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 2012-05-17, 04:39 #1 devarajkandadai     May 2004 4748 Posts Conjecture Is x = l 1 l the only solution to x^2 + 8 = 3^n ? Devaraj
2012-05-17, 04:42   #2
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

11010010000002 Posts

Quote:
 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.

2012-05-17, 22:46   #3
CRGreathouse

Aug 2006

22×3×499 Posts

Quote:
 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.

Last fiddled with by CRGreathouse on 2012-05-17 at 22:48

2012-05-23, 10:59   #4

May 2004

22×79 Posts
conjecture

Quote:
 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.

2012-05-23, 11:29   #5
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

23·3·313 Posts

Quote:
 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?

2012-05-23, 12:47   #6
Gammatester

Mar 2009

468 Posts

Quote:
 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.

 2012-05-23, 15:40 #7 LaurV Romulan Interpreter     "name field" Jun 2011 Thailand 5×112×17 Posts Going for some pythagorean variations: Code: 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: Code: 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.
 2012-05-26, 06:08 #8 devarajkandadai     May 2004 22·79 Posts conjecture I think I had already made it clear that we take only the +ve values of x.
2012-05-26, 09:52   #9
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

101000001011012 Posts

Quote:
 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?

Last fiddled with by LaurV on 2012-05-26 at 09:52

2012-05-26, 10:15   #10
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

26×3×5×7 Posts

Quote:
 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

2012-05-26, 11:54   #11
xilman
Bamboozled!

"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across

32×1,303 Posts

Quote:
 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?

Last fiddled with by xilman on 2012-05-26 at 11:54

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