positive cube = $p^2-p-1$

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photon
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positive cube = $p^2-p-1$

Unread post by photon » Thu Jan 02, 2014 8:53 pm

Find all prime numbers $p$ such that the number $p^2-p-1$ is a cube of some positive integer .
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Nayeemul Islam Swad
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Re: positive cube = $p^2-p-1$

Unread post by Nayeemul Islam Swad » Mon Feb 03, 2014 8:40 pm

Let $p^2 - p - 1 = a^3$. LHS is positive. So, $a \geq 1$. Now, $p^2 - p = a^3 + 1 \Leftrightarrow p(p - 1) = (a + 1)(a^2 + a + 1)$. But, neither $(a + 1)$ nor $(a^2 + a + 1)$ in the RHS can be equal to $1$ or less as $a$ is positive and observe that $a^2 + a + 1$ > $a + 1$ since $a^2 \geq a$. All these together imply that $p - 1 = a + 1$. So, $p = a + 2$ and thus $a^2 + a + 1 = a + 2 \Leftrightarrow a^2 - 1 = 0 \Leftrightarrow a = 1$(since $a$ is positive). So, the only possibility for such $p$ is $(1 + 1) = 2$ which is also a prime. So, $p = 2$
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Re: positive cube = $p^2-p-1$

Unread post by *Mahi* » Mon Feb 03, 2014 8:55 pm

You found $a=1, p=a+2$, and $p=2$, all of which can't be true at once. Check your solution.
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photon
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Re: positive cube = $p^2-p-1$

Unread post by photon » Mon Feb 03, 2014 10:28 pm

Nayeemul Islam Swad wrote:Let $p^2 - p - 1 = a^3$. LHS is positive. So, $a \geq 1$. Now, $p^2 - p = a^3 + 1 \Leftrightarrow p(p - 1) = (a + 1)(a^2 + a + 1)$..........
It will be $p(p-1)=(a+1)(a^2-a+1)$ . for $a=1 , p=2 ; a=2$ has no result for prime $p$ . Following your way $a^2-a+1>a+1$ [for $a>2$]. Now not necessarily $a+1 = p-1$ , it is also possible that $a+1$ is a divisor of $p-1$ as $p-1$ is composite (here $p\not=3$)
Assume $p-1=k(a+1)$ for some $k \in \mathbb{N}$ and go on ...
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asif e elahi
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Re: positive cube = $p^2-p-1$

Unread post by asif e elahi » Tue Feb 04, 2014 12:51 pm

Let $p^{2}-p-1=a^{3}$. Then $p(p-1)=(a+1)(a^{2}-a+1)$. For $a=1,p=2$ is a solution.$a=2$ has no solution.So we assume $a>2$.Then $a^{2}-a+1>a+1$. As $p$ is a prime, $p$ divides either $a+1$ or $a^{2}-a+1$. If $p$ divides $a+1$,then $a^{2}-a+1$ divides $p-1$.
So $p\leq a+1<a^{2}-a+1\leq p-1$ which is impossible.So $p$ divides $a^{2}-a+1$ and $a+1$ divides $p-1$.Let $p-1=(a+1)k$.
Then $p=(a+1)k+1$ divides $a^{2}-a+1$.
$(a+1)k\equiv -1(mod p)$. Again $o\equiv (a^{2}-a+1)(mod p)$
or $0\equiv k^{2}(a^{2}-a+1)(mod p)$
or $0\equiv k^{2}{(a+1)^{2}-3(a+1)k^{2}+4k^{2}}\equiv {(ak+a)^{2}-3(a+1)k^{2}+4k^{2}}\equiv(1+3k+4k^{2}) (mod p)$
So $0\equiv a(4k^{2}+3ak+1)\equiv 4k.ak+3.ak+a(mod p)$
or $0\equiv 4k(-a-1)+3(-a-1)+a\equiv -2(2ak+2k+a+1)(mod p)
or 0\equiv 2(-2a-2+2k+a+1)\equiv 2(2k-a-1)(mod p)$
So $p=ak+a+1$ divides $4k-2a-2$.
So $ak+a+1\leq 4k-2a-2$ or $ak+a+1\leq 2a+2-4k$
For the 1st case $a(k+3)\leq 4k-3=4(k+3)-15<4(k+3)$
So $a\leq 3$.$a=3$ gives no solution
For 2nd case $a(k-1)+4k<1$ which is not possible
So $p=2$ is the only solution
So $k\leq 4$
Checking $k=1,2,3,4$ gives no solution.So $p=2$ is the only solution.

Nayeemul Islam Swad
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Re: positive cube = $p^2-p-1$

Unread post by Nayeemul Islam Swad » Tue Feb 04, 2014 4:05 pm

How silly I am!!! :lol:
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Tahmid Hasan
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Re: positive cube = $p^2-p-1$

Unread post by Tahmid Hasan » Tue Feb 04, 2014 5:04 pm

Similar to a Balkan problem which was in 2012 TST too!
Let $n^3=p^2-p-1 \Rightarrow (n+1)(n^2-n+1)=p(p-1)$.
If $p|n+1$, then $n+1 \ge p \Rightarrow n(n+1) \ge p(p-1)=(n+1)(n^2-n+1) \Rightarrow n \ge n^2-n+1 \Rightarrow n=1,2.$
Plugging them in the original equation we find one solution $(n,p)=(1,2)$.
Now let $p|n^2-n+1$. Let $n^2-n+1=pk[k \in \mathbb N]$. Plugging back we get $k(n+1)=p-1 \Rightarrow p=kn+k+1$.
So $kn+k+1|n^2-n+1 \Rightarrow kn+k+1|n(kn+k+1)-k(n^2-n+1)=2kn+n-k$
$\Rightarrow kn+k+1|2(kn+k+1)-(2kn+n-k)=3k-n+2$.
Now we do some casework.
1.$kn+k+1 \le 3k-n+2 \Rightarrow kn+n \le 2k+1 \Rightarrow n(k+1) \le 2k+1<2(k+1) \Rightarrow n=1$, which we've got before.
2.$kn+k+1 \le n-3k-2 \Rightarrow n(k-1) \le -(4k+3)$, a contradiction.
3.$3k-n+2=0 \Rightarrow n=3k+2$.
Now $n^2-n+1=pk \Rightarrow k|n^2-n+1=$ $(3k+2)^2-(3k+2)+1=9k^2+3k+3$ $ \Rightarrow k|3 \Rightarrow k=1,3$.
So $n=5,11$. Plugging back we get another solution $(n,p)=(11,37)$.
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asif e elahi
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Re: positive cube = $p^2-p-1$

Unread post by asif e elahi » Tue Feb 04, 2014 8:13 pm

My past solution was wrong.Here is another one.
asif e elahi wrote:Let $p^{2}-p-1=a^{3}$. Then $p(p-1)=(a+1)(a^{2}-a+1)$. For $a=1,p=2$ is a solution.$a=2$ has no solution.So we assume $a>2$.Then $a^{2}-a+1>a+1$. As $p$ is a prime, $p$ divides either $a+1$ or $a^{2}-a+1$. If $p$ divides $a+1$,then $a^{2}-a+1$ divides $p-1$.
So $p\leq a+1<a^{2}-a+1\leq p-1$ which is impossible.So $p$ divides $a^{2}-a+1$ and $a+1$ divides $p-1$.Let $p-1=(a+1)k$.
Then $p=(a+1)k+1$ divides $a^{2}-a+1$.
So $(a+1)(a^{2}-a+1)=p(p-1)=k(a+1){k(a+1)+1}$
or $a^{2}-a+1=k^{2}(a+1)+k$
or $a^{2}-a(k^{2}+1)-(k^{2}+k-1)=0$
We consider the equation quadratic in $a$.It has integer solutions if $(k^{2}+1)^{2}+4(k^{2}+k-1)$ is a perfect square
$(k^{2}+1)^{2}+4(k^{2}+k-1)=(k^{2}+3)^{2}+4k-12$
$k=1,2,3$ gives one solution $p=37$
If $k>3$,then $4k>12$
So $(k^{2}+3)^{2}<(k^{2}+3)^{2}+4k-12<(k^{2}+4)^{2}$
So $(k^{2}+3)^{2}+4k-12$ is a positive integer between $2$ consecutive perfect square,hence not a perfect square
So $(p,a)=(2,1),(37,11)$ :mrgreen:

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