Competition problem — Number theory

Number theory Difficulty 5.8 AIME, harder Prove it Brazil

$n$ is a positive integer such that $\frac{n(n+1)}{3}$ is a square. Show that $n$ is a multiple of $3$ , and $n+1$ and $\frac{n}{3}$ are squares.

Solution

Suppose $\frac{n(n+1)}{3} = m^2$ , so $n(n+1) = 3m^2$ . $n$ and $n+1$ are coprime so either

(1) $n = 3a^2$ and $n+1 = b^2$ , or

(2) $n = a^2$ and $n+1 = 3b^2$ .

But (2) implies that $3b^2 - 1$ is a square, whereas squares must be $0$ or $1$ mod $3$ . So we must have (1).

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.