A cube and a triangle?

Number Theory Level 2

What is the sum of all positive integers that are both triangular and a perfect cube?

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2 solutions

David Vreken
Jun 5, 2019

There's a nice solution for this at http://mathworld.wolfram.com/CubicTriangularNumber.html .

An overkill solution. I love it!

Pi Han Goh - 2 years ago

Tom Engelsman
Jun 5, 2019

We are interested in the following relationship:

$\frac{p(p+1)}{2} = q^3$

for $p, q \in \mathbb{N}.$ Solving this equation as a quadratic in $p$ gives:

$p^2 + p - 2q^3 = 0 \Rightarrow p = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2q^3)}}{2} = \frac{-1 \pm \sqrt{8q^3 + 1}}{2} = \frac{-1 \pm \sqrt{(2q+1)(4q^2 - 2q + 1)}}{2}$ .

For $p$ to be an integer, we require the discriminant to be a perfect square. This occurs iff:

$2q+1 = 4q^2 - 2q + 1 \Rightarrow 4q^2 - 4q = 0 \Rightarrow q = 0, 1$ . For $q = 0,$ we obtain $p = -1, 0 \Rightarrow$ NOT ADMISSIBLE. For $q = 1,$ we obtain $p = -2, 1$ . Thus $p = q = 1$ is the only acceptable solution, which makes 1 the only such triangular number & perfect cube.

For $p$ to be an integer, we require the discriminant to be a perfect square. This occurs iff: $2q+1 = 4q^2 - 2q + 1 \ldots$

I disagere with this statement. It could be possible that $2q + 1$ divides $4q^2 - 2q + 1$ . That is, $2q + 1 = a, 4q^2 - 2q + 1 = ab^2$ for some integers $a,b$ .

In which case, $\frac{4q^2 - 2q + 1}{2q + 1} = 2q - 2+ \frac3{2q - 1}$ . So 3 divides $2q - 1$ . Equivalently, $2q - 1= \pm 1, \pm 3$ . Or $q = 0, \pm1, 2$ .

Trial and error show that $q = 1$ is the only solution.

Pi Han Goh - 2 years ago

A cube and a triangle?

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