Conjecture

[devarajkandadai]

Is x = l 1 l the only solution to x^2 + 8 = 3^n ?

Devaraj

[retina]

Originally Posted by devarajkandadai

Is x = l 1 l the only solution to x^2 + 8 = 3^n ?

No.

x=sqrt(19), n=3: is one of an infinite numbers of solutions.

[CRGreathouse]

Originally Posted by devarajkandadai

Is x = l 1 l the only solution to x^2 + 8 = 3^n ?

Did you mean for this to be a Diophantine equation? If so, |x| = 1 are probably the only solutions, yes. Any other solutions have |x| > 10^50000.

[devarajkandadai]

Originally Posted by CRGreathouse

Did you mean for this to be a Diophantine equation? If so, |x| = 1 are probably the only solutions, yes. Any other solutions have |x| > 10^50000.

Yes; I did mean it as a Diophantine eqn. To generalise my conjecture:

Let lxl^2 + c = a^n

Case I: If c is not equal to a nor a multiple of a.

There is only one solution.

CaseII: if c is a multiple of a there can, at most, be two solutions.

Contra examples invited.

[R.D. Silverman]

Originally Posted by CRGreathouse

Did you mean for this to be a Diophantine equation? If so, |x| = 1 are probably the only solutions, yes. Any other solutions have |x| > 10^50000.

It is clear that there are only finitely many solutions. Perhaps
Baker's linear forms in logarithms might apply to bound them?

[Gammatester]

Originally Posted by devarajkandadai

Yes; I did mean it as a Diophantine eqn. To generalise my conjecture:

Let lxl^2 + c = a^n

Case I: If c is not equal to a nor a multiple of a.

There is only one solution.

CaseII: if c is a multiple of a there can, at most, be two solutions.

Contra examples invited.

Either you have not completely specified your generalised conjecture or there is a counter example for Case I! Take a=-2 and c=-9. Then gcd(a,c)=1 but there are at least two solutions for (x,n):

5^2 + c = a^4,
1^2 + c = a^3.

[LaurV]

Going for some pythagorean variations:

7^2 + -57 = -2^3
5^2 + -57 = -2^5
11^2 + -57 = -2^6

5^2 + -33 = -2^3
7^2 + -33 = -2^4
1^2 + -33 = -2^5
17^2 + -33 = -2^8

10^2 + -68 = 2^5
14^2 + -68 = 2^7
18^2 + -68 = 2^8
46^2 + -68 = 2^11

5^2 + -17 = 2^3
7^2 + -17 = 2^5
9^2 + -17 = 2^6
23^2 + -17 = 2^9

on the positive side:

1^2 + 7 = 2^3
3^2 + 7 = 2^4
5^2 + 7 = 2^5
11^2 + 7 = 2^7
181^2 + 7 = 2^15

2^2 + 28 = 2^5
6^2 + 28 = 2^6
10^2 + 28 = 2^7
22^2 + 28 = 2^9
362^2 + 28 = 2^17

1^2 + 15 = 4^2
7^2 + 15 = 4^3

1^2 + 63 = 4^3
31^2 + 63 = 4^5

6^2 + -11 = 5^2
56^2 + -11 = 5^5

17^2 + -73 = 6^3
37^2 + -73 = 6^4

11^2 + 95 = 6^3
529^2 + 95 = 6^7

9^2 + -17 = 8^2
23^2 + -17 = 8^3

These are only few examples when all variables forced to be under 100. I think there should be plenty solutions for given a and c (depends on a and c). And I also think this is kinda miscellaneous topic.

[devarajkandadai]

I think I had already made it clear that we take only the +ve values of x.

[LaurV]

Originally Posted by devarajkandadai

I think I had already made it clear that we take only the +ve values of x.

and... which x is not positive in my examples?

[retina]

Originally Posted by devarajkandadai

Yes; I did mean it as a Diophantine eqn. To generalise my conjecture:

Let lxl^2 + c = a^n

Case I: If c is not equal to a nor a multiple of a.

There is only one solution.

CaseII: if c is a multiple of a there can, at most, be two solutions.

Contra examples invited.

Why is my counter example not valid?

x=sqrt(19), c=8, a=3, n=3

[xilman]

Originally Posted by retina

Why is my counter example not valid?

x=sqrt(19), c=8, a=3, n=3

Possibly because a Diophantine equation requires solutions over Z and x is not an element of that ring?