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 $x,y,z$?
$\begingroup$
$\endgroup$
3
-
12$\begingroup$ Yes. $(x,y,z) = (80,225,320)$ works, giving $25^2$, $340^2$, $2400^2$. There are infinitely many other (non-proportional) examples. $\endgroup$– Noam D. ElkiesApr 17, 2012 at 18:11
-
$\begingroup$ Thanks Noam. How do you find them? $\endgroup$– HejApr 17, 2012 at 19:19
-
$\begingroup$ 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]. $\endgroup$– Noam D. ElkiesApr 18, 2012 at 2:18
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
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.
answered Apr 18, 2012 at 2:16
-
3$\begingroup$ At an appropriate time, I'd love to hear the story of why natural Diophantine equations often give rise to K3 surfaces of maximal or nearly-maximal Picard number. $\endgroup$ Apr 18, 2012 at 2:40
-
$\begingroup$ 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 $20$. $\endgroup$ Apr 18, 2012 at 2:46
-
$\begingroup$ (And often there are also singular points whose resolution contributes to NS.) $\endgroup$ Apr 18, 2012 at 2:58