Mythic Algebra and Other Things

Tuesday, November 5, 2024

Epilogue: More Curves and Fermat

By Michael Griffin, MLS

An alternative statement of Fermat's theorem is that any solutions would have to have coordinates on the Desmos curves (shown in the Prologue) that could be converted into natural number triples. This does happen on the unit circle for n = 2. Note the two coordinate points (0.6, 0.8) and (0.8, 0.6). As fractions these are (3/5, 4/5) and (4/5, 3/5) which convert into the triple 345. Another result occurs below the baseline using fractional values of n. For example the square root is n = 1/2

Zooming into the first quadrant for this we can see that the point (0.25, 0.25) has a solution (0.5, 0.5) on the n = 1 baseline which would convert into the triple 112

That is not quite fitting the rules since ABC are supposed to be distinct. The points on the baseline (0.25, 0.75) and (0.75, 0.25) would become distinct as 134 or 314. Still we might generalize Fermat's theorem as that any natural number solutions would fit the condition of n between or equal to 1/2 and 2, or 1/2 < n < 2. Higher roots would have the same unsolvable problem as higher powers. Or at least intuitively they seem like they should. And the principle would hold that any convertible solution points on the square root curve would never occur on the higher roots curves. However, this is getting away from the blog posts which only dealt with natural numbered powers, not roots.

Addendum

Getting back to the previous blog post about the nth root we may note that there is a valid general principle that any nth root must be smaller than a Euclidean square root. If we then look at the first graph of Desmos curves we see that all of the higher power curves are beyond the second power curve and thus their coordinates are larger than the coordinates of the second power curve. This seems to contradict the principle that higher roots would be smaller than square roots and could thus be mistaken as a graphical proof of Fermat’s Last Theorem. Of course when the decimal coordinates are converted into fractions and then the fractions are converted into triples we would then get smaller roots than the second power, fitting the notion that there are infinitely many solutions for any nth power curve, just none that convert into natural numbers for all three elements of any triple. Which remains to be proved.

Sources

https://www.desmos.com/calculator

Saturday, October 5, 2024

Euclid and Fermat yet again

By Michael Griffin, MLS

This is the fifth article about using ancient Greek formulas with Fermat's Last Theorem. It begins with parts of the previous articles about Euclid's formula:

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

Euclid is credited with a formula to generate all possible Pythagorean triples: for any natural numbers k, m and n, m>n, a triplet of (A,B,C) results from k(m2-n2, 2mn, m2+n2).

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

Does Euclid's formula cover all possible solutions of Fermat’s? It may not matter since it probably does not cover all possible triples anyway. Let's at least explore its root properties.

In general, if A,B,C,N,k,m,n are real numbers then m and n can always be chosen such that the Nth root C = N√(m2-n2)N + (2mn)N. While this may not cover all possible, there are infinitely many solutions in real numbers, just not integers or natural numbers. That is not necessary to comply with Euclid's formula. Restricting these to natural numbers results in Pythagorean triples. If N=2 the result for the Nth root C is Euclid’s definition for the C term, m2+n2:

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

= √(m4+2m2n2+n4 ) = √(m2+n2 )2

If N is beyond 2 we do not get Euclid’s C term and Fermat's Last Theorem means at least one of A,B,C is not an integer. How can we show that?

Consider the nth roots where n is greater than 2. Higher roots would be smaller than the second root which is Euclid's m2+n2. Since they are the C term, higher roots are bigger than the A or B terms. On a number line the terms would arrange as A,B, nth root, Euclid's C. For the A and B terms to be natural numbers so must the m and n terms be.

So we are looking for the nth root C term not to be natural. It's possible domain on the number line goes from the B term to Euclid’s C. Because the nth root is less than Euclid's C we can use some properties of Pythagorean triples. Previous articles discussed the three Greek formulas so we will make a case for each:

If Euclid’s C fits Pythagoras’ formula then C-B=1 and so the domain of the nth root is not large enough after B to make a natural number.

If Euclid’s C fits Plato’s formula then C-B=2 and so the domain of the nth root is not large enough after B to make a natural number except for B+1. In that case C-B=1 and we would have a Pythagorean triple for the second power.

If Euclid’s C fits Euclid’s formula, the domain of the nth root is less than C- B which is m2+n2 - 2mn which factors as (m-n)2. Or if A is the larger term, the domain of the nth root is less than C-A which is m2+n2 - (m2-n2) which is 2n2. For example, for m,n=5,2 then ABC=21,20,29 and the domain of the nth root is less than 8. We have room for natural numbers now, although we can make more restrictions. Since the m and n terms must be natural numbers, so are the domains (m-n)2 and 2n2, so the domain boundaries are only square numbers 1,4,9,16,25, … Higher roots of square numbers are usually not natural numbers. An exception example is like when 2 is the 4th root of 16. Of course the actual C term is below A or B plus 1,4,9,16,25, … and each A or B term is a natural perfect nth root of AN or BN.

There can also be nth roots of combinations of natural numbers like (B+X2). For example, the third root of (2+52) = (2+25) = 27 is 3, a natural number. Euclid's formula is not concluding easily like the other Greek formulas. So this method is not as productive as the prior ways to show Fermat's on Pythagorean triples, using Euclid's formula.

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

Thursday, September 5, 2024

Euclid and Fermat again

By Michael Griffin, MLS

This is the fourth article about using ancient Greek formulas with Fermat's Last Theorem. It begins with parts of the previous article about Euclid's formula:

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

So for AN + BN = CN we use (mN-nN)N + (2mn)N = (mN+nN)N.

First, to see that AN + BN = CN has infinitely many solutions for N, look at the drawing below. The height of the curve CN can be arbitrarily divided into any two parts that will satisfy AN + BN.

The question is now are there any values for m,n that will satisfy:

AN = (mN-nN)N and BN = (2mn)N and CN = (mN+nN)N.

The previous article showed that if we do a binomial expansion for any power of N and then cancel like terms on both sides of the equation what we will have remaining is

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

Putting this solution into Euclid's formula we see:

AN = (mN-mN)N = 0 and BN = (2mm)N = (2m2)N and CN = (mN+mN)N = (2mN)N

So 0 + BN = CN or (2m2)N = (2mN)N so 2Nm2N = 2N(mN)N

Cancelling the 2N on both sides leaves m2N =(mN)N

And the only solution for the exponent is N = 2

So we see that altering Euclid's formula does not change the result when we still used the only solution of an invalid triple where the A term is zero.

We make an intuitive leap to conclude that if the Greek formulas comply with Fermat's, then the solution sets of these formulas will always comply also. While these investigations may then have established Fermat's Last Theorem for all Pythagorean triples, that leaves out all other possible combinations of numbers, that is all ABC’s that do not fit Euclid's formula. These would differ from Euclid's values as:

A = m2-n2 plus or minus 0, 1, 2, ...

B = 2mn plus or minus 0, 1, 2, ...

C = m2+n2 plus or minus 0, 1, 2, ...

If we kept the same factoring of these according to Euclid's formula there would be extra bits resulting in

BN = (2mn)N = 2(+/- leftover terms) plus or minus 1, 2, ... and so m=n is NOT the solution. Also since the ABC’s do not fit Euclid's formula there is no solution for N = 2.

We'd be starting all over, with an infinite endless supply of unproved triples, minus the set of Pythagorean triples.

Does Euclid's formula cover all possible solutions? It does generate all Pythagorean triples if a K factor is put onto its formula for ABC. For example if K is equal to 2 then (3,4,5) and (5,12,13) become (6,8,10) and (10,24,26). And if we interpret Euclid's derivation of the formula as a proof of all possible solutions for N=2 then in fact to prove Fermat's is it sufficient to show that Euclid's has no solutions beyond N=2? That has now been shown.

It would be sufficient if it could be shown that all solutions of AN + BN = CN

must comply with Euclid's. That is easy for many if not all triples, so it is not enough. We can only pursue it as another possible proof just on the set of Pythagorean triples.

In general, if A,B,C,N,k,m,n are real numbers then m and n can always be chosen such that the Nth root C = N√(m2-n2)N + (2mn)N. There are infinitely many solutions in real numbers, just not integers or natural numbers. That is not necessary to comply with Euclid's formula. Restricting these to natural numbers results in Pythagorean triples. If N=2 the result for the Nth root C is Euclid’s definition for the C term, m2+n2. If N is beyond 2 we do not get Euclid’s C term and Fermat's Last Theorem means at least one of A,B,C is not an integer. The next post will discuss that.