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April 26th, 2014, 10:08 AM  #1 
Newbie Joined: Apr 2014 From: USA Posts: 11 Thanks: 0  Bad "Proof" of Fermat's Last explained (in 35 minutes)
I had to comment on the error in a closed post. The argument boiled down to: $\displaystyle z = (C^n  B^n)/(A^{n2})$ where z is an integer. The post claimed this is an immediate contradiction as the GCD of A, B, and C are 1. I can find many examples of (A, B, C, z) that work for n = 3. Simply let A = C  B where gcd(B, C) = 1 Examples: $\displaystyle 9^3  4^3$ is divisible by $\displaystyle 5^1$. $\displaystyle 23^3  7^3$ is divisible by $\displaystyle 16^1$. Suppose you want a counterexample for n > 3. If n = 11, we see that: $\displaystyle (A = 2^9, B = 1001, C = 1001 + 2^9, z = (C^{11}  B^{11})/(2^9))$ is a perfectly good counterexample.  MMG Last edited by mobilemathguy; April 26th, 2014 at 10:11 AM. 

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