, 2010

Assume the contrary. Let $a \ne b$. The condition $ab^2 + ad^2 + cb^2 = ba^2 + bd^2 + ca^2$ we can rewrite in the form: $(a-b)(d^2 - ab - ac - bc) = 0$. Since $a \ne b$ it implies that $d^2 = ab + ac + bc$. Then $a^2 + b^2 + c^2 + d^2 = a^2 + b^2 + c^2 + ab + ac + bc = (a+b+c)^2 - d^2 = (a+b+c+d)(a+b+c-d)$. The number $(a+b+c+d)(a+b+c-d)$ is prime, $a$, $b$, $c$, $d$ are natural numbers, so $a+b+c-d=1$ hence $ab + ac + bc = d^2 = (a+b+c-1)^2$. Removing the brackets in the equality $(a+b+c-1)^2 = ab + ac + bc$, we have: $a^2 + b^2 + c^2 + 1 + 2ab + 2ac + 2bc - 2a - 2b - 2c = ab + ac + bc$ or
$$a(a+b-2) + b(b+c-2) + c(c+a-2) + 1 = 0.$$
But $a$, $b$ and $c$ are natural numbers, so the left side of the last equality is not less than 1. It follows that our assumption was wrong and consequently $a = b$.