Show that there are infinitely many primitive Pythagorean triples ${x,y,z}$ satisfying $z=x+1$. Then show that there are infinitely many primitive Pythagorean triples ${x,y,z}$ satisfying $z=y+2.$
For the first part I have that $x^2 + y^2 = (x+1)^2$ {$4,3,5$} satisfies this but I'm not sure how to prove that there are infinitely many.
For the second part I have $x^2 + y^2 = (y+2)^2$, again {$4,3,5$} satisfies.
On
You can also use that pythagorean triplets are of the form $(r^2-s^2,\ 2rs,\ r^2+s^2)$.
Note that $z$ is always $r^2+s^2$ because this is the greater number.
Since $r\ge s$ by definition, we get $r=s+1$ and find back Fatima's solution.
$\forall s : \begin{cases} x=2rs=2s(s+1)\\ y=r^2-s^2=(s+1)^2-s^2=2s+1 & odd\\ z=r^2+s^2=2s^2+2s+1 \end{cases}$
On
$(x,y,z)= (2n+1, 2n^2+2n, 2n^2+2n+1)$ is a primitive Pythagorean triplet with $z=y+1$ for any $n\in \mathbb N.$
$(x,y,z)=(4m', 4m'^2-1,4m'^2+1)$ is a primitive Pythagorean triplet with $z=y+2 $ for any $m'\in \mathbb N.$
Primitive triplets are all of the form $(m^2-n^2, 2mn,m^2+n^2)$ where $m,n$ are co-prime positive integers, not both odd, with $m>n.$ But we can change the order to $(2mn,m^2-n^2,m^2+n^2).$
For the first case we cannot have $1=(m^2+n^2)-(m^2-n^2)=2n^2$ but we can have $1=(m^2+n^2)-(2mn)=(m-n)^2$ if $m-n=1.$ So the solution in the first case is found by letting $m=n+1$ and $(x,y,z)=(m^2-n^2,2mn,m^2+n^2).$
For the second case we cannot have $2=(m^2+n^2)-(2mn)=(m-n)^2$ but we can have $2=(m^2+n^2)-(m^2-n^2)=2n^2$ if $n=1.$ If $n=1$ then m must be even so the solution in the second case is found by letting $m=2m'$ and $n=1$ and $(x,y,z)=(2mn,m^2-n^2,m^2+n^2).$
hint for the first
$$x^2+y^2=(x+1)^2=x^2+2x+1$$
$$\implies y^2-1=2x $$
take an odd $y$ and get $x=(y^2-1)/2$ and $z=x+1$.
for example,
$$y=3\implies x=8/2=4\implies z=5$$