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Mathematics Forum
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| illoyd Quote | Reply | This message was updated on 6/28/2006 3:11:09 PM by illoyd | dividing powers fermat posted on: 6/28/2006 3:10:31 PM Can anyone help. If x>b and x divides b^n (n odd or may be prime) what conclusions can be drawn about the relationship of x and b. I am particularly interested when b is even and x odd in a derivation of the fermat equation. |
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Euler
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dividing powers fermat
replied on: 6/29/2006 10:52:08 PM Sorry, I'm not very well versed in number theory yet, but I will try to help. If x divides b^n this means: b^n=kx, and since you are interested in b being even and x being odd: (2m)^n=k(2c+1), where c>=m which turns into: (2^n)(m^n)=2kc+k, so m^n=2^(1-n)kc+k(2^-n), but k=(b^n)/x, so m^n=[2^(1-n)][(b^n)/x]c+(2^-n)[(b^n)/x] m^n=[2^(1-n)]([(2^n)(m^n)]/x)c+(2^-n)([(2^n)(m^n)]/x) m^n=(2c/x)(m^n)+(m^n)/x therefore, letting the first RHS term be a^n and the second RHS term be d^n, you get: m^n=d^n+a^n which gives you the diophantine equation that fermat's is based on. I don't know if the last step is validated, as I do not have any number theory experience, but that is what seems logically to me. Hope this helps! |
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illoyd
Quote | Reply | This message was updated on 7/1/2006 4:33:47 AM by illoyd |
dividing powers fermat
replied on: 7/1/2006 4:33:12 AM Thanks very much for your help. Unfortunately the equation: m^n=(2c/x)(m^n) + (m^n)/x has a common non-zero term m^n in each part. It simplifies to x=2c+1, which is where you started. Thats maths. I am often proving x=x! One day we will find the answers! |
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Euler
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dividing powers fermat
replied on: 7/1/2006 2:11:41 PM Ah! I can't believe I didn't catch that one!! Oh well, like you said, "That's math." |
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Euler
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dividing powers fermat
replied on: 7/1/2006 2:26:40 PM Okay- how about this approach: if x divides b^n and b is even then: (2m)^n=(k1)x, now let m^n=[(k1)/2^n]x, so x also divides m^n; therefore: m^n=(k2)x, where k2=[(k1)/2^n]x so then it can be shown that: b^n+m^n=(k1)x+(k2)x, and since they are linearly dependent, the result will be another integer raised to the nth power so: b^n+m^n=c^n, which gives you the diophantine equation which Fermat's is based on. Hopefully this approach is better, but, as I have said, I haven't even taken a single class on number theory, so I don't know how valid it is. Nevertheless, I hope this helps! |
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illoyd
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dividing powers fermat
replied on: 7/2/2006 9:32:24 AM Thank you for that. Your solution certainly works for x odd, which is the case of interest. Your diophantine equation leads directly from fermat's equation and b<X, m<Y and c<Z, which hopefully gives a classic infinite descent situation. Thanks |
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