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Topic: Is this correct?
Replies: 15   Last Post: Jan 20, 2014 11:55 AM

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 sharma.kunapalli@gmail.com Posts: 54 Registered: 1/8/14
Is this correct?
Posted: Jan 17, 2014 6:24 AM

Let n be odd prime and
Let p be a prime factor of a^n + b^n, so a^n + b^n = 0 mod p ------ (1)
Let a = x mod p and b = y mod p and assume x < p and y < p ------ (2)

(a-x)^n + (b-y)^n = 0 mod p^n ------ (3)

Expanding we get, a^n - nC1 a^(n-1) x + nC2 a^(n-1) x^2 - ... - nCn x^n
+ b^n - nC1 b^(n-1) y + nC2 b^(n-1) y^2 - ... - nCn y^n = 0 mod p^n

using (2) we get, a^n - nC1 x^n + nC2 x^n - ... + nC(n-1)x^n - x^n
+ b^n - nC1 y^n + nC2 y^n - ... + nC(n-1)y^n - y^n = 0 mod p^n

Separating the ends from the middle, we get,
a^n + b^n - x^n - y^n + (x^n + y^n)[-nC1 + nC2 -... + nC(n-1)] = 0 mod p^n ------ (4)

Adding nC0 - nCn, we get

a^n + b^n - x^n - y^n + (x^n + y^n)[nC0 -nC1 + nC2 -... + nC(n-1) - nCn] = 0 mod p^n ------ (5)

The expression nC0 -nC1 + nC2 -... + nC(n-1) - nCn = 0 as (1-x)^n = 0 [for x = 1]

There we have a^n + b^n = [x^n + y^n] mod p^n ------ (6)

But x and y are both less than p. So x^n + y^n < 2p^n.

So the only possibility remains whether x^n + y^n = p^n. ------ (7)

We can easily show that (7) has no solution.

Hence p^n cannot be a factor of (a^n + b^n) if p is a factor of (a^n + b^n).

Hence FLT!