=m^2-n^2$ is divisible by 3. In the latter case, we have $m^2\equiv\pm1\pmod3$ and $n^2\equiv\mp1\pmod3.$ $\therefore$ 3 divides $m^2+n^2=z.$

If one of $m$ and $n$ is divisible by 5, then $y$ would also be divisible by 5. Suppose 5 does not divide $m$ or $n.$ By Fermat’s little theorem, $m^4\equiv1\pmod5$ and $n^4\equiv1\pmod5.$ Hence 5 divides $m^4-n^4=(m^2-n^2)(m^2+n^2)=xz.$ As 5 is prime, this means 5 divides either $x$ or $z.$