Let x,y,z be a primitive Pythagorean Triple with y even.
a)Prove that exactly one of x and y is divisible by 3. (Hint: Proof by contradiction)
b)Prove that exactly one of x and y is divisible by 4.
The square of a number not divisible by $\displaystyle 3$ is congruent to $\displaystyle 1$ modulo $\displaystyle 3$ (prove this). And no square is congurent to $\displaystyle 2$ modulo $\displaystyle 3$ (prove this also)
Suppose $\displaystyle x$ and $\displaystyle y$ are not divisible by $\displaystyle 3$, then the left hand side of:
$\displaystyle
x^2+y^2=z^2
$
is congruent to $\displaystyle 2$ modulo $\displaystyle 3$ while the right hand side cannot be congurent to $\displaystyle 2$ modulo $\displaystyle 3$, a contradiction, hence at least one of $\displaystyle x$ and $\displaystyle y$ is divisible by $\displaystyle 3$.
If both $\displaystyle x$ and $\displaystyle y$ are divisible $\displaystyle 3$ then the left hand side is divisible by $\displaystyle 3$ but the right hand side cannot be divisivle by $\displaystyle 3$ because $\displaystyle \{x,y,z\}$ is a primitive Pythagorean triple and so $\displaystyle x,y,x$ share no common factor. This is a contradiction, hence both $\displaystyle x$ and $\displaystyle y$ cannot be divisible by $\displaystyle 3$.
Hence exactly one of $\displaystyle x,y$ is divisible by $\displaystyle 3$.
(note we have not used the fact that $\displaystyle y$ is even)
CB