# Primitive integral solutions to $x^2+y^3=z^2$

The Diophantine equation $$x^2+y^3=z^2$$ has solutions $$(\pm 1,2,\pm 3)$$ and $$(\pm 13,3,\pm 14)$$.

Brown [Int. Math. Res. Not. IMRN 2012, no. 2, 423–436; MR2876388] states that "there are infinitely many parameterized solutions with $$xyz \neq 0$$ and $$z^2 \neq 1$$ when $$n \leq 5$$" to the equation $$x^2 +y^3 = z^n,$$ but does not give a reference. I am looking for a reference that catalogs all primitive solutions.

• You can treat this as a special case of factoring, e.g. $Y=(z+x)(z-x)$, where you can have one of the factors (say $z-x$) be $y$ to a power. If you want everything integral, make sure the factors have the same parity. Gerhard "Difference Of Squares Is Easy" Paseman, 2017.03.14. – Gerhard Paseman Mar 15 '17 at 2:33
• The last, and by far the hardest, case is $n=5$, which is solved in: J. Edwards: A Complete Solution to $X^2 + Y^3 + Z^5 = 0$, Journal f. d. reine und angew. Math. (Crelle's Journal) 571 (2004), 213-236. (Change $(X,Y,Z)$ to $(x,y,-z)$ for your form of the equation.) That paper should also give references for $n<5$. – Noam D. Elkies Mar 15 '17 at 3:21
• @NoamD.Elkies: there are references but they concern the cases when $3 \leq n \leq 5$. – Pietro Paparella Mar 15 '17 at 3:57
• The parameterization equation $x^2+y^3=z^2$ can record such. artofproblemsolving.com/community/c3046h1217955_the_cube_view For $x^2+y^3=z^4$ artofproblemsolving.com/community/… As already said - it is better for each individual degree to solve directly the equation. – individ Mar 15 '17 at 4:44

All solutions of the Diophantine equation $x^2-y^2=z^3$ can be found in the book "Number Theory. Volume II: Analytic and Modern Tools" (Springer Science, 2007) by Henri Cohen. Below the relevant page from this book is reproduced.

As for general case of the Generalized Fermat equation, see https://www.staff.science.uu.nl/~beuke106/Fermatlectures.pdf The generalized Fermat equation, by Frits Beukers), http://homepages.warwick.ac.uk/~maseap/papers/bealconj.pdf (The Generalized Fermat Equation, by Michael Bennett, Preda Mihailescu and Samir Siksek) and http://people.math.sfu.ca/~ichen/pub/BeChDaYa.pdf (Generalized Fermat equation: a miscellany, by M.A. Bennett et al.).

A warm-up:

$$\left(\frac{s^n-t^n}2\right)^2 +\ (s\cdot t)^n\ =\ \left(\frac{s^n+t^n}2\right)^2$$

provides integer solutions for $\,\ s\equiv t \mod 2$.

(Sorry, I couldn't help it).

REMARK   More generally, going in the abc direction:

$$\left(\frac{s^m-t^n}2\right)^2 +\ s^m\cdot t^n\ =\ \left(\frac{s^m+t^n}2\right)^2$$

for $\ s\equiv t\equiv 1 \mod 2,\$ and $\ \gcd(s\ t) = 1$.

• In the remark, should it be $s \equiv t \equiv 1 \bmod 2$? – Pietro Paparella Mar 15 '17 at 16:46
• @PietroPaparella, yes, mod 2--thank you, I'll fix it. – Włodzimierz Holsztyński Mar 15 '17 at 17:11