IMO 1963 - Problem 1

Problem: Determine all real solutions of the equation $\sqrt{x^2-p}+2\sqrt{x^2-1}=x$, where $p$ is a real number.
Solution: Obviously, $x\ge 0$; hence squaring the given equation yields an equivalent equation $5x^2-p-4+4\sqrt{(x^2-1)(x^2-p)}=x^2$, i.e, $4\sqrt{(x^2-1)(x^2-p)}=(p+4)-4x^2$. If 4x^2\le (p+4), we may square the equation once again to get $-16(p+1)x^2+16p=-8(p+4)x^2+(p+4)^2$, which is equivalent to $x^2=\frac{(4-p)^2}{4(4-2p)}$,i.e $x=\frac{4-p}{2\sqrt{4-2p}}$. For this to be a solution we must have $p\le 2$ and $\frac{(4-p)^2}{4-2p}=4x^2\le p+4$. Hence $\frac{4}{3}\le p\le 2$. Otherwise there is no solution.