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Number Theory Level 2

What is the number of distinct positive integers $n$ such that $n+3^2$ and $n^2 + 3^3$ are both perfect cubes?

The answer is 0.

3 solutions

Pi Han Goh
May 14, 2014

Suppose there is a solution, then their product is also a perfect cube, we have a positive integer $m$ such that

$\large m^3 = (n+3^2)(n^2 + 3^3) = n^3 + 9n^2 + 27n + 243 = (n+3)^3 + 6^3$

Which contradicts Fermat's Last Theorem.

Same method used. Great application of Fermat's (for once)! :D

Finn Hulse - 7 years ago

nice one although i got it

Niladri Dan - 6 years, 12 months ago

The best way to do it. Fermat's conjecture hidden in plain sight.

Pratyush Pandey - 4 years, 4 months ago

Sudhamsh Suraj
Aug 26, 2016

(n+3^2)=k and (n^2+3^3)=l ....now kl = n^3+9n^2+27n+243=(n+3)^3+6^3

So what? Why is the answer 0?

Pi Han Goh - 4 years, 3 months ago

I.e by Fermat's last theorem. You mentioned in your solution. @Pi Han Goh

Sudhamsh Suraj - 4 years, 3 months ago

I'm reading your solution right now, I don't care if my solution have written FLT, how should I know that your solution also applies FLT? You should make that explicit.

Pi Han Goh - 4 years, 3 months ago

Chenyang Sun
May 4, 2015

Simpler solution: a^3-b^3=(a-b)(a^2+ab+b^2)=n^2-n+18=unfactorable.

You need to show that there's no integer solution for $n^2-n+18 = (a-b)(a^2+ab+b^2 )$ . It's not sufficient to prove that it's "unfactorable". You could say that $a-b=\pm 1$ or $a^2+ab+b^2 = \pm 1$ has no solution, but it's easier albeit an overkill to apply Fermat's Last Theorem. I actually don't have to use Fermat's Last Theorem, I just need to show that no perfect cubes differ by $6^3$ which is an easy task.

Pi Han Goh - 6 years, 1 month ago

Not Taxicab number!

The answer is 0.

3 solutions

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