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?

#NumberTheory #DiophantineEquations #PrimeNumbers

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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`\boxed{123}`	$\boxed{123}$

Comments

[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.

Fahim Shahriar Shakkhor - 7 years, 1 month ago

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

Can you find out what's wrong with your reasoning?

Mursalin Habib - 7 years, 1 month ago

I was totally wrong. :(

Fahim Shahriar Shakkhor - 7 years, 1 month ago

It's only true that $p$ must divide one of the two factors

Daniel Remo - 7 years, 1 month ago

Thanks. It was really helpful.

Siam Habib - 7 years, 1 month 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 ?

Pradeep Ch - 7 years, 1 month ago

Try rewriting it as $(n-1)(n^2+n+1) = p(p-1)$

Josh Rowley - 7 years, 1 month 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?

Siam Habib - 7 years, 1 month ago

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.

Patrick Corn - 7 years, 1 month ago

Thank you very much. It was great help. I have been trying this problem for almost two weeks. You are awesome.

Siam Habib - 7 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$