Note: The original formulation of this problem was: Quote: Find all solutions of the system \[x + y + z = 3,\]\[x^3 + y^3 + z^3 = 15,\]\[x^5 + y^5 + z^5 = 83.\] It was adapted from this to the version in the post above for the shortlist.

amparvardi wrote: Knowing that the system \[x + y + z = 3,\]\[x^3 + y^3 + z^3 = 15,\]\[x^4 + y^4 + z^4 = 35,\] has a real solution $x, y, z$ for which $x^2 + y^2 + z^2 < 10$, find the value of $x^5 + y^5 + z^5$ for that solution. Solution Setting $x+y+z=p;xy+yz+xz=q;xyz=r$ we have: $ \begin{cases} p=3 \\ p(p^2-3q)+3r=15 \\ p^4 +4pr-4p^2q+2q^2=35 \end{cases} $ Easy to find $p,r,q$ with $ x^2+y^2+z^2 <10 $ Note $x^5+y^5+z^5=(x+y+z)(x^4+y^4+z^4)-(x^3+y^3+z^3)(xy+yz+xz)+(x^2+y^2+z^2)xyz $ Done.

$x^5+y^5+z^5=(x^4+y^4+z^4)(x+y+z)-(x^3+y^3+z^3)(xy+yz+zx)+xyz(x^2+y^2+z^2)$ this is equivalent to $35.3-15.1+7(-1)=83$

The system can easily be solved for all complex triples $(x,y,z)$. We will have to only keep the real triples within radius $\sqrt{10}$. Basically, if $a=x+y+z$, $b=xy+xz+yz$, and $c=xyz$, we get a system that is easy to solve. The new system is found using Newton sums. $a=3$ $a^3-3ab+3c=15$ $a^4-4a^2b+4ac+2b^2=35$ The only complex solutions for $(a,b,c)$ are the integer solutions $(3,1,-1)$ and $(3,-1,-7)$. So the original system has $12$ solutions consisting of all permutations of the roots of $u^3-3u^2+u+1=0$ and all permutations of the roots of $u^3-3u^2-u+7=0$. Since $x^2+y^2+z^2=a^2-2b$ which is $7$ for the former case and $11$ for the latter case, we use the former case to answer the problem. For the tuple $(a,b,c)=(3,1,-1)$ we have $x^5+y^5+z^5=a^5-5a^3b+5a^2c+5ab^2-5bc=83$.

Let $x+y+z=a$, and $xyz=c$ and $xy+xz+yz=b$, a=3 $a(a^2-3b)+3c=15$ $a^4+4ac+2b^2-4ba^2=35$ and $x^5+y^5+z^5=3\cdot35-15-7=\boxed{83}.$