IMO 1961 · Problem 3

Solve the equation \(\cos^n x-\sin^n x=1\), where \(n\) is a natural number.

Official text of the problem. Official problem paper (PDF)

Answer

For positive integers \(n\), all real solutions are \[ \boxed{ \begin{cases} x=k\pi, & n\text{ even},\\[2mm] x=2k\pi\quad\text{or}\quad x=2k\pi-\dfrac{\pi}{2}, & n\text{ odd}, \end{cases} \qquad k\in\mathbb Z. } \] If \(0\) is included among the natural numbers, \(n=0\) has no solutions wherever the expression is defined.

Solution

Suppose first that \(n\) is positive and even. Since \[ \cos^n x\le 1,\qquad \sin^n x\ge 0, \] the equation holds if and only if \(\cos^n x=1\) and \(\sin^n x=0\). These conditions are equivalent to \(x=k\pi\), and every such value satisfies the equation.

Now suppose that \(n\) is odd. The equation gives \[ \cos^n x=1+\sin^n x\ge 0, \qquad \sin^n x=\cos^n x-1\le 0. \] Because \(n\) is odd, we obtain \(\cos x\ge0\) and \(\sin x\le0\). Set \[ u=\cos x,\qquad v=-\sin x. \] Then \(u,v\ge0\), and \[ u^2+v^2=1,\qquad u^n+v^n=1. \]

If \(n=1\), then \(u+v=1\), so \[ 1=(u+v)^2=u^2+v^2+2uv=1+2uv. \] Thus \(uv=0\).

If \(n\ge3\) and both \(u,v\) were positive, the identity \(u^2+v^2=1\) would imply \(0<u,v<1\). Hence \[ u^n+v^n<u^2+v^2=1, \] a contradiction. Thus again \(uv=0\).

Consequently, in every odd case, \[ (u,v)=(1,0)\quad\text{or}\quad(u,v)=(0,1). \] These correspond respectively to \[ x=2k\pi\quad\text{or}\quad x=2k\pi-\frac{\pi}{2}, \qquad k\in\mathbb Z. \] Direct substitution verifies both families for every positive odd \(n\).

Finally, if \(n=0\) is admitted, the left-hand side equals \(1-1=0\) wherever both powers are defined, so there are no solutions.

The solution was checked against John Scholes' IMO solution archive (Kalva).