Diophantine Equation
-
mutasimmim
- Posts:107
- Joined:Sun Dec 12, 2010 10:46 am
Does there exist primes $p,q$ such that $p^2+p+1=q^3$ ?
-
Raiyan Jamil
- Posts:138
- Joined:Fri Mar 29, 2013 3:49 pm
Re: Diophantine Equation
I don't think such two prime numbers exist .
A smile is the best way to get through a tough situation, even if it's a fake smile.
-
Phlembac Adib Hasan
- Posts:1016
- Joined:Tue Nov 22, 2011 7:49 pm
- Location:127.0.0.1
- Contact:
Re: Diophantine Equation
Rewrite $p(p+1)=q^3-1\quad (1)$.
Note that $p\not |q-1$, because if it did, then $q>p\Longrightarrow q^3-1>p^3-1>p(p+1)$ Impossible.
So $ord_p(q)=3\Longrightarrow 3|p-1$ But from the main equation $q|p^3-1\Longrightarrow ord_q(p)\in \{1,3\}$. So if $ord_q(p)=3,$ then $3|q-1$. Therefore RHS of (1) is divisible by $3$ while LHS is not. Absurd!
So $q|p-1\Longrightarrow p=kq+1$. Plunging it back in (1) gives $q^2k^2+3qk=q^3-3\Longrightarrow q=3$. But a simple checking shows this is not possible.
So there is no such prime pair.
Note that $p\not |q-1$, because if it did, then $q>p\Longrightarrow q^3-1>p^3-1>p(p+1)$ Impossible.
So $ord_p(q)=3\Longrightarrow 3|p-1$ But from the main equation $q|p^3-1\Longrightarrow ord_q(p)\in \{1,3\}$. So if $ord_q(p)=3,$ then $3|q-1$. Therefore RHS of (1) is divisible by $3$ while LHS is not. Absurd!
So $q|p-1\Longrightarrow p=kq+1$. Plunging it back in (1) gives $q^2k^2+3qk=q^3-3\Longrightarrow q=3$. But a simple checking shows this is not possible.
So there is no such prime pair.
Welcome to BdMO Online Forum. Check out Forum Guides & Rules