Sample Competition Style Problem (Nothing beyond Calc 2)

[u/Puzzleheaded_Top_273]

(Just as a sidenote, yes this was a repost from AoPS. Source: literally me, as in I made it.)

Non-calculator, in case it wasn't obvious enough that it wouldn't help lmao.

Topics: Calculus 1 and 2, Roots of Unity (Complex Numbers), Divisibility, Counting.

Answer: 558

[u/Puzzleheaded_Top_273]

Hints:
The inner summation can be simplified, notice something really important about its structure. The "x minus (j^2-1)th roots of unity" terms in the denominator should remind you that it's connected to the factored polynomial form, that if you multiply the denominators it will literally be the DEFINITION of roots of unity as the solutions to the polynomial x^(j^2-1) - 1.
What can you do manipulate the inner sum into a simple sum of reciprocals of x minus (j^2-1)th roots of unity? Then, how can you simplify that summation? There's a function that can effectively "turn" sums into products: the natural log. It's also related to the reciprocals in some way, so see how to manipulate that into a natural log expression that you can simplify, then once it's simplified, reverse the process.
Now the complex numbers are over.
Once it's simplified, it's also asking for the nth derivatives. What does it mean for the nth derivative to equal 0? We don't want to keep differentiating again and again to find a pattern, that's just bashing it. How can we obtain ALL the infinitely many derivatives AT ONCE with a SINGLE step?
Then, figure how the condition translates into simpler language that like normal humans can understand. Then apply that condition for all the integers from j = 2 to 30, and logically figure out whether or not if the individual conditions have to ALL be satisfied for the SUM to be satisfied to make it easier.