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What is known about Diophantine equations of the form $z^2=x^4+m^2 y^4$ for various values of $m$? Of course we can use infinite descent to show that when $m=1$ there are no solutions. Also when $m=2$ there are no nontrivial solutions, since this would imply that there exists solutions to $a^2=b^4-c^4$. However when $m=3$ there are infinitely many solutions. For example $(x,y,z)=(2,1,5)$ is a primitive solution from which infinitely many non-primitive solutions can be produced. Is there any way to determine which values of $m$ allow for solutions?

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  • $\begingroup$ The elliptic curve of the related congruent number problem is here. $\endgroup$ – Tito Piezas III Jan 17 '18 at 3:20
  • $\begingroup$ ${{\left( {{s}^{4}}-24 {{h}^{2}} n\, {{s}^{2}}+16 {{h}^{4}}\, {{n}^{2}}\right) }^{4}}+{{h}^{2}} n\, {{s}^{2}}\, {{\left( {{s}^{2}}-4 {{h}^{2}} n\right) }^{2}}\, {{\left( 4 {{s}^{2}}+16 {{h}^{2}} n\right) }^{4}}$ $\endgroup$ – AlexSam Jan 17 '18 at 10:37
  • $\begingroup$ AlexSam's identity can be simplified as $$ (a^2 - 2 b^2)^4 + (a^2 - b^2)b^2\,(2a)^4 = (a^4 + 4 a^2 b^2 - 4 b^4)^2$$ and one can then easily solve $a^2-b^2=c^2$. $\endgroup$ – Tito Piezas III Mar 7 '18 at 12:36
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Standard methods show that the quartic $z^2=x^4+m^2y^4$ is birationally equivalent to the elliptic curve \begin{equation*} V^2=U^3-4m^2U = U(U-2m)(U+2m) \end{equation*} with \begin{equation*} \frac{x}{y}=\frac{V}{2U} \end{equation*}

This is the congruent number elliptic curve for triangle area $N=2m$. Thus the quartic has a non-trivial solution when $2m$ is a congruent number.

As an example, there will be a solution when $m=7$ since $14$ is a congruent number. The generator of the elliptic curve is $(U,V)=-(7/4,147/8)$ which gives $x=21$, $y=4$ and $z=455$.

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  • $\begingroup$ Maybe you also know the answer to this similar question. $\endgroup$ – Tito Piezas III Oct 15 '16 at 0:54

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