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# How many cubes

Recently I came across this problem from my friend in which I got stumped. Can anyone please help?

How many positive integer solutions $$(n, p)$$ are there to the equation $$n^3=p^2-p+1$$ where $$p$$ is a prime number?

Note by Siam Habib
3 years, 3 months ago

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I haven't seen a full answer yet. Mine is kind of ugly but I think it works.

As in the previous solutions, reduce to the case where $$p | (n^2+n+1)$$. Write $$pk = n^2+n+1$$, so $$p-1 = k(n-1)$$. Solve for $$p$$ to get $$p = k(n-1) + 1$$. Plug in to get $$k^2(n-1) + k = n^2+n+1$$. Write as a quadratic in $$n$$: $$n^2 + (1-k^2) n + (k^2-k+1) = 0$$.

Use the quadratic formula: $$n = \frac12\left( k^2-1 \pm \sqrt{(k^2-1)^2-4(k^2-k+1)} \right) = \frac12 \left( k^2-1 \pm \sqrt{(k^2-3)^2+4k-12} \right)$$.

If $$k = 1$$ or $$k = 2$$ then the quantity under the square root is negative. If $$k = 3$$ then we get $$n = \frac12(8 \pm 6) = 1,7$$. The former solution is no good ($$p = 1$$) but the latter solution yields $$(7,19)$$. If $$k \ge 4$$ then the quantity inside the square root is greater than $$(k^2-3)^2$$. But since $$k^2-k+1 > 0$$ for all $$k$$, the quantity under the square root is less than $$(k^2-1)^2$$. So the square root itself is strictly between $$k^2-3$$ and $$k^2-1$$.

So if we choose the minus sign for $$n$$, we get that $$n$$ is strictly between $$0$$ and $$1$$, which is no good. If we choose the plus sign for $$n$$, we get that $$n$$ is strictly between $$k^2-2$$ and $$k^2-1$$, which is also no good. So $$(7,19)$$ is the only solution. · 3 years, 3 months ago

Thank you very much. It was great help. I have been trying this problem for almost two weeks. You are awesome. · 3 years, 3 months ago

Try rewriting it as $$(n-1)(n^2+n+1) = p(p-1)$$ · 3 years, 3 months ago

I tried that and considered it from multiple angle but, failed to go anywhere from there. Can you be more specific about what to do from there? · 3 years, 3 months ago

[EDITED]

$$n^3 = p^2 - p +1$$

$$\Rightarrow n^3 - 1 = p^2 - p$$

$$\Rightarrow (n-1)(n^2+n+1)=p(p-1)$$



Case 1: $$p$$ is a factor of $$(n-1)$$.

Let $$\frac{n-1}{p} = k$$;where $$k$$ is a natural number

$$\Rightarrow n-1 = pk$$

Now, $$\frac{(n-1)}{p} (n^2+n+1) = p-1$$

$$\Rightarrow k(n^2+2+n-1) = p-1$$

$$\Rightarrow k(n^2+2+pk) = p-1$$

$$\Rightarrow kn^2 + 2k + pk^2 = p-1$$

$$\Rightarrow kn^2 + 2k + 1 = p-pk^2$$

$$kn^2+2k+1=p(1-k^2)$$

$$(kn^2+2k+1) > 0$$. So, $$(1-k^2) > 0$$ or $$-1<k<1$$. There exist no such value for $$k$$ which satisfies the equation as $$k$$ is a natural number.



Case 2:$$p$$ is a factor of $$(n^2+n+1)$$.

I'm still working on it. · 3 years, 3 months ago

That is not true, is it? Try $$(n, p)=(7, 19)$$.

Can you find out what's wrong with your reasoning? · 3 years, 3 months ago

I was totally wrong. :( · 3 years, 3 months ago

Thanks. It was really helpful. · 3 years, 3 months ago

It's only true that $$p$$ must divide one of the two factors · 3 years, 3 months ago

what if n-1 = kp or $$n^{2} + n+ 1$$ = kp , then you cannot assert that p = n-1 or $$n^{2} + n+ 1$$ like mursalin habib's case where kp = $$n^{2} + n+ 1$$ with k =3 ? · 3 years, 3 months ago