# I have to solve this Diophantine equation , possibly through the use of Infinite Descents or going back to Pythagorean triples

Note that I am working with real integers and I should be possibly using infinite descents and modular arithmetics.

Given the following diophantine equation $x+y^{2}-z^{2}$.

1.Show that for every positive odd integer x there exist integers y and z such that $(x,y,z)$ is a solution 2.For which positive even integers x do there exist integers $y$ and $z$ such that $(x,y,z)$ is a solution.

1. factorizing i have $x =(y-z)(y+z)$ and since x is odd we need both $(y-z)$ and $(y+z)$ odd. Suppose $(y+z)=1$ and $(y-z)=2n+1$ then $x = (2n + 1)*1 = (n + n + 1)*(n + 1 – n) = (n + 1) 2 - n2$. Is that proved for all? not sure 2.for the even case I am stuck. I tried take mod 4 both parts and therefore I know that my even x must be a multiple of 4 but then I cannot go on.

If someone can help would be appreciated.

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A very similar question was asked a few hours ago. Perhaps you will find the answers useful. Pythagorean triples are not relevant, and descent is not the best way to proceed. –  André Nicolas Feb 1 '13 at 0:15
Note that $1.$ and $2.$ do not ask for a general solution of the equation. If you are looking for the general solution, then factorization as in your attempt is a good idea. But it may be that the only way to get someone to write out the general solution is to specifically ask for it. –  André Nicolas Feb 1 '13 at 0:25