Cube from another cube

Number Theory Level 2

Given that $n$ is a cube of a positive integer, for how many values of $n$ is $\large{n^2+3n+3}$ a perfect cube?

3 1 Infinitely many 2 0

1 solution

Sathvik Acharya
Jul 11, 2019

Assume that there exists a value of $n$ such that both $n$ and $n^2+3n+3$ are perfect cubes.

This implies that their product, $n(n^2+3n+3)=n^3+3n^2+3n=(n+1)^3-1$ is also a perfect cube.

Since $(n+1)^3-1$ cannot be a perfect cube, we obtain a contradiction. Therefore, if $n$ is a perfect cube then $n^2+3n+3$ cannot be a perfect cube.

Great solution! It's also a constructive proof for the problem generalised to all integers - you've shown there are at most two solutions for that problem, and it's trivial to check that only one works.

Chris Lewis - 1 year, 11 months ago

One solution is n=-1. What is the other Chris?

A Former Brilliant Member - 1 year, 11 months ago

My mistake, I solved a slightly different problem! I've corrected my comment.

Chris Lewis - 1 year, 11 months ago

One little mistake in the solution. It should be mentioned that n is a positive integer.

A Former Brilliant Member - 1 year, 11 months ago

Well, the question does mention that n is a positive integer.

Sathvik Acharya - 1 year, 11 months ago

But not in the solution.

A Former Brilliant Member - 1 year, 11 months ago

Very similar to this problem .

Pi Han Goh - 1 year, 11 months ago

Ah, yes, that was really nice!

Sathvik Acharya - 1 year, 11 months ago

Cube from another cube

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