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Regional Competition

Number theory Difficulty 5.0 AIME, harder Prove it Austria

Determine all positive integers kk and nn satisfying the equation
k2−2016=3nk^2 - 2016 = 3^n

Solution

We immediately see that n=1n = 1 does not lead to a solution, while n=2n = 2 yields the solution (k,n)=(45,2)(k, n) = (45, 2).

We show that there is no solution with n≥3n \ge 3. In that case 3n3^n is divisible by 99 and thus k2k^2 is divisible by 99 which implies that k=3lk = 3l for some positive integer ll. After division by 99, the equation reads l2−224=3n−2l^2 - 224 = 3^{n-2}. Modulo 33, this yields l2−2≡0(mod3)l^2 - 2 \equiv 0 \pmod{3}, a contradiction because 22 is not a quadratic residue modulo 33.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.