Baltic Way 1995 · Problem 2

Number Theory

Let $a$ and $k$ be positive integers such that $a^{2}+k$ divides $(a-1) a(a+1)$ . Prove that $k \geq a$ .

When you’re ready

Review material becomes available with the next Daily.

Review

Divisibility and factorization

Solution

Solution:

We have $(a-1) a(a+1) = a(a^{2}+k) - (k+1)a$ . Hence $a^{2}+k$ divides $(k+1)a$ , and thus $k+1 \geq a$ , or equivalently, $k \geq a$ .

Results from Baltic Way 1995

9 teams

Score distribution

All team scores