\[(x,y,z)\]
us a Pythagorean triple with \[x^2+y^2=z^2\]
then at least one of \[x, \; y, \; z\]
must be divisible by 3, at least one must be divisible by 4 and at least one must be divisible by 5.We can show this by working modulus 3, 4 and 5 respectively. The only quadratic residues of 3 and 4 are 0 and 1, so that
\[z^2 \equiv a \; (mod 3 \; or \; 4)\]
only has a solution if \[a \equiv 0 \; or \; 1 \; (mod \; 3 \; or \; 4)\]
. Suppose that noe of \[x, \; y, \; z\]
are divisible by 3 or 4, then \[x^2, \; y^2, \; z^2 \equiv 1 \; (mod \; 3 \; or \; 4)\]
and \[x^2+y^2 \equiv 2 \equiv z^2 \; (mod \; 3 \; or \; 4)\]
which is impossible since 2 is not a quadratic residue of 3 or 4.The quadratic residues of 5 are 0, 1 and 4 and we must have two quadratic residues add to give a quadratic residue. Adding any two of these gives 0, 1, 2, 3. 2 and 3 are not quadratic residues of 5, so we can only add to get 0 or 1 and can only do this as
\[0+0, \equiv 0, \; 0+1=1+0 \equiv 1, \; 1+4=4+1 \equiv 0 \; (mod \; 5) \]
. At least one of term in each sum must be zero modulus 5 so at least one is divisible by 5.