IMO 1967 LL POL38

Origin: POL

Problem

Does there exist an integer such that its cube is equal to 3n2 + 3n + 7, where n is integer?

Solution

Suppose that there exist integers n and m such that m3 = 3n2 + 3n + 7. Then from m3 \equiv1 (mod 3) it follows that m = 3k + 1 for some k \inZ. Substituting into the initial equation we obtain 3k(3k2 + 3k + 1) = n2 + n + 2. It is easy to check that n2 + n + 2 cannot be divisible by 3, and so this equality cannot be true. Therefore our equation has no solutions in integers.