text: Q:
  
  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
  
  A:
  
  $$z^2=x^2+y^2=(y+2)^2$$
  $$=y^2+4 (1+y) $$
  $$\implies x^2=2^2 (1+y)$$
  we need $1+y=t^2$
  thus $$(x,y,z)=(2t,t^2-1,t^2+1) $$
  for $t=10,$ it gives
  $$(x,y,z)=(20,99,101) $$
  
  A:
  
  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. 
  
  