# All integer solutions to diophantine equation: $x^2+p y^2=z^2$?

I would like to find all integer solutions to the Diophantine equation $$x^2+p y^2=z^2$$ where $p\ge2$ is a given prime number. Also prove that my (probably parametric similar to Pythagorean triples $x=a^2-b^2,\ y=2ab,\ z=a^2+b^2$) solution gives all of the.

I tried to move $x^2$ to the right hand side and write $py^2=(z-x)(z+x)$. then assume two cases: (1) $\ \ p|z-x$ , (2) $\ p|z+x$. then I wrote some conditional relation depending on $y$. but I am not satisfied with that. Any help would be appreciated?

• What is your conditional relation on $y$ that you are not satisfied with? And what is your propably parametric solution, and why are you not sure whether it is parametric or not? Commented Aug 1, 2015 at 11:17

This is no different from the Pythagorean case. To classify integer solutions, look instead for rational solutions of $$x^2+py^2=1$$ Thus we want rational points on an ellipse. One solution is $P =(-1,0)$. Stereographically project from that point (that is, given a point $(0,h)$ we extend the line connecting it to P and find its intersection with the ellipse). Elementatry algebra shows that that the point on the ellipse we find this way is rational if and only if h is rational. Explicitly, we get the solution $$\left(\frac{1-ph^2}{1+ph^2},\frac{2h}{1+ph^2}\right)$$ Clearing denominators and letting $h=\frac ab$ we get the general solution $$\left(b^2-pa^2,2ab,b^2+pa^2\right)$$ By construction, this gives all rational solutions up to an integer multiplier. As a commenter( WillJagy) correctly remarked, this method can generate solutions with common factors which must then be canceled to obtain primitive solutions.

• stereographic projection does give all rational solutions, but you still need to check the possible gcd. When $p=5,$ your recipe does not give $2^2 + 5 \cdot 1^2 = 3^2.$ The recipe does give the imprimitive $(-4,2,6)$ with $a=1,b=1$ so $\gcd(a,b) = 1.$ Commented Aug 1, 2015 at 19:35
• @WillJagy You are absolutely correct. I will edit accordingly.
– lulu
Commented Aug 1, 2015 at 19:59

$$py^2=z^2-x^2=(z+x)(z-x)$$ $p$ is an odd prime, then it divides one of the factors. Then,WLOG: $$\dfrac{z+x}{p}=g^{2-t}a^2$$ $$z-x=g^tb^2$$ This yields $$\pm z=\dfrac{pg^{2-t}a^2+g^tb^2}{2}$$ $$\pm x=\dfrac{pg^{2-t}a^2-g^tb^2}{2}$$ $$\pm y=gab$$ Where $a,b,g$ are all integers.

The term $$\,y=2ab\,$$ cannot be correct because $$\,y\,$$ must be odd as shown below.

\begin{align*} x^2+py^2&=z^2\\ \implies py^2&=z^2-x^2\\ \implies y^2&=\dfrac{(z-x)(z-x)}{p}\\ \implies &\space p\vert (z-x)\lor p\vert (z+x) \end{align*}

If $$\,z\,$$ and $$\,x\,$$ are both odd, their sums and differences are both even. Since $$\,p\,$$ must be an odd prime number, then $$\,y\,$$ must be odd and one of $$\,z\,$$ or $$\,x\,$$ must be even if $$\,p\,$$ is to divide the sum or difference of $$\,z\,$$ and $$\,x$$.

e.g. $$\,6^2+5\cdot3^2=9^2,\,$$ i.e. $$\,5\vert(9+6).$$