<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] $x^2 + y^2 = z^2$ and $y = x + 1 $   $\implies   (3x+2z+1)^2 + (3x+2z+2)^2 = (4x+3z+2)^2$
And this series goes on forever by $\cases{x := 3x+2z+1 \\
y := 3x+2z+2 \\
z := 4x+3z+2}$
$\ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \  $ (here $:=$ stands for assignment)
It is not very difficult to check that $\gcd(x,y,z) = 1$ at each step, thus ensuring the primitivity of the triplets.  
EDIT:
An example as to how a triplet is generated from the triplet in the previous step.
Suppose we start off from the triplet $3,4,5$ as indeed $4 = 3 + 1$.
Then the next triplet would be $(3 \times 3 + 2 \times 5 + 1, 3 \times 3 + 2 \times 5 + 2, 4 \times 3 + 3 \times 5 + 2) = (20,21,29)$.
And at the next step $(x,y,z) = (20,21,29)$.  
Hence the next triplet now would be $(3 \times 20 + 2 \times 29 + 1, 3 \times 20 + 2 \times 29 + 2, 4 \times 20 + 3 \times 29 + 2) = (119,120,169)$.
And at the next step $(x,y,z) = (119,120,169)$.
And this goes on, if you will, forever. 
</s>