resembles good at some bmo question? anyway: $(a-2004)(b-2004)=2004^2$ a,b must be larger than 2004 (elseway the initial equality can never hold with positive integers) $2004^2 = 2^4.3^2.167^2$ so that makes $5.3.3 = 45$ options, did I make a mistake? 20 secs for solving isn't enough so I probably missed something...

Peter VDD wrote: Indeed, the given that it are integers is pretty handy resembles good at some bmo question? anyway: $(a-2004)(b-2004)=2004^2$ a,b must be larger than 2004 (elseway the initial equality can never hold with positive integers) $2004^2 = 2^4.3^2.167^2$ so that makes $5.3.3 = 45$ options, did I make a mistake? 20 secs for solving isn't enough so I probably missed something... Yeah you are right. the objective is to make $b-2004\mid2004^2$. Always your friend, 3X.lich.

Use SFFT to simplify to $(a-2004)(b-2004) = 2004^2 = 2^4 \cdot 3^2 \cdot 167^2 \implies (4+1)(2+1)(2+1) = \boxed{45}$. Both $a,b$ are larger than $2004$, or else one of $\frac{1}{a},\frac{1}{b}$ will already be $\geq \frac{1}{2004}$.

OlympusHero wrote: Use SFFT to simplify to $(a-2004)(b-2004) = 2004^2 = 2^4 \cdot 3^2 \cdot 167^2 \implies (4+1)(2+1)(2+1) = \boxed{45}$. You should also mention why we can't have $a-2004\le0$ and $b-2004\le0$.

Yet again, me and mathmax12 have the same solution. We use SFFT to get that $(a-2004)(b-2004)=2004^2=2^4 \cdot 3^2 \cdot 167^2$, so we have $5\cdot3\cdot3=\boxed{45}.$