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 $x,y,z$?
 A: 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 is actually quite common for
natural Diophantine equations to give rise to K3 surfaces of
maximal or nearly-maximal Picard number, but that's a story for
another time.
