49th International Mathematical Olympiad Spain

If two of $a,b,c$ are equal, it is immediate that all the three are equal. So we may assume that $a\ne b\ne c\ne a$. Subtracting the equations we get $a^n-b^n=-p(b-c)$ and two cyclic copies of this equation, which upon multiplication yield $$\frac{a^n-b^n}{a-b}\cdot \frac{b^n-c^n}{b-c}\cdot \frac{c^n-a^n}{c-a}=-p^3.$$ If $n$ is odd then the differences $a^n-b^n$ and $a-b$ have the same sign and the product on the left is positive, while $-p^3$ is negative. So $n$ must be even. Let $d$ be the greatest common divisor of the three differences $a-b,b-c,c-a$, so that $a-b=du,b-c=dv,c-a=dw;\gcd (u,v,w)=1,u+v+w=0$. From $a^n-b^n=-p(b-c)$ we see that $(a-b)\mid p(b-c)$, i.e., $u\mid pv$; and cyclically $v|pw,w|pu$. As $\gcd (u,v,w)=1$ and $u+v+w=0$, at most one of $u,v,w$ can be divisible by $p$. Supposing that the prime $p$ does not divide any one of them, we get $u|v,v|w,w\mid u$, whence $|u|=|v|=|w|=1$; but this quarrels with $u+v+w=0$. Thus $p$ must divide exactly one of these numbers. Let e.g. $p\mid u$ and write $u=p{u}_1$. Now we obtain, similarly as before, $u_1|v,v|w,w\mid u_1$ so that $\left|u_1\right|=|v|=|w|=1$. The equation $p{u}_1+v+w=0$ forces that the prime $p$ must be even; i.e. $p=2$. Hence $v+w=-2u_1=\pm 2$, implying $v=w(=\pm 1)$ and $u=-2v$. Consequently $a-b=-2(b-c)$. Knowing that $n$ is even, say $n=2k$, we rewrite the equation $a^n-b^n=-p(b-c)$ with $p=2$ in the form $$\left(a^k+b^k\right)\left(a^k-b^k\right)=-2(b-c)=a-b.$$ The second factor on the left is divisible by $a-b$, so the first factor ( $a^k+b^k$ ) must be $\pm 1$. Then exactly one of $a$ and $b$ must be odd; yet $a-b=-2(b-c)$ is even. Contradiction ends the proof.

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Alternative solution.

The beginning is as in the first solution. Assuming that $a,b,c$ are not all equal, hence are all distinct, we derive equation (1) with the conclusion that $n$ is even. Write $n=2k$. Suppose that $p$ is odd. Then the integer $$\frac{a^n-b^n}{a-b}=a^{n-1}+a^{n-2}b+\cdots +b^{n-1},$$ which is a factor in (1), must be odd as well. This sum of $n=2k$ summands is odd only if $a$ and $b$ have different parities. The same conclusion holding for $b,c$ and for $c,a$, we get that $a,b,c,a$ alternate in their parities, which is clearly impossible. Thus $p=2$. The original system shows that $a,b,c$ must be of the same parity. So we may divide (1) by $p^3$, i.e. $2^3$, to obtain the following product of six integer factors: $$\frac{a^k+b^k}{2}\cdot \frac{a^k-b^k}{a-b}\cdot \frac{b^k+c^k}{2}\cdot \frac{b^k-c^k}{b-c}\cdot \frac{c^k+a^k}{2}\cdot \frac{c^k-a^k}{c-a}=-1.$$ Each one of the factors must be equal to $\pm 1$. In particular, $a^k+b^k=\pm 2$. If $k$ is even, this becomes $a^k+b^k=2$ and yields $|a|=|b|=1$, whence $a^k-b^k=0$, contradicting (2). Let now $k$ be odd. Then the sum $a^k+b^k$, with value $\pm 2$, has $a+b$ as a factor. Since $a$ and $b$ are of the same parity, this means that $a+b=\pm 2$; and cyclically, $b+c=\pm 2,c+a=\pm 2$. In some two of these equations the signs must coincide, hence some two of $a,b,c$ are equal. This is the desired contradiction.