Do there exist real numbers a,b,c such that the system of equations x+y+z=ax2+y2+z2=bx4+y4+z4=chas infinitely many real solutions (x,y,z)?
Problem
Source: Polish MO Finals P4 2024
Tags: algebra, system of equations, parameterization, algebra proposed
05.04.2024 13:15
Only if, the the equation (b−z2)2−[(a−z)2+z2−b]22=c−z4 have infinite many solutions z! As the solution below shows, this is possible if, for example, a=0,b=c=2. (Remark: I denoted x+y=S,xy=P so we have S=a−z,S2−2P=b−z2,(S2−2P)2−2P2=c−z4and I replaced S and P.)
05.04.2024 13:20
It should be (b−z2)2 and +z2−b instead of +z2−1 on your LHS. And actually your conclusion is quite wrong.
05.04.2024 13:29
Not "quite" because you quoted my computations! Now I reformulated! Thank you for the honor!
02.06.2024 02:35
It turns out one can actually describe all triples (a,b,c) for which there are infinitely many solutions. (Credits to @Archimedes3.14159) A direct way to obtain these is as follows. Denote p=xy+yz+zx, q=xyz. Then b=(x+y+z)2−2(xy+yz+zx)=a2−2p and c=(x2+y2+z2)2−2(x2y2+y2z2+z2x2)=b2−2(xy+yz+zx)2+4xyz(x+y+z)=b2−2p2+4aq. If a≠0, then given (a,b,c) one expresses p and q uniquely and hence x, y, z are uniquely determined as roots of the polynomial X3−aX2+pX−q (either they exist as real numbers, with only one possibility for them as a set, or the system has no solutions). Thus for a≠0 we have (up to permutations) at most one possible (x,y,z). On the other hand, if a=0, then b=−2p=2(x2+xy+y2) and c=b2−2p2=2p2=2(x2+xy+y2)2. These insist on b,c>0 (as x2+xy+y2≥0 with equality only for x=y=0) and as long as c=b22 we have infinitely many solutions, since x2+xy+y2−b2=0 has a real root for x precisely when D=2b−3y2, i.e. |y|≤√2b3 (which is an interval of infinitely many y).