Show that there are infinitely many primitive Pythagorean triples satisfying $z=x+1$

Question

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.

Answer 1

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,

$$y=3\implies x=8/2=4\implies z=5$$

Answer 2

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.

• $x=r^2-s^2$ gives $r^2+s^2=r^2-s^2+1\iff 2s^2=1$ not possible.
• $x=2rs$ gives $r^2+s^2=2rs+1\iff (r-s)^2=1\iff r=s\pm 1$

Since $r\ge s$ by definition, we get $r=s+1$ and find back Fatima's solution.

$\forall s : \begin{cases} x=2rs=2s(s+1)\\ y=r^2-s^2=(s+1)^2-s^2=2s+1 & odd\\ z=r^2+s^2=2s^2+2s+1 \end{cases}$

Answer 3

$(x,y,z)= (2n+1, 2n^2+2n, 2n^2+2n+1)$ 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 $(x,y,z)=(2mn,m^2-n^2,m^2+n^2).$

Answer 4

A variation of Euclid's formula generates a subset of Pythagorean triples where $GCD(X,Y,Z)$ is an odd square. This includes all primitives.

$$X=2(2m-1+n)n\quad Y=(2m-1+n)^2-n^2\quad Z=(2m-1+n)^2+n^2$$

We can expand terms and replace $m,n$ by $n,k$. Now we let $X^2+Y^2=Z^2$ where

\begin{align*} &X=\phantom{(2n-1)^2+{}} 2(2n-1)k+2k^2\\ &Y=(2n-1)^2+2(2n-1)k\\ &Z=(2n-1)^2+2(2n-1)k+2k^2 \end{align*} If we now let $\,n-1,\,$ we get

$$X=2k+2k^2\qquad Y=1+2k\qquad Z=1+2k+2k^2$$