We are given
2ⁿ = tⁿ - wⁿ
Now, we transpose wⁿ to the left
2ⁿ + wⁿ = tⁿ
Now 2, w, t and n are all natural numbers, and n > 1.
Now, we can prove that it is impossible for n > 1 by proving 2 things:
a) It is impossible for n > 2.
b) It is impossible for n = 2.
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a) Now, if n > 2, then according to Fermat's Last Theorem, there are no integer solutions to that equation, QED.
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b) Now, if n = 2, then the equation becomes
2² + t² = w²
or
4 + t² = w²
Now, t² is positive, and t² + 4 needs to be a perfect square.
We can easily prove that t and w are either both odd or both even.
Thus, their difference is a multiple of 2.
Let it be 2x.
Now, 2x is positive.
Thus,
w - t = 2x
We transpose
w = t + 2x
We square both sides
w² = t² + 4tx + 4x²
We substitute w²
4 + t² = t² + 4tx + 4x²
Now, we cancel t²
4 = 4tx + 4x²
We divide by 4
tx + x² = 1
We solve for t
t = 1/x - x
Now, we know that both 1/x and x must be positive natural numbers.
The only possible value for x is 1.
Thus, x = 1, and 2x = 2.
Thus, the difference of t and w is 2.
But if their difference is 2, then
4 + t² = t² + 4t + 4
And
4 = 4t + 4
and
4t = 0
Thus,
t = 0
Which is not a natural number. Therefore, it is impossible for n = 2, QED.
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Therefore, it is impossible to write
2ⁿ = tⁿ - wⁿ
for n > 1 having natural solutions for t and w. QED
^_^
2006-11-07 22:13:38
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answer #1
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answered by kevin! 5
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you would need to solve for a root of 2.
2006-11-08 06:08:14
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answer #2
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answered by Barabas 5
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When its a fact why strain our brains prooving it.
2006-11-08 08:13:19
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answer #3
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answered by Anonymous
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2^(n) = t^(n) - w^(n)
write as
2^(n) + w^(n) = t^(n)
and we have smoe statement that is proved by wiles a coulpe of ears ago.
2006-11-08 08:05:15
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answer #4
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answered by gjmb1960 7
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go to ur schhol and bore ut teacher
2006-11-08 05:45:49
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answer #5
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answered by pp 2
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Sorry, this stumps me! wow
2006-11-08 05:36:36
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answer #6
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answered by Anonymous
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im good at math , but cant solve this~!
2006-11-08 05:55:47
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answer #7
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answered by koogii 3
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Cant get it
2006-11-08 05:43:53
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answer #8
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answered by Varunjay 2
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