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 2007-08-09, 16:25 #34 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 22·33·19 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^(n-1) = B^n when n is greater than 2 and A and B are integers which means that A^(n-1) - 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 ^ (n-1) = 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
2007-08-09, 17:08   #35
axn

Jun 2003

25×5×31 Posts

Quote:
 Originally Posted by mfgoode A and B are integers preferably not equal to 0 or 1 to rule out the obvious solution 1 ^ (n-1) = 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.
Quote:
 Originally Posted by alpertron For the case A^(n-1) = B ^n there are two cases: * n=2, A can be any natural number and A = B^2 * B=1, n=1 and A can be any natural number.
Quote:
 Originally Posted by R.D. Silverman You missed, e.g. A = 5^6, B = 5^5 and n = 6, i.e. A = x^n, B = x^n-1 for any x.
As the two posters have demonstrated, there are only trivial solutions. Pick two integers x, and k, both > 1. Let A = x^k, B=x^(k-1), n=k. This will satisfy your condition.

Numerical example. Let x=2, k=3. Therefore, A=8, B=4, n=3.

8^2 = 4^3. QED

2007-08-10, 07:44   #36
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

1000000001002 Posts
A Flash!

Quote:
 Originally Posted by R.D. Silverman You missed, e.g. A = 5^6, B = 5^5 and n = 6, i.e. A = x^n, B = x^n-1 for any x. The original question clearly meant a^n - b^n = 1.

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 2007-08-10 at 07:49 Reason: Add line.

2007-08-10, 07:59   #37
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

80416 Posts

Quote:
 Originally Posted by axn1 As the two posters have demonstrated, there are only trivial solutions. Pick two integers x, and k, both > 1. Let A = x^k, B=x^(k-1), n=k. This will satisfy your condition. Numerical example. Let x=2, k=3. Therefore, A=8, B=4, n=3. 8^2 = 4^3. QED

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

 2007-08-11, 09:38 #38 mgb   "Michael" Aug 2006 Usually at home 2×41 Posts I believe the Penelope Crux Theorem is, as yet, unsolved...
2007-08-11, 11:27   #39
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22·33·19 Posts
Crux of the matter!

Quote:
 Originally Posted by mgb I believe the Penelope Crux Theorem is, as yet, unsolved...

Yeah! Penelope Cruz is still a conjecture depending on the Salma Hayek conjecture before she can be proved into a thigh-ram~.

Shades of Andrew Wiles' proof of FLT.?

Mally

Last fiddled with by mfgoode on 2007-08-11 at 11:30

 2007-09-21, 08:19 #40 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 205210 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 e-mail jimgoode@blueyonder.co.uk Sincerely, Bruce & Warren
2008-03-21, 19:26   #41
CRGreathouse

Aug 2006

598510 Posts

Quote:
 Originally Posted by mfgoode [BEAL'S CONJECTURE: If A^x + B^y = C^z, where A, B, C, x, y and z are positive integers and x, y and z are all greater than 2, then A, B and C must have a common prime factor.] Well for a start William how about this; 756^3 + 945^3 = 189^4 There are an infinite amount of solutions. Unbelievable ? yes its true! Please note the terms cannot be co prime and will always have a common prime factor. In this case there are two 3 , 7.
Your examples show cases where gcd(A, B, C) > 1.

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 one-sentence note! Usually those are reserved for things like "Note that we can assume without loss of generality that A > B."

2008-03-22, 02:01   #42
davieddy

"Lucan"
Dec 2006
England

2×3×13×83 Posts

Quote:
 Originally Posted by CRGreathouse Your examples show cases where gcd(A, B, C) > 1. 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 one-sentence note! Usually those are reserved for things like "Note that we can assume without loss of generality that A > B."
I'm afraid Mally passed away 6 months ago (see the previous post).
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 2008-03-22 at 02:07

 2009-04-23, 04:29 #43 flouran     Dec 2008 72·17 Posts If I may so inquire, who here on Mersenneforum attended the late Malcolm's funeral?
2009-05-04, 00:19   #44
flouran

Dec 2008

34116 Posts

Quote:
 Originally Posted by mfgoode Yeah! Penelope Cruz is still a conjecture depending on the Salma Hayek conjecture before she can be proved into a thigh-ram~. Shades of Andrew Wiles' proof of FLT.? Mally
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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