Problem
ALG-B1-M01-P024 Recover a Pair from Powers
#24
★★★★★ Level 5 of 5
Real numbers \(x,y\) satisfy \(x^2+y^2=10\), \(x^3+y^3=10\). Find all possible values of \(x+y\).
Set \(s=x+y\), \(p=xy\). The conditions will give an equation only in \(s\).
Let \(s=x+y\), \(p=xy\). From \(x^2+y^2=s^2-2p=10\), we get \(p=\frac{s^2-10}{2}\).
Also \(x^3+y^3=s^3-3ps=10\). Substitute \(p\): \[ s^3-\frac{3s(s^2-10)}{2}=10. \] Multiplying by \(2\), \(2s^3-3s^3+30s=20\), so \(s^3-30s+20=0\) after changing sign.
Additionally, real \(x,y\) require \(s^2-4p=s^2-2(s^2-10)=20-s^2\ge0\). Thus the possible values of \(x+y\) are precisely the real roots of \(s^3-30s+20=0\) lying in \([-\sqrt{20},\sqrt{20}]\).
This problem deliberately shows that the answer need not be a nice integer.