FERMAT'S LAST THEOREM, SHIMURA TANIYAMA, IS EASY

Fermat was a 17th-century mathematician who wrote a note in the margin of his book stating a particular proposition and claiming to have proved it. His proposition was about an equation which is closely related to Pythagoras' equation. Pythagoras' equation gives you:

x2 + y2 = z2
You can ask, what are the whole number solutions to this equation, and you can see that:

32 + 42 = 52


and

52 + 122 = 132


And if you go on looking then you find more and more such solutions. Fermat then considered the cubed version of this equation:

x3 + y3 = z3


He raised the question: can you find solutions to the cubed equation? He claimed that there were none. In fact, he claimed that for the general family of equations:

xn + yn = zn where n is bigger than 2


it is impossible to find a solution. That's Fermat's Last Theorem.

NOW LET'S SOLVE THAT DAMN PROBLEM!!!

Let:

(p + p1S(1/2))(1/m) + (p - p1S(1/2))(-1/m)

be a function on 5 parameters y1, ..., y5

where p,p1,S are symmetric on the same 5 parameters and m ≥ 2.

Then:

this function returns m distinct values when its parameters are permuted

Proof:

(1) By Lemma 1, here, S(1/2) takes 2 distinct return values when its parameters are permuted and we can call these two values s1, s2 and we know that s1 = -s2

(2) So, it follows that the two cases gives us two possible expressions:

(p + p1s1)(1/m) + (p - p1s1)(1/m)

and

(p + p1s2)(1/m) + (p - p1s2)(-1/m)


(3) But, clearly:

(p + p1s1)(1/m) + (p - p1s1)(1/m) =(p + [-p1s2])(1/m) + (p - [-p1s2])(1/m) =

= (p - p1s2)(1/m) + (p + p1s2)(1/m)

(4) So, it follows that swapping s1 and s2 does not change the value, so the number of return values when the m parameters are permuted is m.

QED

Theorem 2: A solution to the general quintic equation is not expressible by radicals.

Proof:

(1) Assume that a solution to the general quintic equation can be expressed in radicals. That is we can state an equation for each root y in terms of a,b,c,d,e such that:


y5 - ay4 + by3 - cy2 + dy - e = 0


(2) If a solution to this equation exists, it can be represented as follows (see Theorem 5, here):



where m is a prime number and R, p2, ..., pm-1 are all functions of this same form finitely nested at the deepest level each p,pi,R is a function of the coefficients of the general quintic equation.

(3) Using this form [see Theorem 5, here], we further note that the distinct roots of the quintic equation are expressible in terms of R:





...



where α is an mth primitive root of unity [See here for a review of primitive roots of unity if needed]

(4) The Fundamental Theorem of Algebra gives us that the quintic equation has 5 roots. [See here for proof of the Fundamental Theorem of Algebra]

(5) Since this is the general solution, we can assume that each of the roots are distinct and that therefore, at the top level, m = 5.

(6) We can also state R(1/5) in terms of the roots (this was the gap that Ruffini assumed but did not prove). Using Corollary 5.1 here, we note that:

R(1/5) = (1/5)(y1 + α4y2 + ... + αy5)

(7) Now, by our result in step #2, we can assume that this same equation holds true at all levels. So at the lowest levels, we can assume that R is a rational function of the coefficients a,b,c,d,e.

(8) We can also show that at the bottom level, R is a function of the 5 roots. [See Corollary 3.1, here]

(9) Then, from step #7 and step #8, using Corollary 6.1, here, we can conclude that at the depeest level, m = 2.

(10) Now, we have two states to consider. Either the expression in step #2 consists of only two levels (at the top level m=5 and at the bottom level m=2) or its consists of more than two levels.

(11) So, if there exists a level above the lowest level but below the top level, then by step #2 above, it must be characterized by a prime m and by step #8 above we can assume that it is a function of the 5 roots.

(12) From Lagrange's Theorem (See Theorem, here), we can assume that if we permuted the 5 roots, the number of return values will divide 5! = 120.

(13) From Corollary 1.1, here and step #9 above, we know that the number of return values will be 2*m. Since 2*m must divide 120 and m must be a prime, we conclude that m=2, 3, or 5.

(14) But m ≠ 5, since from the reasoning of Corollary 1.1 in step #13, if m=5, then ultimately we would have 2*5*5*n return values which does not divide 120.

(15) We also know that m ≠ 2 since this gives us 2*2=4 return values and by Cauchy's Theorem (see Theorem 2, here), if the return values are less than 5, then it is 1 or 2.

(16) Further, we know that m ≠ 3 since:

(a) Let r = (p + p1S(1/2))(1/m)

where p,p1,S are symmetric with respect to the 5 roots of a quintic equation.

(b) Let r1 = (p - p1S(1/2))(1/m)

(c) So, we define a function on the 5 roots of the quintic equation such that:

f(y1, ..., y5) = r + r1

(d) From Lemma 1 above, such a function will return m distinct values when its m parameters are permuted.

(e) But, by Cauchy's Thereom used in step #15 above, if m is less than 5, it can only be 1 or 2.

(k) Hence, this means m ≠ 3.

(17) But, then it follows from steps #13, #14, #15, #16 above that the solution to the general quintic equation has only two levels and therefore the solution has the following form:



where m=5 and p,R,p2, ... p4 are functions of the form s + s1T(1/2) where s, T, s1 are rational functions of the coefficients of the quintic equation: a,b,c,d,e.

(18) Now, from step #6 above, we have:

R(1/5) = (1/5)(y1 + α4y2 + ... + αy5)

where α is a primitive 5th root of unity and y1, ..., y5 are the 5 distinct roots of a quintic equation.

(19) Now, from #17 above, we have:

R(1/5) = (s + s1T(1/2))(1/5)

where s, s1, T are symmetric on y1, ..., y5

(20) So, putting these two steps together gives us:

(1/5)(y1 + α4y2 + ... + αy5) = (s + s1T(1/2))(1/5)

(21) But this is clearly impossible since (1/5)(y1 + α4y2 + ... + αy5) can take 5!=120 distinct return values when its the parameters are permuted and (s + s1T(1/2))(1/5) can only take 2*5=10 distinct return values.

(22) Therefore, we have a contradiction and we reject our assumption in step #1.

QED

Corollary 2.1: A solution to the general n-th equation where n ≥ 6 is not expressible by radicals.

Proof

(1) Assume that a solution to a general n-th equation where n ≥ 6 is expressible by radicals.

(2) Let:

y5 - ay4 + by3 - cy2 + dy - e = 0

be a general quintic equation.

(3) Let i = n - 5 so that i ≥ 1.

(4) If we multiply (y - c1)*...*(y - ci)*(y5 - ay4 + by3 - cy2 + dy - e ), then we have an n-th equation.

(5) By assumption, we now solve this n-th equation so that our roots are y1, ..., y5, y6, ..., yn

(6) But then we have also solved the quintic equation using radicals which by Theorem 1 above is impossible.

(7) So, we have a contradiction and we reject our assumption in step #1.

QED

SIMPLE!