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Problem 1101

AIME late
Number theory Difficulty 5.0 Prove it Iranian Mathematical Olympiad · Iran

Let nn be an integer and 0mn210 \le m \le n^2 - 1. Prove that the total number of mm such that
xn+ynm(modn2) x^n + y^n \equiv m \pmod{n^2}
has no integer solution, is at least n(n1)2\frac{n(n-1)}{2}.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

It is known that if nn divides aba - b, then anbna^n - b^n is divisible by n2n^2. Now, choose zx(modn)z \equiv x \pmod{n} and yt(modn)y \equiv t \pmod{n}. Then
xn+yn(zn+tn)=(xnzn)+(yntn)0(modn2) x^n + y^n - (z^n + t^n) = (x^n - z^n) + (y^n - t^n) \equiv 0 \pmod{n^2}
Thus, for z,tz, t belonging to {0,1,,n1}\{0, 1, \dots, n-1\}, we must calculate maximal number of different values of zn+tn(modn2)z^n + t^n \pmod{n^2}. Then, there are x,yx, y such that xn+ynzn+tn(modn2)x^n + y^n \equiv z^n + t^n \pmod{n^2}. These numbers are at most n(n+1)2\frac{n(n+1)}{2}. Therefore, there are at least n2n(n+1)2=n(n1)2n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2} different values for mm.

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