Question 25
Prove that there are no positive integers $x$ and $y$ such that
$$
x^{2008} + 2008! = 21^y.
$$
$$
x^{2008} + 2008! = 21^y.
$$
MathNet 2008-part-01 — 165 problems (EN–VI) Question 25
Math Olympiad Vault · MathNet 2008-part-01 — 165 problems (EN–VI)