2004 Pan African MO Problems/Problem 1

Problem

Do there exist positive integers $m$ and $n$ such that: $3n^2+3n+7=m^3$

Solution

Note that $n^3$ can only be congruent to $0, 1, -1$ modulo 9. So we can check the remainder when $3n^2 + 3n + 7$ is divided by 9. There are only nine cases to check, so we can do it manually. To make computation easier, we can factor the left-hand side as $3n(n+1)+7$ .

$n \pmod{9}$	0	1	2	3	4	5	6	7	8
$3n(n+1)+7 \pmod{9}$	7	4	7	7	4	7	7	4	7

There are no cases of $n$ where $3n(n+1)+7 \equiv -1 \pmod{9}$ or $3n(n+1)+7 \equiv 1 \pmod{9}$ , so there are no positive integers $m, n$ that satisfies the original equation.

2004 Pan African MO (Problems)
Preceded by First Problem	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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2004 Pan African MO Problems/Problem 1

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