<s>[INST] prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$ Here is my question: Prove that there are infinitely many primitive pythagorean triples $x,y,z$  such that $y=x+1$.
They gave me a hint that I should consider the triple $3x+2z+1, 3x+2z+2, 4x+3z+2$, but I honestly do not know how to start. 
Do I need contradiction?
 [/INST] Start with the hint . . .

Let $a,b,c$ be given by
\begin{align*}
a &= 3x+2z+1\\[4pt]
b &= 3x+2z+2\\[4pt]
c &= 4x+3z+2
\end{align*}
where $x,z$ are unknown positive integers.

If we were lucky, the equation $a^2 + b^2 = c^2$ would hold identically, for all $x,z$. 

Let's try . . .
\begin{align*}
a^2 + b^2 - c^2 &= (3x+2z+1)^2 + (3x+2z+2)^2 - (4x+3z+2)^2\\[4pt]
&= 2x^2 + 2x + 1 - z^2\qquad\text{[by expanding and then combining like terms]}\\[4pt]
\end{align*}
So unfortunately, we were not that lucky, since $a^2 + b^2 - c^2$ simplifies to $2x^2 + 2x + 1 - z^2$, which is not identically zero. 

But what if it was the case that $2x^2 + 2x + 1 - z^2 = 0\,$ for some $x,z$? 

With that idea, let's look more closely at the equation $2x^2 + 2x + 1 - z^2 = 0$, to see if we can find a way to make it happen.

The key observation is 
\begin{align*}
&2x^2 + 2x + 1 - z^2 = 0\\[4pt]
\iff\; &2x^2 + 2x + 1 = z^2\\[4pt]
\iff\; &x^2 + (x^2 + 2x + 1) = z^2\\[4pt]
\iff\; &x^2 + (x+1)^2 = z^2\\[4pt]
\end{align*}
Given that all transitions above are of the form "if and only if", it follows that

$$(x,x+1,z)$$
$$\text{is a pythagorean triple}$$
$$\text{if and only if}$$
$$(3x+2z+1,3x+2z+2,4x+3z+2)$$
$$\text{is a pythagorean triple}$$

Thus, starting with any pythagorean triple of the form $(x,x+1,z)$ we can get a new one which is strictly larger (e.g., larger perimeter).

Note also that any pythagorean triple of the form $(x,x+1,z)$ is automatically a primitive pythagorean triple since $\gcd(x,x+1) = 1$.

Thus, all we need is one such triple to get started. Fortunately, that's easy$\,{\large{-}}\,$just start with the triple $(3,4,5)$. 

So we weren't so unlucky after all!
</s>