Fermat's Last Theorem Proved and Award Offered for Refutation. (by Shafi Ahmed.)
Here we will look at another method of simple proof of Fermat's Last Theorem (FLT) which was published in the booklet "Fermat's Last Theorem Proved and Award Offered for Refutation" in 1990 with a supplement in 1994 which discussed many invalid but interesting criticisms. (This page will be available on the Internet for interest of mathematicians in seeing the proof, however for the Award conditions ( award valid till the end of 2003) and full discussion of criticisms readers are requested to Purchase the book.)
ANOTHER SIMPLE PROOF OF FERMAT'S LAST THEOREM (FLT)
Assume the equation:
an= bn + cn for integer values of (a,b,c) >0 and n any integer >2...Equation (A1). First of all we need a Lemma for this proof which we will call Lemma 2.
Lemma 2 "If a homogeneous second degree expression of the form Ax 2 + By2 +Cxy is a perfect square, then C must equal 2
AB"Proof: As the expression is a perfect square, it must be of the form (Px + Qy)2. Expanding this and equating coeffients: A= P2 , B= Q2 and C= 2PQ = 2
AB as required.Important Note: The Lemma is only Valid and applicable when the expression quoted is given or found to be a perfect square without any necscessity to manipulate values of A, B, C, x and y to "cook" the result a perfect square.
For example manipulating values for (x,y) as (3,4) and A=B=1 we get for the simplified expression 12C + 25
We can then make this equal to a a perfect square say p 2 which is the cooking step. We can then manipulate further values for p so that C is an integer different to 2, for instance if p=19 we can get C = 28, where C is not 2 but the Expression is a perfect square i.e 19 2. Such manipulation of values of the Lemma Expression is totally fallacious because of the COOKING STEP mentioned above and is not applicable for any logical argument to establish or refute anything. Therefore a "cooked" result is not the subject of the Lemma but is irrelevant for our purpose.
Now Observations on equation (A1)
a > b...........(1)
a > c...........(2)
a < (b + c)......(3)
because if a >= (b + c), then LHS of (A1) will be always greater than RHS.
From (1) ,(2), and (3) it follows that a triangle can be constructed with sides representing (a,b,c) in value, and a the largest side since the sum of any two sides is greater than the third.
So let us construct the triangle as shown below and cosider the equation (A2) which is the equation for such a triangle.
BC=a, AC=b, and AB=c Then the equation of the triangle is: a2=b2+c2-2bc Cos A...equation(A2)
where A is the angle between sides b and c.
Now as n > 2 (see A1), Cos A cannot be zero...... (4)
Also it will be seen that a2 < b2 + c2 Because if a2 > = b2 + c2 , then (A1) would be contradicted in having other terms on the RHS if both sides were raised to the power n/2. Therefore Cos A cannot be negative.............(5)
Also from (1) and (2), angle A > 600 So Cos A < ½............(6)
From (4), (5) and (6), 0 < Cos A < ½.........(7)
Now let us subject the equation (A2) to Lemma 2 for test. As the LHS of (A2) is a perfect square, the RHS of (A2) must also be a perfect square and if RHS of (A2) is also to be a homogeneous expression of second degree, by Lemma 2 the coefficient of bc, 2 Cos A should be 2
1.1 = 2.But the value of 2 Cos A is obviously not 2 as we observe from (7). So the possible conclusions are:
Conclusion 1: The RHS of (A2) is homogeneous but not a perfect square denoting either b or c or both as irrational. 0r, Conclusion 2: The RHS of (A2) is not homogeneous and so integral values of (a,b,c) can be found.
As Conclusion 1 immediately establishes FLT we need to look at the Conclusion 2 in detail as follows:
The equation (A1) is the premise from which (A2) is derived by mathematical arguments, and (A1) is a homogeneous equation in (a,b,c) of index n.
So if Conclusion 2 is true then there is a contradiction between (A1) (premise) and (A2) (its mathematical derivation) as to homogeneity which is an important mathematical property of equations.
This contradiction between (A1) and its derivative as to homogeneity shows that equation (A1) is not possible in integers as assumed. QED.
Note 1: If (A2) is homogeneous it falls foul of Lemma 2 and integrity of (a,b,c) is lost. If (A2) is not homogeneous it directly conflicts with (A1) its logical parent as to the property of homogeneity.
So by either conclusion FLT is proved. Readers may try to construct a triangle with three integer sides and see whether homogeneity of (A2) can also be retained at the same time, (i.e. the term bc remaining in tact in the term 2bc CosA) because if so, then, this proof will be refuted. For instance with (a,b,c) = (8,5,7) the equation of the triangle is a 2 = b2 - 2/7 bc which apperas to be a homogeneous second degree equation with bc in tact. However as c=7, 2/7 bc is in effect 2b when the fraction is converted into integers showing homogeneity is lost in the last term. Another way of refuting the proof is to show by starting with a valid homogeneous equation and by further valid logical argument establish that a homogenous derivative equation does not necessarily follow from it. For instance taking (a,b,c) as (5,8,6) we have a valid second degree homogeneous equation 4a2 = b2 + c2. Now by argument as a2 is less than b2 + c2 we have a2 = b2 + c2 - m, where m is any other integer. The equation with m does not seem homogeneous whereas the parent from which it is derived clearly is homogeneous! Establishment of such contradiction is totally spurious and fallacious. Becasue comparing the m equation and its parent, m = 3 a2 i.e the m equation is also homogeneous in (a,b,c) but is deliberately disguised by fallacious symbolism.
Another type of attack by numerical counterexamples to show that homogeneity is irrelevant was done as follws:
Take a homogeneous equation (x + Y) 3 = z3 +6 z x2 as Premise which is obviously true for (x,y,z) = (7,8,9) respectively. As 6zx2 is greater than 0, (x + y)3 is greater than z3, hence (x+y) is greater than z, and hence (x, y, z) forms a triangle (conclusion).
The triangle conclusion can further be used to derive an expression like (A2) in text to show that there is a contradiction in homogeneity with the premise but the premise is not false! Readers with training in logic would note that in general terms it is only shown that (x+y) is greater than z. The full requirements that the sum of any two sides is greater than the third, is NOT shown in general terms but only by the use of the numericals for a particular set of values when (x,y,z) = (7,8,9) respecively. This is the fallacy of 'establishing the general from a particular' and is invalid to obtain the conclusion of the argument. No jiggerey-pokery with numbers can establish any general argument or conclusion, so readers must remain vigilant of this fallacy if they want to find any other example like this to refute the relevance of homogeneity in mathematical derivation. Good Luck!
Note 2: In the book published in 1990 I overlooked to mention the contradiction with Lemma 2 ( as it was so blatantly obvious that I had mentally discarded the use of this Lemma about 40 years ago), so much invalid but seemingly authoritative criticisms on this point (i.e RHS of (A2) is homogeneous, so there is no contradiction on homogeneity!) were rceived from many eminent mathematicians which are discussed in the 1994 supplement.)
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