Find all prime numbers $p$, for which $$2x^{p-1} + 2009= y ^{p-1}$$ has an infinte number of solutions in non-negative integers? This is a contest problem. Any ideas?
EDIT:
Here is the progress i made. Help would be welcomed.
Let's rewrite the equation as
$$2x^{p-1} + 2009-y ^{p-1}=0$$
From Fermat's theorem we have that
$0\equiv 2x^{p-1} + 2009-y ^{p-1}\equiv 2008,\ 2009$ ,$2010$ or $2011\pmod p$
So obviously $p$ can only be a prime divisor of $2008,\ 2009$ ,$2010$ or $2011$, $=>p=2, 3, 5, 7, 41,67, 251$ or $2011$
- $p=2$ we get $2x+2009-y=0$ which has infinite number of solutions.
$p=5\ => 2x^4+2009=y^4$. $y^4\equiv {0,1,15} \pmod {16},\ 2x^4+2009\equiv{8,9,10} \pmod {16} =>$ no solutions.
$p=7\ => 2x^6+7.7.41=y^6$ $=> 2x^6-y^6\equiv 0\pmod 7$.
From Fermat's theorem $2x^6\equiv 0$ or $2 \pmod 7$, $y^6 \equiv 1$ or $0 \pmod 7$ =>$x=7k\ ,\ y=7l $ => $2009\equiv 0 \pmod {7^6}$ which is false => no possible solutions for $p=7$.- The case for $p=41$ is solved analogously to $p=7$.
I have no ideas for $p=3, 5 ,67, 251$ or $ 2011$ yet.
The primes are $p=2$ and $p=3.$ Any larger exponent forces a finite number of solutions, Thue's Theorem. There are infinitely many solutions to $x^2 - 2 y^2 = 2009,$ every sixth pair related by $$ x_{n+2} = 6 x_{n+1} - x_n, $$ $$ y_{n+2} = 6 y_{n+1} - y_n. $$ That is, there are six orbits of solutions under the action of the (oriented) automorphism group of the quadratic form.
as in $$ (47,10); (181,124); (1039, 734); (6053, 4280);... $$ and five others.