Integer solutions of $ x^4+34x^2 y^2+y^4=z^2$

Question

Now I'm trying to solve one problem, and already twice an expression of the form $a x^4+b x^2 y^2+a y^4 \,\,(x > y > 1)$ appears under the square root.

First was: $9 x^4 - 14 x^2 y^2 + 9 y^4$

Second: $x^4 + 34 x^2 y^2 + y^4$

I suppose that there will be other equations of this type.

Question: Are there integer solutions or a parametrization of equations$ a x^4+b x^2 y^2+a y^4=z^2$ in general, or of the variants proposed above in particular? Or maybe can the absence of solutions be proven?

Upd: No solutions

Second case: according to https://www.cambridge.org/core/services/aop-cambridge-core/content/view/0B14DEDE386776126F3B5A36CA325ECF/S0017089500007862a.pdf/x4-dx2y2-y2-z2-some-cases-with-only-trivial-solutions-and-a-solution-euler-missed.pdf (Thanks Will Jagy for link)

First case: $9 x^4 - 14 x^2 y^2 + 9 y^4$ can be written as $(3(x^2-y^2))^2 +(2xy)^2=z^2$ Pythagorean triple. So $2xy=2ab, 3(x^2-y^2)=a^2-b^2$

From this: $3y^4+(a^2-b^2)y^2-3a^2b^2=0$ can be obtain. Discriminant $a^4+34a^2b^2+b^4$ (Second equation of my question. Strange coincidence). So there is no rational roots.

Answer 1

At first, it looks like the equation can be rewritten as a Pythagorean triple $\space A^2+B^2=C^2\quad$ here shown by Euclid's formula: $\quad A=(m^2-k^2) \quad B=2mk\quad C=(m^2+k)^2\space $ and a triple may be found given any legitimate value of $A,\space B, \space \text{or}\space C.$

$\text{Here we have}\quad(x^2-y^2)^2 + (6xy)^2 =z^2\space$ and $\space B= 6xy\ne 2mk\space$ so it is not a Pythagorean triple because it cannot quite be a right triangle. The best approach may be experimentation and observation.

\begin{align*} x=0,\space y=0 &\implies z\in\mathbb{Z}^2 \\ x=y,\space\space \space x,y>0 &\implies z=6n^2\space ,n\in\mathbb{N}\\ \not\exists \space x>y>1 \\ x=0,\space y=4 &\implies z=\pm8 \\ x=4,\space y=0 &\implies z=\pm8 \end{align*}

Answer 2

Your equations are examples of Elliptic Curves. More precisely you can dehomogenise the LHS of your equation so that you are looking for rational points on the (affine) genus $1$ curves $$ C_1 : z^2 = 9x^4 - 14x^2 + 9$$ $$C_2 : z^2 = x^4 + 34x^2 + 1$$

Now if you look for a few seconds you'll realise there are the points $(0,3)$ and $(0,1)$ on $C_1$ and $C_2$ respectively -- i.e., $C_1$ and $C_2$ are elliptic curves, just not in Weierstrass form. To resolve the rational points we put them in Weierstrass form and note that they are both isomorphic to the elliptic curves in the LMFDB $48a3$.

This curve has precisely $8$ rational points. On your original models, these correspond to the points $(x : z : y)$ (note the coordinate rearrangement) $$\{ (1 : -3 : 0), (1 : 3 : 0), (-1 : -2 : 1), (-1 : 2 : 1), (0 : -3 : 1), (0 : 3 : 1), (1 : -2 : 1), (1 : 2 : 1) \}$$ and $$\{ (1 : -1 : 0), (1 : 1 : 0), (-1 : -6 : 1), (-1 : 6 : 1), (0 : -1 : 1), (0 : 1 : 1), (1 : -6 : 1), (1 : 6 : 1) \}.$$

Here the colon means that these are the only solutions up to rescaling of the form $(ux, u^2z, uy)$.