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(For the case of solutions (a,b,c) to $a^2+b^2 = x$, $b^2+c^2 = y$, see Number of x such that $m-x^2$ and $n-x^2$ are both squares)

I'm interested in how many solutions $(a,b,c,d)$ where $a,b,c,d$ are all nonnegative integers there are to the set of equations $a^2+b^2 = x, b^2+c^2 = y, c^2+d^2 = z$ for given $x,y,z$ such that $x \neq y$ and $y \neq z$. In particular, is the number of solutions bounded above by a constant independent of $x,y,z?$

My previous question was done by using the positive rank elliptic curve $a^2+b^2 = 1370, b^2+c^2 = 2210.$ I'm under the impression that this cannot be done here, because of the Mordell conjecture, though I don't know the Mordell conjecture well enough to say for sure.

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    $\begingroup$ Those equations define a curve of genus $5$. Caporaso, Harris, and Mazur proved that for any $g>1$ a uniform (albeit ineffective) bound on the number of rational points of a genus-$g$ curve would follow from the Bombieri-Lang conjectures on rational points on varieties of general type. (The question at hand asks for a uniform bound on the number of integral points, but that's equivalent by scaling.) But even for genus-$5$ curves of this restricted form there's likely no technique available at present to prove this uniform boundedness. $\endgroup$
    – Noam D. Elkies
    Commented May 7, 2012 at 5:35

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Certainly there can't be a bound independent of $x,y,z$. For example if $x=y=z = u^2 + v^2$, you can take $a = c = u$ and $b = d = v$. If you choose $x$ to be a product of many primes $\equiv 1$ mod $4$, you can have as many solutions as you like.

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  • $\begingroup$ Sorry, I forgot to write that I wanted $x \neq y$ and $y \neq z.$ $\endgroup$
    – David
    Commented May 7, 2012 at 4:32

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