4 $[$ Parity and Imparity $]$ $[$ Residue Arithmetic (Miscellaneous) $]$

Solve the equation $x^{2}+y^{2}+z^{2}=2 x y z$ in integers.

#
If there is a non-zero solution to this equation in integers, then there is also a solution in natural numbers. Let's prove that there are no solutions in natural numbers.

Suppose this is not the case. Then consider the smallest natural number \(a\), for which there exist such natural numbers \(b, c, n\) that \(a^{2}+b^{2}+c^{2}=2^{n} a b c\). Note that all terms on the left side are even (otherwise, the left side would not be divisible by 4, while the right side is). Substituting \(a=2 u, b=2 v, c=2 w\) we get \(4\left(u^{2}+v^{2}+w^{2}\right)=2^{n+3} u v w\), or \(u^{2}+v^{2}+w^{2}=2^{n+1} u v w\). Since \(u<a\), this contradicts the choice of the number \(a\).

## Answer

\((0,0,0)\)

Around a round table, \(2 n\) deputies were in a meeting. After a break, these same \(2 n\) deputies sat around the table, but in a different order.

Prove that there are two deputies between whom the same number of people sat both before and after the break.

## Hint

Assign to each deputy the number of seats by which he moved clockwise after the break. If the condition of the problem is not satisfied, then all these numbers must be different.

## Solution

Number the deputies \(1, 2, \ldots, 2 n\) in the order (clockwise) in which they sat before the break; number the seats around the table in the same way. Let \(a_{i}\) be the number of seats by which the \(i\)-th deputy moved clockwise after the break. Thus, each of the numbers \(a_{1}, a_{2}, \ldots, a_{2 n}\) takes some value from the set \(\{0, 1, \ldots, 2 n-1\}\).

If some two deputies moved the same number of seats after the break, then the same number of people sat between them both before and after the break, so the statement of the problem is satisfied.

Otherwise, the numbers \(a_{1}, a_{2}, \ldots, a_{2 n}\) form a permutation of the numbers \(0, 1, \ldots, 2 n-1\). The number \(i+a_{i}\) is congruent modulo \(2 n\) to the number of the seat on which the \(i\)-th deputy sat after the break. Since after the break each of the \(2 n\) seats is again occupied by one of the deputies, the sum \(\left(1+a_{1}\right)+\left(2+a_{2}\right)+\ldots+\left(2 n+a_{2 n}\right)\) is congruent modulo \(2 n\) to the sum of the numbers of all seats \(1+2+\ldots+2 n\). It follows that the sum \(a_{1}+a_{2}+\ldots+a_{2 n}\) is divisible by \(2 n\). However, \(a_{1}+a_{2}+\ldots+a_{2 n}=0+1+2+\ldots+(2 n-1)=n(2 n-1)\), which is not divisible by \(2 n\). Contradiction.