# Symmetric functions on three parameters being perfect squares

Is it possible for $x+y+z, xy+yz+zx$, and $xyz$ to be perfect squares at the same time for positive integer values of $x,y,z$?

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Yes. $(x,y,z) = (80,225,320)$ works, giving $25^2$, $340^2$, $2400^2$. There are infinitely many other (non-proportional) examples. – Noam D. Elkies Apr 17 '12 at 18:11
Thanks Noam. How do you find them? – Hej Apr 17 '12 at 19:19
You're welcome. The initial solution is found by exhaustive search (scaling from $(16,45,64)$, though this turns out not to be minimal). For further solutions and details, see the answer I just posted [which also explains why I'm editing to include the "k3-surfaces" tag]. – Noam D. Elkies Apr 18 '12 at 2:18

Yes. By straightforward search the smallest example is $\lbrace x,y,z \rbrace = \lbrace 45,64,180 \rbrace$, with $$(t+45) (t+64) (t+180) = t^3 + 17^2 t^2 + 150^2 t + 720^2.$$ Given any solution $(x,y,z)$ we may produce infinitely many others (other than the trivial scaling $(c^2 x, c^2 y, c^2 z)$) by using the theory of elliptic curves to find rational $z'$ such that $x+y+z'$, $xy+yz'+z'x$, and $xyz'$ are all squares, at which point $(d^2 x, d^2 y, d^2 z')$ works for any integer $d>0$ such that $d^2 z' \in {\bf Z}$. For example, in $\lbrace 45,64,180 \rbrace$ we may replace $64$ by $(460163992/28591599)^2$, and then multiply through by $28591599^2$ to obtain the new solution $$\lbrace 28591599^2 \cdot 45, \phantom{+} 460163992^2, \phantom{+} 28591599^2 \cdot 180 \rbrace .$$
A complete parametrization is not possible, because it would be tantamount to a rational parametrization of the surface $$S: xy + yz + zx = r^2, \phantom{and} (x+y+z)xyz = s^2$$ in projective $(1:1:1:1:2)$ space, and that surface is K3. If I did this right, $S$ is a "singular" K3 surface, i.e. has Picard number $20$ which is maximal for a K3 surface in characteristic zero, and the Néron-Severi group ${\rm NS}(S)$ has rank $20$ and discriminant $-48$, and consists of (classes of) divisors defined over ${\bf Q}(i)$.
It's mostly a surprisingly long collection of examples; I don't claim to have a structural "why". Heuristically what seems to happen is that — assuming the Diophantine equation is of the right dimension and complexity to yield a K3 surface in the first place — there's enough divisors coming from trivial solutions, and often enough symmetry, that there's barely enough room for the Néron-Severi lattice to accommodate them all under the constraint of rank at most $20$. – Noam D. Elkies Apr 18 '12 at 2:46