Euclid and Fermat

By Michael Griffin MLS

This is the second article about using ancient Greek formulas with Fermat's Last Theorem.

In Fermat's problem:

AN + BN = CN  has infinitely many solutions for N but the highest whole number solution is 2, as with the sides of a right triangle in the Pythagorean theorem: A2 + B2 = C2 

As with Fermat’s theorem, this is easily true if we make no restrictions on the values of A,B, or C.  For the sake of argument and the rules of the game, A,B,C, and N in Fermat’s,  are natural numbers only. 

Euclid is credited with a formula to generate all possible Pythagorean triples: for any natural numbers m and n, m>n, a triplet of (A,B,C) results from

(m2-n2, 2mn, m2+n2).  

For example, (m,n) of (2,1) makes (3,4,5), the first such triple, for 9+16=25.

Let us begin by expanding Euclid’s formula in the power of two:

 (m2-n2)2 + (2mn)2 = (m2+n2)2   is   A2 + B2 = C2   and we get:

 m4-2m2n2+n4 + 4m2n2 =  m4+2m2n2+n4

Cancelling terms leaves  -2m2n2 + 4m2n2 =  2m2n2  or  4m2n2 =  4m2n2  

A trivial result that we see has any possible solutions for m and n.

When we go to higher powers we get instead binomial expansions, with alternating +/- terms on the left, and only positive terms canceling on both sides of the equation.

If N=3

 (m2-n2)3 + (2mn)3 = (m2+n2)3   is   A3 + B3 = C3   and we get:

(m2-n2)( m4-2m2n2+n4 ) + 8m3n3=  (m2+n2) (m4+2m2n2+n4 )   then distributing:

m6-2m4n2+m2n4 - n2m4+2m2n4 - n6 + 8m3n3  = m6+2m4n2+m2n4+n2m4+2m2n4+n6

Cancelling identical terms on both sides leaves:

-2m4n2 - n2m4 - n6 + 8m3n3 = 2m4n2+n2m4+ n6

 Except for the term 8m3n3 everything on the left is just the negative of what is on the right so let us combine like terms:

8m3n3 -4m4n2 -2n2m4 -2n6 = 0   and then we can divide all by 2

4m3n3 -2m4n2-n2m4 -n6  = 0  

then what remains after expansion and cancellation is:

4m3n3 - 3m4n2 - n6 = 0    and m=n is the solution of this. That is not allowed since the only resulting triple would be (0, 2m2, 2m2) which also shows why we need m>n.

   

N = 3 is the first case where  N is odd. What about when N is even? Let's examine the expansion of the A term with its subtraction components

if N = 4: 

(m2-n2)4 = (m2-n2)2(m2-n2)2  = ( m4-2m2n2+n4 )( m4-2m2n2+n4 ) =  

m8-2m6n2+m4n4 -2m6n2+4m4n4 -2m2n6+ m4n4 -2m2n6+n8 

The positive terms would be cancelled with their matches in the C term on the other side of the equation, leaving the negative terms:  -4m6n2 -4m2n6

If we bring the remaining C terms over to the left side and write out the full equation we now have:

 (2mn)4 -8m6n2 -8m2n6 = 0     Then dividing by 8:

 2m4n4 -m6n2 -m2n6 = 0       And once again m=n is the natural number solution.

In general, a binomial expansion and cancellation leaves:

(2mn)N  =  2(+/- leftover terms)       and so m=n is the solution.

By intuition, any higher powers would result in the same balancing of +/- terms and a final result with m=n the only solution. So if N>2 there is no m>n solution to Euclid’s formula.

∎

QED

Note, it is an unfounded assumption that just because Euclid’s formula applies to N=2 it should generalize to all powers. What this exercise shows is that it cannot.

Sources

Fermat's Last Theorem

From Wikipedia, the free encyclopedia

This page was last edited on 31 August 2020

https://en.wikipedia.org/wiki/Fermat's_Last_Theorem

Formulas for generating Pythagorean triples

From Wikipedia, the free encyclopedia

This page was last edited on 24 June 2020

https://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples

Proof of Fermat's Last Theorem for specific exponents

From Wikipedia, the free encyclopedia

This page was last edited on 19 October 2020

https://en.wikipedia.org/wiki/Proof_of_Fermat%27s_Last_Theorem_for_specific_exponents#Two_cases

    

Pythagorean triple

From Wikipedia, the free encyclopedia

This page was last edited on 3 September 2020

https://en.wikipedia.org/wiki/Pythagorean_triple#Special_cases_and_related_equations

 Pythagoras and Fermat

By Michael Griffin MLS

This is the first article in a series about using ancient Greek formulas with Fermat's Last Theorem.

Although it has been solved by modern methods, Fermat's Last Theorem will continue to draw anyone curious as to a method Fermat himself might have used to solve it. To that end,  a few idle investigations are reported here, with some conclusions that by no means make a rigorous proof.

In Fermat's problem:

An + Bn = Cn  has infinitely many solutions for n but the highest whole number solution is 2, as with the sides of a right triangle in the Pythagorean theorem:

A2 + B2 = C2 

And Pythagoras found that you can always generate solutions from

A2 =  B + C where C - B = 1

This is almost a midpoint formula since the length of A2 is divided by a midpoint which is added to length B to make length C. The clear property of this type of Pythagorean midpoint triple is that every single  odd  digit is the beginning of a set of Pythagorean triples. 3, 5, 7, … etc  all fit the same formula that generates triples. We get (3,4,5) (5,12,13) (7,24,25) … etc. The same cannot be said for the second digit B. We cannot say that every even digit B is part of a set of triples, but there is something we can say. There is a clear relation between subsequent B digits proceeding off of an increasing sequence of triples:  the next B digit that fits a triple pattern is 4n more than the previous: 

Bn = Bn--1 + 4n

Where n is the ordinal value of that triple. So after 7 is 9 for the 4th triple, n=4, 24+4(4)=24+16=40 so the next triple is (9,40,41) and 81+1600=1681=412.

 It doesn't take much work to see this pattern. 

Set up a row of the first 6 even members of Pythagorean midpoint triples:

4, 12, 24, 40, 60, 84

Then make a row of the increasing distance between members:

8, 12, 16, 20, 24

The common factor is 4 so divide it out:

2, 3, 4, 5, 6

Which gives the n-ordinal value of that member’s triple.

So each next B value is 4n more than the previous one. 

Again, Bn = Bn--1 + 4n  formula notation, however that is as far as that goes for now.

We shall set it aside and instead make a generalization of the Pythagorean midpoint formula for triple solutions (A,B,C) where n is any positive whole digit: 

An  = Bn--1 + Cn--1 

As with Fermat’s theorem, this is easily true if we make no restrictions on the values of A,B,C, or n.  For the sake of argument and the rules of the game, we make the unfounded assumption that A,B,C, and n are natural numbers only. 

This unfounded notion leads to a result that there are no solutions for this kind of Pythagorean midpoint triple for n  greater than 2:

If  A3 + B3 = C3  then 

A3  = B2 + C2  and substituting into the first equation:

 B2 + C2 + B3 = C3  rearranging:

B2 +  B3 = C3 - C2  and factoring:

B2 (1 +  B) = C2 (C - 1)

For the smallest pair of C - B = 1  (B,C) is (1,2)

12 (1 +  1) = 22 (2 - 1)

1(2) = 4(1) = 4   2 = 4 false

The larger (B,C) gets, the worse the inequality gets

Take the next pair (2,3) 

22 (1 +  2) = 32 (3 - 1)

4(3) = 9(2) = 18   12 = 18 false

The inequality has grown from a difference of 4 - 2 = 2  to 18 - 12 = 6 

And it will just get bigger. 

Also the larger n gets, the worse the inequality gets:

Instead of n = 3 let n = 4

If  A4 + B4 = C4  then 

A4  = B3 + C3  and substituting into the first equation:

 B3 + C3 + B4 = C4  rearranging:

B3 +  B4 = C4 - C3  and factoring:

B3 (1 +  B) = C3 (C - 1)

For the smallest pair of C - B = 1  (B,C) is (1,2)

13 (1 +  1) = 23 (2 - 1)

1(2) = 8(1) = 8   2 = 8   false

 The inequality has grown from a difference of 4 - 2 = 2  to 8 - 2 = 6 

And it will just get bigger. 

So as pairs (B,C) or the power n increases, so does the inequality.

So if  C - B = 1  there is no A,B,C for n > 2 such that 

An  = Bn--1 + Cn--1

∎

QED

This only applies to Pythagorean midpoint triples based on the unfounded: 

An  = Bn--1 + Cn--1 Other articles will address formulas made by Euclid and Plato.

Sources

Fermat's Last Theorem

From Wikipedia, the free encyclopedia

This page was last edited on 31 August 2020

https://en.wikipedia.org/wiki/Fermat's_Last_Theorem

Formulas for generating Pythagorean triples

From Wikipedia, the free encyclopedia

This page was last edited on 24 June 2020

https://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples

Proof of Fermat's Last Theorem for specific exponents

From Wikipedia, the free encyclopedia

This page was last edited on 19 October 2020

https://en.wikipedia.org/wiki/Proof_of_Fermat%27s_Last_Theorem_for_specific_exponents#Two_cases

    

Pythagorean triple

From Wikipedia, the free encyclopedia

This page was last edited on 3 September 2020

https://en.wikipedia.org/wiki/Pythagorean_triple#Special_cases_and_related_equations

 Prologue: Curves and Fermat

By Michael Griffin, MLS

In Fermat's Last Theorem, An + Bn = Cn  has infinitely many solutions for n but the highest whole number solution is 2, as with the sides of a right triangle in the Pythagorean theorem. These solutions are called Pythagorean triples; for example the first natural numbers triple ABC is 345 since 9 + 16 = 25. Although it has been solved by modern methods, Fermat's Last Theorem will continue to draw anyone curious as to a method Fermat himself might have used to solve it. To that end,  a few idle investigations are reported here in future posts, with some conclusions that by no means make any proofs except on the set of Pythagorean triples. While these results might make for some interesting algebra, they are so trivial that it is not surprising if no one has published them before. To show just how trivial they are, we will look at some graphs that show the Captain Obvious nature of why any  Pythagorean triple would not work at a higher power. 

We begin with a graph from an online textbook (schoolbag) which gives a geometric set-up for a proof of Euclid's formula to generate Pythagorean triples:

What matters to us is that the general Pythagorean theorem is converted Into a unit circle here by division:

If  A2 + B2 = C2  then    (A2 + B2 = C2) / C2   and   A2 / C2 +  B2 / C2 =  C2 / C2 

and    A2 / C2 +  B2 / C2 = 1   so we define A2 / C2 = X2  and  B2 / C2 = Y2 

So we have a unit circle of X2 +  Y2 = 1 shown in the first quadrant above.

The natural number solutions for Pythagorean triples will lie on the unit circle. 

It is easy enough to show that these points will stay on the unit circle and not appear in higher curves.

With the same kind of division process we can turn higher powers into unit equations and graph them. So A3 + B3 = C3 becomes X3 +  Y3 = 1, A4 + B4 = C4 becomes X4 +  Y4 = 1, A5 + B5 = C5 becomes X5 +  Y5 = 1, etc. Using the online Desmos graphing calculator we can show these equations and their curves. Even powers give closed figures starting with a circle n = 2 whereas odd powers have unbounded lines that in the first quadrant curve out near the even powers. 

We zoom in on the first quadrant because it would be the positive values to fit Fermat's theorem. Zooming in also shows that as n increases the curvature moves further out from the circle in an orderly progression. The straight line is our beginning  baseline when n = 1 so

X + Y= 1.

 

The curves never intersect except at the points (0,1) and (1,0). So any natural number Pythagorean triples never happen on higher power curves, which is a visual depiction that these triples would never provide solutions to Fermat's Last Theorem. All of these graphs simply serve as alternative proofs of the algebra to come.

Sources

https://www.desmos.com/calculator

EUCLID'S METHOD FOR FINDING PYTHAGOREAN TRIPLES

https://schoolbag.info/mathematics/numbers/79.html