<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] Given the odd number $x$,
$$x^2=2y^2-1$$
which is just a Pell equation, then we get the Pythagorean triple,
$$\Big(\frac{x-1}2\Big)^2+\Big(\frac{x+1}2\Big)^2=y^2$$
The difference $d$ between the addends, of course, is $d=1$. This yields,
$$3^2+4^2=5^2\\20^2+21^2=29^2\\119^2+120^2=169^2$$
and so on.

P.S. If interested, a nice infinite counterpart is $a^2+b^2=(b+1)^2$ with solution,
$$(2m+1)^2+(2m^2+2m)^2 = (2m^2+2m+1)^2$$
which yields,
$$3^2+4^2=5^2\\5^2+12^2=13^2\\7^2+24^2=25^2$$
ad infinitum.
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