20120517, 04:39  #1 
May 2004
2^{2}·79 Posts 
Conjecture
Is x = l 1 l the only solution to x^2 + 8 = 3^n ?
Devaraj 
20120517, 04:42  #2 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}·1,499 Posts 

20120517, 22:46  #3 
Aug 2006
3·1,987 Posts 
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 20120517 at 22:48 
20120523, 10:59  #4  
May 2004
2^{2}×79 Posts 
conjecture
Quote:
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. 

20120523, 11:29  #5  
Nov 2003
2^{2}×5×373 Posts 
Quote:
Baker's linear forms in logarithms might apply to bound them? 

20120523, 12:47  #6  
Mar 2009
2·19 Posts 
Quote:
5^2 + c = a^4, 1^2 + c = a^3. 

20120523, 15:40  #7 
Romulan Interpreter
Jun 2011
Thailand
3·3,049 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 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 
20120526, 06:08  #8 
May 2004
100111100_{2} Posts 
conjecture
I think I had already made it clear that we take only the +ve values of x.

20120526, 09:52  #9 
Romulan Interpreter
Jun 2011
Thailand
3×3,049 Posts 
and... which x is not positive in my examples?
Last fiddled with by LaurV on 20120526 at 09:52 
20120526, 10:15  #10  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}·1,499 Posts 
Quote:
x=sqrt(19), c=8, a=3, n=3 

20120526, 11:54  #11 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
3^{3}×389 Posts 
Possibly because a Diophantine equation requires solutions over Z and x is not an element of that ring?
Last fiddled with by xilman on 20120526 at 11:54 
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