# Russian Math Olympiad Problem (2001) — Question and Answer

(Russia 2001).

Let a and b be distinct positive integers such that $ab(a+b)$ is divisible by $a^{2}+ab+b^{2}$. Prove that $|a-b|> \sqrt{ab}$.

Proof.

Let $g=gcd(a,b)$ and write $a=xg$ and $b=yg$ with $gcd(x,y)=1$. Then, $\frac{ab(a+b)}{a^{2}+ab+b^{2}}=\frac{xy(x+y)g}{x^{2}+xy+y^{2}}$

is an integer. N^ote that $gcd(x^{2}+xy+y^{2}, x)=gcd(y^{2},x)=1$. Similarly, $gcd(x^{2}+xy+y^{2},y)=1$.

Because $gcd(x+y,y)=1$, we have $gcd(x^{2}+xy+y^{2},x+y)=gcd(y^{2}, x+y)=1$

Now, we apply the following lemma: Let a and b be two coprime numbers. If c is an integer such that $a|c$, then $ab|c$.

Hence, we get, $x^{2}+xy+y^{2}|g$ implying that $g \geq x^{2}+xy+y^{2}$. Therefore, $|a-b|^{3}=|g(x-y)|^{3}=g^{2}|x-y|^{3}.g \geq g^{2}.1.(x^{2}+xy+y^{2})>g^{2}xy=ab$.

It follows that $|a-b|>\sqrt{ab}$. QED.

Note that the key step $x^{2}+xy+y^{2}$ divides g can also be obtained by clever algebraic manipulations such as $a^{3}=(a^{2}+ab+b^{2})a-ab(a+b)$.

More Olympiad problems later,

Nalin Pithwa