I can show that there is a high probability that Fermat knew his famous "last" proof. The reason for this is that the closer you look at the other Diophantine proofs that Fermat produced, the more obvious his last proof becomes. At the heart of the logic which uncovers the final proof is the discovery of the rule that any whole integer taken to a power greater than two can be represented as the difference between two whole integer squares. Once this substitution is made, there are many proofs involving squares that don't apply to higher powers. In this way I will show how you can logically arrive at a simple proof of Fermat's famous problem.
(Lemma 1) The sum of a consecutive series of odd positive integers is divisible by both the amount of terms and it’s arithmetic mean.
This is not hard to discover even if you are new to math. Try it! Take a series of consecutive odd numbers and divide the sum by the number of terms, or by the arithmetic mean. A series of odd numbers starting at 17 and ending at 25 is the following example.
[display]\overbrace{17+19+21+23+25}^{5}=105[/display]
There are five terms and 21 is the arithmetic mean. So therefore,
[display]5*21=105[/display]
If we know the series of consecutive odd numbers that sum to a particular number, then we know at least two of that number's factors.
(Lemma 2) Every positive integer square can be represented as the consecutive series of odd positive integers starting at 1 and containing one factor equal to both the amount of terms and the arithmetic mean of the series.
Therefore, the arithmetic mean will always equal the number of terms.
I'll give you a few quick examples.
[math]1+3+5+7+9=5^2=25[/math]
[math]1+3+5+7+9+11+13=7^2=49[/math]
[math]1+3+5+7+9+11+13+15+17+19+21+23=12^2=144[/math]
(Lemma 3) Every positive integer, taken to a power cubed or greater, can be represented as the difference between two positive integer squares.
Consider the identity [math]p^n=x^2-y^2[/math], by the distributive rule [math]p^n=(x+y)(x-y)[/math], where [math]p^n[/math] contains two factors of the form [math]p^n=x+y[/math] and [math]p^k=x-y[/math], where [math]n=2d+3[/math], and [math]k=d+2[/math] for any positive integer [math]d \ge 0[/math]. If the difference between any two positive integer squares can be represented as a consecutive series of odd numbers, the sum of which should be divisible by the initial whole number [math]p[/math], then [display]p^n=\sum_y^x 2y-1=(\sum_{a=1}^x 2a-1) - (\sum_{a=1}^y 2a-1)[/display] or [display]p^n = (2y + 1) + \ldots + (p^{n-k}) + \ldots + (2x - 1)[/display]
where [math]p^n[/math] always equals the sum of [math]p^k[/math] consecutive odd positive integers from [math](2y+1)[/math] to [math](2x-1)[/math] with [math](p^{n-k})[/math] as the arithmetic mean in the sequence.
Therefore, solving for [math]x[/math] and [math]y[/math] for any odd [math]n[/math] gives us [math]x=\frac{1}{2}(p+1)p^{\frac{n-1}{2}}[/math] and [math]y=\frac{1}{2}(p-1)p^{\frac{n-1}{2}}[/math] showing that [math]x[/math] and [math]y[/math] will always contain one factor relatively prime to [math]p[/math] and each other. It should now be obvious that every positive integer taken to a power greater than two can be represented as a consecutive series of odd positive integers.
(Lemma 4) A positive integer cannot be evenly divided by a proper factor greater than half of itself.
This too should be an obvious statement, but it is relevant in the following way.
Calculating [math]x[/math] and [math]y[/math] for any even [math]n[/math] only changes a common factor, and is trivial, so assume [math]n[/math] is odd. For any [math]q[/math], then [math]p > q > 0[/math] is assumed, as reversing the inequality is trivial. We find that the entire range of odd numbers
[display]m=x_{p+q}^2 - y_{p}^2=\frac{1}{2}(p+q+1)p^{\frac{n-1}{2}} - \frac{1}{2}(p-1)p^{\frac{n-1}{2}}[/display]will always have a gap[display]t=y_{p+q}^2 - x_{p}^2=\frac{1}{2}(p+q-1)p^{\frac{n-1}{2}} - \frac{1}{2}(p+1)p^{\frac{n-1}{2}}[/display]
If we consider , then for any [math]c=p+q+i[/math] where [math]p \gt q \gt i \gt 0[/math], we then find that [math]c^n \ne m-t[/math] and therefore [math]a^n + b^n \ne c^n[/math] because [math]m-t[/math] will always contain the factor [math]2p+q[/math]. Consequently, any positive integer [math]c[/math] cannot contain [math]2p+q[/math] as a factor because [math]2p+q \gt \frac{a+b}{2}[/math].
Q.E.D.