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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
2 years, 9 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. Patrick Corn · 2 years, 9 months ago

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@Patrick Corn Thank you very much. It was great help. I have been trying this problem for almost two weeks. You are awesome. Siam Habib · 2 years, 9 months ago

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Try rewriting it as \( (n-1)(n^2+n+1) = p(p-1) \) Josh Rowley · 2 years, 9 months ago

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@Josh Rowley 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 · 2 years, 9 months ago

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[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 · 2 years, 9 months ago

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@Fahim Shahriar Shakkhor That is not true, is it? Try \((n, p)=(7, 19)\).

Can you find out what's wrong with your reasoning? Mursalin Habib · 2 years, 9 months ago

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@Mursalin Habib I was totally wrong. :( Fahim Shahriar Shakkhor · 2 years, 9 months ago

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@Fahim Shahriar Shakkhor Thanks. It was really helpful. Siam Habib · 2 years, 9 months ago

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@Fahim Shahriar Shakkhor It's only true that \(p\) must divide one of the two factors Daniel Remo · 2 years, 9 months ago

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@Fahim Shahriar Shakkhor 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 · 2 years, 9 months ago

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