I want to solve $24x^4-5y^4=z^2$ in integers not all zero, and to fix ideas, I want to find them such that x,y are coprime.
I've tried plugging in small values of $x$ and $y$ and they don't return a square value for $24x^4-5y^4$. So I guess that no (nontrivial) solutions exist.
How would I prove this? I can't seem to derive a contradiction in $\mathbb{R}$ nor in $\mathbb{Q}_p$. So I guess that this equation violates Hasse's local-global principle. That is, I have to derive a contradiction by some other method than arguing in $\mathbb{R}$ or mod $p$ for some prime $p$.
The only example I know of a method for solving an equation that violates Hasse's principle is for the example $2a^2+17b^4=c^4$. I shall now describe the method, and my question is, does this method generalize to my equation $24x^4-5y^4=z^2$? (And if not, how else would I go about solving it?)
Here is the method I know for showing that no integers solutions exist to $2a^2+17b^4=c^4$, where $a,b,c$ are not all zero, and $(b,c)=1$. For any prime $p$ dividing $a$, since $17b^4 \equiv c^4$ modulo $p$, we get that $17$ is a quadratic residue mod $p$. By quadratic reciprocity, we get that $p$ is a quadratic residue mod $17$, where $p$ was any prime factor of $a$.
This implies that $a$ is a quadratic residue mod $17$. Write $a \equiv x^2$ mod $17$. Then $2x^4 \equiv c^4$ mod $17$, and then $2$ is a fourth power mod $17$. This is a contradiction.
Best Answer
Here's a rudimentary proof. Suppose $z^2=24x^4-5y^4$. Then looking modulo $4$ gives \begin{align} z^2 &= 24x^4-5y^4\\ &\equiv -y^4\mod 4 \end{align}
Since the only squares modulo $4$ are $0$ and $1$, this implies $z^2\equiv y^4\equiv 0\mod 4$, so $2|z,y$. Then setting $\overline{y}=\frac{y}{2}$ and $\overline{z}=\frac{z}{2}$, our original equation becomes \begin{align} 4\overline{z}^2 &= 24x^4-16\cdot5\overline{y}^4\\ \overline{z}^2 &= 6x^4-4\cdot5\overline{y}^4\\ \end{align} This then immediately implies that $2|\overline{z}$, so we can again set $w=\frac{\overline{z}}{2}$ and find \begin{align} 4w^2 &= 6x^4-4\cdot5\overline{y}^4\\ 2w^2 &= 3x^4-2\cdot5\overline{y}^4\\ \end{align} Now this gives $2|x$, so we finally set $\overline{x}=\frac{x}{2}$ and get \begin{align} 2w^2 &= 16\cdot3\overline{x}^4-2\cdot5\overline{y}^4\\ w^2 &= 24\overline{x}^4-5\overline{y}^4. \end{align} Thus we get a new solution with every term strictly smaller, so by infinite descent no solutions can exist.