\section*{Problem 7}

Find all real \(x, y\) such that \((1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)=1+y^{7},(1+y)\left(1+y^{2}\right)\left(1+y^{4}\right)=1+x^{7}\) ?

Answer \(\quad(\mathrm{x}, \mathrm{y})=(0,0)\) or \((-1,-1)\)


\section*{Solution}

If \(x=y\), then clearly \(x \neq 1\), so we have \(\left(1-x^{8}\right)=(1-x)\left(1+x^{7}\right)=1-x^{8}-x+x^{7}\), so \(x=0\) or \(x^{6}=1\), whose only real root (apart from the \(x=1\) we have discarded) is \(x=-1\). That gives the two solutions above. So assume \(x \neq y\). wlog \(x>y\).

So \((1+\mathrm{x})>(1+\mathrm{y})\) and \(\left(1+\mathrm{x}^{7}\right)>\left(1+\mathrm{y}^{7}\right)\). So we must have \(\left(1+\mathrm{x}^{2}\right)\left(1+\mathrm{x}^{4}\right)0\), then \((1+\mathrm{x})\left(1+\mathrm{x}^{2}\right)\left(1+\mathrm{x}^{4}\right)>1>1+\mathrm{y}^{7}\), so \(\mathrm{x}0\) and each term on rhs \(<0\). Contradiction. So there are no more solutions.

