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Old 2007-08-09, 16:25   #34
mfgoode
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Question 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
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Old 2007-08-09, 17:08   #35
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Quote:
Originally Posted by mfgoode View Post
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 View Post
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 View Post
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
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Old 2007-08-10, 07:44   #36
mfgoode
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Quote:
Originally Posted by R.D. Silverman View Post
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.
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Old 2007-08-10, 07:59   #37
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Quote:
Originally Posted by axn1 View Post
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
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Old 2007-08-11, 09:38   #38
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I believe the Penelope Crux Theorem is, as yet, unsolved...
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Old 2007-08-11, 11:27   #39
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Thumbs up Crux of the matter!



Quote:
Originally Posted by mgb View Post
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
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Old 2007-09-21, 08:19   #40
mfgoode
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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
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Old 2008-03-21, 19:26   #41
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Quote:
Originally Posted by mfgoode View Post
[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."
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Old 2008-03-22, 02:01   #42
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Quote:
Originally Posted by CRGreathouse View Post
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
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Old 2009-04-23, 04:29   #43
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If I may so inquire, who here on Mersenneforum attended the late Malcolm's funeral?
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Old 2009-05-04, 00:19   #44
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Quote:
Originally Posted by mfgoode View Post






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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