Solve the following equation:

$$
\cos ^{n} x-\sin ^{n} x=1,
$$

where $n$ is any given natural number.
I. solution. By writing $\cos ^{2} x+\sin ^{2} x$ on the right side, (1) can be transformed as follows:

$$
\sin ^{2} x\left(1+\sin ^{n-2} x\right)+\cos ^{2} x\left(1-\cos ^{n-2} x\right)=0
$$

Postponing the examination of the case $n=1$, in all other cases $n-2 \geq 0$, so $\sin ^{n-2} x \geq-1$, and $\cos ^{n-2} x \leq 1$, hence $1+\sin ^{n-2} x \geq 0$, and $1-\cos ^{n-2} x \geq 0$, which means that neither product on the left side of (2) is negative. Thus, (2) can only be satisfied if both products vanish:

$$
\begin{gathered}
\sin ^{2} x\left(1+\sin ^{n-2} x\right)=0, \text { and } \\
\cos ^{2} x\left(1-\cos ^{n-2} x\right)=0
\end{gathered}
$$

simultaneously. - (3) is satisfied in two cases: if
a) $\sin ^{2} x=0$, i.e., $x=k \pi(k$ is an integer);
b) $\sin ^{n-2} x=-1$, which has a real solution only for odd $n$: $\sin x=-1$, i.e.,

$$
x=\frac{3}{2} \pi+2 k \pi
$$

In case a), $\cos x= \pm 1$, so (4) can only be satisfied if its second factor is 0. For even $n$, $\cos ^{n-2} x=1$, so the second factor is always 0; for odd $n$, however, the second factor is 0 only if $\cos x=+1$, i.e., when $x=2 k \pi$.

In case b), $\cos x=0$, so (4) is satisfied.

Finally, for $n=1$, (1) can be written as:

$$
\frac{1}{\sqrt{2}} \cos x-\frac{1}{\sqrt{2}} \sin x=\cos \frac{\pi}{4} \cos x-\sin \frac{\pi}{4} \sin x=\cos \left(\frac{\pi}{4}+x\right)=\frac{1}{\sqrt{2}}
$$

from which

$$
\frac{\pi}{4}+x= \pm \frac{\pi}{4}+2 k \pi, \text { i.e., } x=\left\{\begin{array}{l}
2 k \pi \\
\frac{3}{2} \pi+2 k \pi
\end{array}\right.
$$

We arrived at the same result by another route, as above for odd $n$ greater than 1. According to these, the solution is:

$$
\begin{array}{ll}
\text { if } n=2 j(j>0, \text { integer }): & x=k \pi(k \text { integer }) \\
\text { if } n=2 j-1(\text { integer }): & x=\left\{\begin{array}{l}
2 k \pi, \\
\left(\frac{3}{2}+2 k\right) \pi
\end{array}\right.
\end{array}
$$

László Gillemot (Budapest, József A. g. IV. o. t.)

II. solution. For even $n$, $\cos ^{n} x$ and $\sin ^{n} x$ are numbers between 0 and 1 - including the two limits - , hence their difference is maximized when $\cos ^{n} x$ is at its maximum and $\sin ^{n} x$ is at its minimum:

$$
\cos ^{n} x-\sin ^{n} x \leq 1
$$

The given equation precisely describes the attainment of this maximum value, so

$$
\cos ^{n} x=1, \quad \sin ^{n} x=0, \text { from which } \cos x= \pm 1, x=k \pi .
$$

For odd $n$, $\cos x=1, x=2 k \pi$ also provides a solution; otherwise, $\sin ^{n} x=\cos ^{n} x-1>0$, so $\sin y>0$, and (1) becomes

$$
\cos ^{n} y+\sin ^{n} y=1
$$

From this,

$$
\cos ^{n} y=1-\sin ^{n} y \geq 0
$$

so $\cos y \geq 0$, hence only angles $y$ in the range $\pi / 2 \geq y>0$ are possible. It is obvious that $y=\pi / 2$ is a solution, and thus $x=3 \pi / 2$ is also a solution to (1).

We will show that for $0O P, \text { i.e., } \cos y+\sin y>1
$$

For $n \geq 3$, however, $\cos ^{n} y<\cos ^{2} y, \sin ^{n} y<\sin ^{2} y$, and thus $\cos ^{n} y+\sin ^{n} y<\cos ^{2} y+\sin ^{2} y=1$.

Our results are in complete agreement with the summary at the end of the first solution.