@SohamChowdhury Yes, I know those are grad school. I haven't got anything planned really. I know of a professor from HRI, though. (have taken his classes)
I am not sure. I was never looking for persons who can help me in my study of math -- it was a lucky chance that one of my school professors know the head of the department of the uni I go to.
try going into your nearby universities, I guess. i have no idea, really.
but i am not thinking about admissions in unis/papers/etc right now :P what else can a person want if he has time to do math?
i haven't tried, but IMO isn't really hard. you just need some practice to cope up with the time limit.
i have seen some of the questions -- most of them are manipulations. there's a positive proportion of them where you need some creative thinking, however.
it'd be nice if you can do a lot of number theory while you're at it, though. it helps, because most of IMO is number theory.
no complicated stuff needed. with my method, you can not only prove that $q$ is a square, but also that $q$ is an integer iff $(a, b) = (t, t^3)$ or $(a, b) = (t^3, t^5 - t)$
Since $a$ is an integer, discriminant must be a perfect square.
i.e., $b^2 k^2 - 4b^2 k^2 - 4b^2$. multiply by $b^2$ without loss of perfect square-ness : $b^4k^2 - 4b^4 k^2 - 4b^4 = (b^2k + 2)^2 - (4 + 4b^4) \leq (b^2k + 2)^2$
Case I : if $k \geq b^2$, then $4b^2k - 4b^4 \geq 0$, which implies $b^4k^2 - 4b^4k^2 - 4b^4 \geq (b^2k)^2$.
so this thing can either be $(b^2k)^2, (b^2k + 1)^2, (b^2k + 2)^2$. the last case is not possible, as $4 + 4b^4 \geq 0$. that said, the middle case is not possible as otherwise $k$ wouldn't be an integer (verify this).
hence, $4b^2k - 4b^4 = 0$, hence $k = b^2$, a perfect square. manipulation gives $b = a^3$, proving a part of our claim.
Case II : $k < b^2$. this is harder, but no more than some manipulation. you'd end up with $a$ being a cube and $b = a^{5/3} - a^{1/3}$, proving the second part of our claim.
@Soham I have typos in there. The discriminant is $k^2b^2 - 4(b^2 - k)$. but my overall calculation holds.
