50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010)

Assume the contrary. Let $a\ne b$. The condition $a{b}^2+a{d}^2+c{b}^2=b{a}^2+b{d}^2+c{a}^2$ 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 $a=b$.