20070809, 16:25  #34 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Ques: rephrased.
I'm terribly sorry I have caused you all a lot of confusion, so I will rephrase my question and make it plain and simple. Can A^(n1) = B^n when n is greater than 2 and A and B are integers which means that A^(n1)  B^n = 0 Eg: n=3 : Can A^2 = B^3 or in general when n > 2 , give an integral solution ? A and B are integers preferably not equal to 0 or 1 to rule out the obvious solution 1 ^ (n1) = 1 ^n regardless of the value of n > 2 It will be better if you can illustrate any solutions numerically to make it plain and simple. Thank you, Mally 
20070809, 17:08  #35  
Jun 2003
2·3^{2}·269 Posts 
Quote:
Quote:
Quote:
Numerical example. Let x=2, k=3. Therefore, A=8, B=4, n=3. 8^2 = 4^3. QED 

20070810, 07:44  #36  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
A Flash!
Quote:
Hats off to you Dr. Silverman! Your extra ordinary flash of brilliance has settled the matter once and for all. I was struggling for days tryng to crack this out and thanks for this insight. However, and I may be wrong, but to make your equation correct I think the values of A and B should be juxtaposed/ interchanged for n = 6 This may be a natural error but I have got the gist of it and more. The constant 1 makes no difference as it is still an integral solution. Your generalisation of x to be any number was excellent! This is what I call Mathematics! Thank you once again. Mally . Last fiddled with by mfgoode on 20070810 at 07:49 Reason: Add line. 

20070810, 07:59  #37  
Bronze Medalist
Jan 2004
Mumbai,India
100000000100_{2} Posts 
Quote:
Thank you axn1. You are right but Dr. Silverman has generalised x which is what I wanted. Please refer to my reply and his post and you will see why. Thanks, Mally 

20070811, 09:38  #38 
"Michael"
Aug 2006
Usually at home
120_{8} Posts 
I believe the Penelope Crux Theorem is, as yet, unsolved...

20070811, 11:27  #39 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Crux of the matter!
Yeah! Penelope Cruz is still a conjecture depending on the Salma Hayek conjecture before she can be proved into a thighram~. Shades of Andrew Wiles' proof of FLT.? Mally Last fiddled with by mfgoode on 20070811 at 11:30 
20070921, 08:19  #40 
Bronze Medalist
Jan 2004
Mumbai,India
2052_{10} Posts 
FYI
We, the sons of Malcolm Goode, wish to inform you of his untimely death.
He suffered a sudden heart attack. He took ill on the evening of Tuesday, the 18th, was rushed to the hospital, but unfortunately didn't make it. The funeral is set for 1030 AM on Saturday, the 22nd, at St. John the Evangelist Church, Marol. Should you require any further information, kindly ring 9869255301 or email jimgoode@blueyonder.co.uk Sincerely, Bruce & Warren 
20080321, 19:26  #41  
Aug 2006
3×1,987 Posts 
Quote:
Your statement "Please note the terms cannot be co prime and will always have a common prime factor." is precisely the claim that Beal's conjecture holds. That seems like a lot for a onesentence note! Usually those are reserved for things like "Note that we can assume without loss of generality that A > B." 

20080322, 02:01  #42  
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Quote:
In my book (if not necessarily his:) he may therefore miss out on this attempt at enlightenment (the latest of many) :( David Last fiddled with by davieddy on 20080322 at 02:07 

20090423, 04:29  #43 
Dec 2008
7^{2}×17 Posts 
If I may so inquire, who here on Mersenneforum attended the late Malcolm's funeral?

20090504, 00:19  #44 
Dec 2008
7^{2}×17 Posts 
So was this Malcolm's last post? Kinda weird I am reading the posts of a dead man... creeps me out. Maybe there is ghost on this forum!!!!!!!!

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