Fermat's Little Theorem

Ok, Fermat's Little Theorem states that the formula A^n=B^n+C^n can only be true with whole numbers for n less than or equal to 2. It was proven about a decade ago, but the proof is like a hundred pages long and involves some really complex math and is hard to verify. So I felt motivated to try for a simpler one.

 


In the picture above, I took A^n=B^n + C^n, solved for A, then took two derivatives with respect to B, then I called what that right side equals k. Then I divided the right side by k so I had 1 on the left side. Then I substituted into that formula everywhere I saw a nth root of (B^n + C^n)  an A. Then I solved that formula for A.  That gave me the last line in the above picture. But I like the way it looks better if I pull a b^2 out of what's in the parenthesis. So here is the final formula I'll be using:

Ok, so look at the above solved for k:

Let's start by writing the A^(2n-1) as A/A^(2n) . Also multiply the b^(n-2) by b^2/2 to get (b^n)/2 for reasons that will be explained later. 

  Let's by convention say c is larger than b. 

  Now for k*(b^2)/2  to be a whole number every factor in 2*A^(2n) would have to cancel, so assume to factor the most out we replace c^n*b^(n) with 2*c^n*c^(n)=2*c^2n and all of the factors in that go towards canceling out factors in 2*A^(2n). The 2 multiple in each cancel.  Ok, now on the bottom of the fraction we will when all those factors cancel have p^(2n) left on the bottom where p was the one factor not in c that was in A. We know there must have been at least one because A divides c completely cancels the most terms from A but something is left over because A is larger than c. Then raise that one factor p to the 2n power because there was one left over for each A in A^(2n).  

  So we have now k*b^2/2 = A*(1-n)/p^2n if the denominator is as small as possible, remember that A and p^2n have only one factor in common because dividing A^2n by c^2n left only one factor for each A term p. So we have w*(1-n)/p^(2n-1) where w is A/p. 

So can (-1+n) cancel p^(2n-1)? No because for n> 2 even 2^(2n-1) is larger than (-1+n), and certainly for any larger p even more so. 

  So k*b^2/2 is a fraction with a different numerator from denominator in reduced form because the denominator can never completely cancel.   Now let's substitute and call k by itself x/y. 

Remember k is actually shorthand for the second derivative of A=(b^n + c^n)^(1/n) with respect to b. So we can integrate x/y twice with respect to b and get back to A. 

x/y twice integrated with respect to b is x*b^2 / 2*y. But, x/y=k when multiplied by b^2/2 can't be a whole number as we've proven.  So since x*b^2/2*y = A is not a whole number, A^n must not be a whole number. 

  Notice that this proof only works because the second derivative is non-zero and non constant, so it would not work on A=B+C or A^2 = B^2 + C^2,as required.