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)(n, p) are there to the equation n3=p2p+1n^3=p^2-p+1 where pp is a prime number?

Note by Siam Habib
5 years, 6 months ago

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[EDITED]

n3=p2p+1n^3 = p^2 - p +1

n31=p2p\Rightarrow n^3 - 1 = p^2 - p

(n1)(n2+n+1)=p(p1)\Rightarrow (n-1)(n^2+n+1)=p(p-1)

Case 1: pp is a factor of (n1)(n-1).

Let n1p=k\frac{n-1}{p} = k;where kk is a natural number

n1=pk\Rightarrow n-1 = pk

Now, (n1)p(n2+n+1)=p1\frac{(n-1)}{p} (n^2+n+1) = p-1

k(n2+2+n1)=p1\Rightarrow k(n^2+2+n-1) = p-1

k(n2+2+pk)=p1\Rightarrow k(n^2+2+pk) = p-1

kn2+2k+pk2=p1\Rightarrow kn^2 + 2k + pk^2 = p-1

kn2+2k+1=ppk2\Rightarrow kn^2 + 2k + 1 = p-pk^2

kn2+2k+1=p(1k2)kn^2+2k+1=p(1-k^2)

(kn2+2k+1)>0(kn^2+2k+1) > 0. So, (1k2)>0(1-k^2) > 0 or 1<k<1-1<k<1. There exist no such value for kk which satisfies the equation as kk is a natural number.

Case 2:pp is a factor of (n2+n+1)(n^2+n+1).

I'm still working on it.

Fahim Shahriar Shakkhor - 5 years, 6 months ago

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

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

Mursalin Habib - 5 years, 6 months ago

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I was totally wrong. :(

Fahim Shahriar Shakkhor - 5 years, 6 months ago

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

Daniel Remo - 5 years, 6 months ago

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Thanks. It was really helpful.

Siam Habib - 5 years, 6 months ago

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what if n-1 = kp or n2+n+1n^{2} + n+ 1 = kp , then you cannot assert that p = n-1 or n2+n+1n^{2} + n+ 1 like mursalin habib's case where kp = n2+n+1n^{2} + n+ 1 with k =3 ?

Pradeep Ch - 5 years, 6 months ago

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Try rewriting it as (n1)(n2+n+1)=p(p1) (n-1)(n^2+n+1) = p(p-1)

Josh Rowley - 5 years, 6 months ago

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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 - 5 years, 6 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(n2+n+1) p | (n^2+n+1) . Write pk=n2+n+1 pk = n^2+n+1 , so p1=k(n1) p-1 = k(n-1) . Solve for p p to get p=k(n1)+1 p = k(n-1) + 1 . Plug in to get k2(n1)+k=n2+n+1 k^2(n-1) + k = n^2+n+1 . Write as a quadratic in n n : n2+(1k2)n+(k2k+1)=0 n^2 + (1-k^2) n + (k^2-k+1) = 0 .

Use the quadratic formula: n=12(k21±(k21)24(k2k+1))=12(k21±(k23)2+4k12) 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 k = 1 or k=2 k = 2 then the quantity under the square root is negative. If k=3 k = 3 then we get n=12(8±6)=1,7 n = \frac12(8 \pm 6) = 1,7 . The former solution is no good (p=1 p = 1 ) but the latter solution yields (7,19) (7,19) . If k4 k \ge 4 then the quantity inside the square root is greater than (k23)2 (k^2-3)^2 . But since k2k+1>0 k^2-k+1 > 0 for all k k , the quantity under the square root is less than (k21)2 (k^2-1)^2 . So the square root itself is strictly between k23 k^2-3 and k21 k^2-1 .

So if we choose the minus sign for n n , we get that n n is strictly between 0 0 and 1 1 , which is no good. If we choose the plus sign for n n , we get that n n is strictly between k22 k^2-2 and k21 k^2-1 , which is also no good. So (7,19) (7,19) is the only solution.

Patrick Corn - 5 years, 6 months ago

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

Siam Habib - 5 years, 6 months ago

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