6 Step Fermat's Last Theorem Proof

[jhon13]

1. aⁿ+bⁿ=cⁿ
2. Dividing both sides by cⁿ yields (aⁿ/cⁿ)+(bⁿ/cⁿ)= (a/c)ⁿ+(b/c)ⁿ =1
3. This forces a and b to assume some fractional value of c where a ≤c and b≤c
4. Therefore (afraction/c )ⁿ + (bfraction/c)ⁿ = 1
5. Restoring Fermat’s aⁿ+bⁿ=cⁿ we multiply both sides by cⁿ yielding (afraction )ⁿ + (bfraction)ⁿ = cⁿ
6. Therefore no three positive integers a, b, and c can satisfy the equation aⁿ + bⁿ = cⁿ for any integer value of n greater than two.

 

[HallsofIvy]

1. aⁿ+bⁿ=cⁿ
2. Dividing both sides by cⁿ yields (aⁿ/cⁿ)+(bⁿ/cⁿ)= (a/c)ⁿ+(b/c)ⁿ =1
3. This forces a and b to assume some fractional value of c where a ≤c and b≤c

Why are you assuming \(\displaystyle a\le c\) and \(\displaystyle b\le c\).

4. Therefore (afraction/c )ⁿ + (bfraction/c)ⁿ = 1

What? What does "afraction/c" mean? What does "afraction" mean?

5. Restoring Fermat’s aⁿ+bⁿ=cⁿ we multiply both sides by cⁿ yielding (afraction )ⁿ + (bfraction)ⁿ = cⁿ

No, it just puts you back to \(\displaystyle a^n+ b^n= c^n\) again. I learned that multiplying a number by another and then dividing by that same number puts you right back where you started when I was 7 or 8 year old.

6. Therefore no three positive integers a, b, and c can satisfy the equation aⁿ + bⁿ = cⁿ for any integer value of n greater than two.

I see. Fermat's terrible mistake was learning arithmetic! He could have proven the theorem easily if only his grasp of arithmetic had been as weak as yours.


(It occurs to me now that jhon13 may have intended this as a joke, a spoof on real "Fermatist" who at least make less obvious errors.)