$S_1=x_1^2-x_1x_2+x_2^2\\ S_2=x_2^2-x_2x_3+x_3^2\\ S_3=x_3^2-x_3x_4+x_4^2\\ S_4=x_4^2-x_4x_5+x_5^2\\ S_5=x_5^2-x_5x_1+x_1^2$ And $WLOG$ $x_1=max\{x_1,x_2,x_3,x_4,x_5\}$. From $S_1=S_2$ we get that $(x_1-x_3)(x_1+x_3-x_2)=0$ and since $x_1=max\{x_1,x_2,x_3,x_4,x_5\}$ we get $(x_1+x_3-x_2)>0$. So $x_1=x_3$. So we get that $x_3=x_1=max\{x_1,x_2,x_3,x_4,x_5\}$. Similarly from $S_3=S_4$ we get $x_5=x_3$. Again from $S_5=S_1$ we get $x_2=x_5$. Finally, from $S_2=S_3$ we get $x_4=x_2$. So we get $x_1=x_2=x_3=x_4=x_5$.

$x_1^2$$-x_1x_2+$$x_2^2$$=$$x_2^2$ $-x_2x_3+$ $x_3^2$ $\implies$ $x_1^2$ $-x_1x_2$$=$ $x_3^2$ $-x_2x_3 $$\implies$ $(x_3-x_1)(x_3+x_1-x_2)=0$ with this we have $x_3=x_1 or x_2=x_3+x_1$, let's choose $x_2=x_3+x_1$. $(1)$ We can get analogous equalities because of the simetry, therefore: $x_3=x_4+x_2$,$(2)$ $x_4=x_3+x_5$ $(3)$ and $x_5=x_4+x_1$$(4)$ $(3)+(4)= x_4+x_5=x_3+x_4+x_5+x_1$ $\implies$ $x_1=-x_3$ $\implies$ $x_1+x_3=0$ which is a contradiction because $x_1$ and $x_3$ are positive real numbers so it's sum must be strictly bigger than $0$ and that means one in of the equalities that we didn't consider must be true but because of the simetry that immediately implies $x_1=x_2=x_3=x_4=x_5$ . $\square$.