# Formula for the the integers of all Pythagorean triples with $z = y + 2$

Find formulas for the integers of all Pythagorean types $x,y,z$ with $z = y + 2$

I know $z = m^2 + n^2$ and $y = 2mn$ so substituting I get...

$m^2 + n^2 = 2mn + 2 \Rightarrow (m-n)^2 = 2$

which now I feel I'm stuck, cuz im going to be carrying around a $\sqrt{2}$ and I need integer solutions.

I must be making some kind of silly mistake, but I dont see it, would love some help, thanks

• How about $(6,8,10)$? May 3, 2018 at 17:40
• Or $(4,3,5){}$? May 3, 2018 at 17:46

$$z^2=x^2+y^2=(y+2)^2$$ $$=y^2+4 (1+y)$$

$$\implies x^2=2^2 (1+y)$$

we need $1+y=t^2$

thus $$(x,y,z)=(2t,t^2-1,t^2+1)$$

for $t=10,$ it gives $$(x,y,z)=(20,99,101)$$

There are two things you have to bear in mind: first of all, the triples of the form $$(m^2 - n^2, 2mn, m^2 + n^2)$$ where $m > n$ are coprime integers of opposite parity, give all the primitive triples. If you want a general formula for all triples, you need to write $$k \cdot (m^2 - n^2, 2mn, m^2 + n^2)$$ where $k \in \mathbb{Z}$ and for $m, n$ the same conditions as before.

Secondly, it is not necessarily the case that $y = 2mnk$. You also have to consider the case where $y = k(m^2 - n^2)$. Combining these two, you should be able to fix your solution.

This is not an answer but a clarification of where the OP went wrong. I like the answer given by Salahamam_ Fatima. Instead of involving the variables $m$ and $n$, and attempting to first identify all satisfying primitive triplets and then (somehow) attempting to identify all satisfying triplets (primitive or non-primitive), the problem probably intended for the OP to realize that

$E_1:\;\;x^2 = (z-y)(z+y) = 2(2y + 2).$

Once $E_1$ is gleaned, the problem drastically simplifies.

These functions will generate only and all Pythagorean triples where $$C-B=2:$$ $$A=4n\qquad B=4n^2-1\qquad C=4n^2+1$$ The first $$6$$ of them are: $$4,3,5\quad 8,15,17\quad 12,35,37\quad 16,63,65\quad 20,99,101\quad 24,143,145$$