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?

You may just ignore the hint and prove that $x^2+(x+1)^2=y^2$ has an infinite number of solutions in $\mathbb{N}\times\mathbb{N}$. Such identity is equivalent to
$$ (2x+1)^2-2 y^2 = -1 $$
hence it is enough to show that there are infinite natural solutions to $a^2-2b^2=-1$ with $a$ being odd. But, wait, if $a,b\in\mathbb{Z}$ and $a^2-2b^2=-1$ then $a$ has to be odd. And $a^2-2b^2$ is just the norm over $\mathbb{Z}[\sqrt{2}]$, where $3\pm2\sqrt{2}$ is an invertible element (unit norm). It follows that by raising $(3+2\sqrt{2})$ to some power we still get an invertible element, and a solution of $a^2-2b^2=1$. Since the norm of $1+\sqrt{2}$ is $-1$, in such a case $(a+2b)^2-2(a+b)^2=-1$ and we get a solution of $x^2+(x+1)^2=y^2$. For instance,


*

*$(3+2\sqrt{2})^3 = 99+70\sqrt{2}$

*$99^2-2\cdot 70^2 = 1$

*$239^2-2\cdot 169^2 = -1$

*$119^2+120^2=169^2$


or


*

*$(3+2\sqrt{2})^4 = 577+408\sqrt{2}$

*$577^2-2\cdot 408^2 = 1$

*$1393^2-2\cdot 985^2 = -1$

*$696^2+697^2=985^2$


In particular, if we define the Pell number $P_n$ as $\frac{(1+\sqrt{2})^n-(1-\sqrt{2})^n}{2\sqrt{2}}$, we have that

$$ \left(\frac{P_{2n-1}+3 P_{2n}-1}{2}\right)^2+\left(\frac{P_{2n-1}+3 P_{2n}+1}{2}\right)^2 = \left(P_{2n-1}+2\,P_{2n}\right)^2.$$

Such well-known identity can be simplified a bit.
