IMO-Auswahlklausur

Solution:

If two of the three numbers $a, b, c$ are equal, then, because of the cyclic interchangeability, we may assume without loss of generality that $a = b$. Then the left given equation becomes $a^{2} + p b = a^{2} + p c$, from which, since $p \neq 0$, it follows directly that $b = c$, so that the claim is fulfilled. In the following we may therefore assume $a \neq b \neq c \neq a$.

Rearranging the equations gives
$$p = \frac{b^{2} - a^{2}}{b - c} = \frac{c^{2} - b^{2}}{c - a} = \frac{a^{2} - c^{2}}{a - b}.$$
Multiplying the three terms and cancelling gives
$$p^{3} = - (a + b)(b + c)(c + a) \tag{1}$$
Of the three given numbers, by the pigeonhole principle at least two are even or at least two are odd; their sum is therefore divisible by $2$. Hence the product on the right-hand side of (1) is even. It follows that $p = 2$ and therefore $(a + b)(b + c)(c + a) = -8$. $\text{2}$

If two of the brackets in (2) are equal, then without loss of generality let $a + b = b + c$. It follows directly that $a = c$ and therefore the equality of all three given numbers.

If exactly one of the brackets is odd (equal to $\pm 1$), then $(a + b) + (b + c) + (c + a)$ is on the one hand odd, and on the other hand equal to $2(a + b + c)$, hence even – a contradiction!

There remain therefore (up to cyclic interchangeability) the cases ($\pm 1; \mp 1; 8$) to be examined.

From $a + b = 1$, $b + c = -1$, $c + a = 8$ it follows that $a = 5$, $b = -4$, $c = 3$, which, with $25 - 8 \neq 16 + 6$, gives a contradiction to the first given equation.

From $a + b = -1$, $b + c = 1$, $c + a = 8$ it follows that $a = 3$, $b = -4$, $c = 5$, which, with $9 - 8 \neq 16 + 10$, gives a contradiction to the first given equation.

Therefore only $a = b = c$ is possible. Indeed, $a^{2} + p a = a^{2} + p a$ holds.