# Find all prime numbers $p$, for which $2x^{p-1} + 2009= y ^{p-1}$ has an infinite number of solutions in non-negative integers?

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.

• If $x$ and $p$ are coprime, you can apply Fermat theorem, if not, $x^{p-1}\equiv0\pmod p$. The same thing with $y$. Give it a try. Try also saying that $gdc(x,y)=d$, which yeilds $\exists u,v:\; xu+yv=d$. Aug 5, 2016 at 15:18
• Did you find any prime numbers $p$ for which the relation admits an infinite number of solutions $x,y$? Generally you are expected to supply more context than a bare problem statement. Why is the problem interesting or difficult? What do you already know about approaches to this problem? Aug 5, 2016 at 22:52

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.
jagy@phobeusjunior:~$./Pell_Target_Fundamental Automorphism matrix: 3 4 2 3 Automorphism backwards: 3 -4 -2 3 3^2 - 2 2^2 = 1 x^2 - 2 y^2 = 2009 Fri Aug 5 13:57:33 PDT 2016 x: 47 y: 10 ratio: 4.7 SEED KEEP +- x: 49 y: 14 ratio: 3.5 SEED KEEP +- x: 53 y: 20 ratio: 2.65 SEED KEEP +- x: 79 y: 46 ratio: 1.71739 SEED BACK ONE STEP 53 , -20 x: 91 y: 56 ratio: 1.625 SEED BACK ONE STEP 49 , -14 x: 101 y: 64 ratio: 1.57812 SEED BACK ONE STEP 47 , -10 x: 181 y: 124 ratio: 1.45968 x: 203 y: 140 ratio: 1.45 x: 239 y: 166 ratio: 1.43976 x: 421 y: 296 ratio: 1.4223 x: 497 y: 350 ratio: 1.42 x: 559 y: 394 ratio: 1.41878 x: 1039 y: 734 ratio: 1.41553 x: 1169 y: 826 ratio: 1.41525 x: 1381 y: 976 ratio: 1.41496 x: 2447 y: 1730 ratio: 1.41445 x: 2891 y: 2044 ratio: 1.41438 x: 3253 y: 2300 ratio: 1.41435 x: 6053 y: 4280 ratio: 1.41425 x: 6811 y: 4816 ratio: 1.41424 x: 8047 y: 5690 ratio: 1.41424 x: 14261 y: 10084 ratio: 1.41422 x: 16849 y: 11914 ratio: 1.41422 x: 18959 y: 13406 ratio: 1.41422 x: 35279 y: 24946 ratio: 1.41421 x: 39697 y: 28070 ratio: 1.41421 x: 46901 y: 33164 ratio: 1.41421 x: 83119 y: 58774 ratio: 1.41421 x: 98203 y: 69440 ratio: 1.41421 x: 110501 y: 78136 ratio: 1.41421 x: 205621 y: 145396 ratio: 1.41421 x: 231371 y: 163604 ratio: 1.41421 x: 273359 y: 193294 ratio: 1.41421 x: 484453 y: 342560 ratio: 1.41421 x: 572369 y: 404726 ratio: 1.41421 x: 644047 y: 455410 ratio: 1.41421 x: 1198447 y: 847430 ratio: 1.41421 x: 1348529 y: 953554 ratio: 1.41421 x: 1593253 y: 1126600 ratio: 1.41421 x: 2823599 y: 1996586 ratio: 1.41421 x: 3336011 y: 2358916 ratio: 1.41421 x: 3753781 y: 2654324 ratio: 1.41421 x: 6985061 y: 4939184 ratio: 1.41421 x: 7859803 y: 5557720 ratio: 1.41421 x: 9286159 y: 6566306 ratio: 1.41421 x: 16457141 y: 11636956 ratio: 1.41421 x: 19443697 y: 13748770 ratio: 1.41421 Fri Aug 5 13:58:13 PDT 2016 x^2 - 2 y^2 = 2009 jagy@phobeusjunior:~$