Scatterplot of first pythagorean triples inside 4500

I had the pleasure of filling in for a colleague in her Discrete Structure class, which is basically our introduction to proof class. Currently they are covering some introductory concepts from Number Theory. One of my favorite results had been covered in the class just before I needed to fill in: The Euclidean Algorithm. Anyways, toward the end of class we introduced the concept of Pythagorean Triples.

Def: An ordered triple, [tex](a, b, c)[/tex] of positive integers such that [tex]a^2 + b^2 = c^2[/tex] is called a pythagorean triple.

EX: (3, 4, 5)

EX: (6, 8, 10)

EX: (9, 12, 15)

EX: (5, 12, 13)

Notice that the first three examples are of the same “type.” That is, the second and the third are multiples of the first. In fact, we could state the following theorem.

Thm: If [tex](a, b, c)[/tex] is a Pythagorean triple then for any positive integer [tex]k[/tex], [tex](ka, kb, kc)[/tex] is also a Pythagorean triple.

Now, the last of my examples is fundamentally different from the other three. In fact, we could create a new definition.

Def: If the greatest common divisor of the pythagorean triple [tex](a, b, c)[/tex] is [tex]1[/tex], then it is called a primitive pythagorean triple. (In other words, if they have no common factors it is a primitive pythagorean triple)

I left the class with an question and recommended that if they were interested they should do a little research to find the answer.

Question: How many primitive pythagorean triples are there? That is, are there infinitely many or only some finite number of them?

Here are some interesting facts from Wikipedia about Pythagorean triples:

- Exactly one of a, b is odd, c is odd.
- Exactly one of a, b is divisible by 3
- Exactly one of a, b is divisible by 4
- Exactly one of a, b, c is divisible by 5
- Exactly one of a, b, (a+b), (a?b) is divisible by 7
- At most one of a, b is a square
- The hypotenuse, c, is an odd number
- Every integer greater than 2 is part of a Pythagorean triple
- The area (A=ab/2) is not an integer
- For any Pythagorean triple, the product of the two nonhypotenuse legs is always divisible by 12, and the product of all three sides is divisible by 60.
- There exist infinitely many primitive Pythagorean triples whose hypotenuses are squares of natural numbers.
- There exist infinitely many primitive Pythagorean triples of which one of the arms is the square of a natural number.
- For each natural number n, there exist n Pythagorean triples with different hypotenuses and the same area.
- For each natural number n, there exist at least n different Pythagorean triples with the same arm a, where a is some natural number
- For each natural number n, there exist at least n different triangles with the same hypotenuse.
- In every Pythagorean triple, the radius of the in-circle and the radii of the three ex-circles are natural numbers.
- There is no Pythagorean triple of which the hypotenuse and one arm are the arms of another Pythagorean triple.

An actual proof is required here, not a circuitous restatement of the fact that (2p)^n has an nth root.

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Fermat’s Last Theorem can be proved by expressing the equation as the difference binomial expression (p +q)^n -(p- q)^n which always has 2 as a common factor, so that the nth root of this expression will contain the irrational number 2^(1/n). This irrationality will only be corrected if p=q. This proves Fermat’s Last Theorem.

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Well, if you can briefly prove that what remains of the expression cannot reconcile that irrational root, then you’ll go down in history. But I must emphasize that the common factor of two does not surprise me, since that difference has to be even.

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What I think you are pointing out is that it is just possible for the irrational 2^(1/n) to be corrected if p and q are unequal. If p and q are unequal then the result should be rational not the irrational one required to correct the 2^(1/n). An example is needed if this is to be refuted.

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For example: For the case n = 3, how would you prove that 6qp^2 + 2q^3 cannot be a perfect cube?

According to you it is simply obvious that once you pull the common factor of 2 out and take the cube root, 2^(1/3) cannot be reconciled by any of the factors within the radicand of (3qp^2 + q^3)^(1/3). In other words, 4 cannot be a factor of 3qp^2 + q^3, unless more 2’s but not 8’s are factors. This IS obvious when p is even and q is odd, since 3qp^2 + q^3 would have to be odd. But what about when q is even and p is odd?

Now, I’m not quite sure what you mean by, “An example is needed if this is to be refuted.”

If you are requesting a counterexample, I can tell you it doesn’t exist. As I hope you know,

Fermat’s Last Theorem was proved almost 20 years ago. A counterexample would debunk the proof and the theorem. BTW there is a relatively short proof for the case n=3, but I don’t believe it involves your line of reasoning.

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Actually, it seems there are many proofs for n=3. Here is one.

Theorem: Euler’s Proof for FLT: n = 3

x3 + y3 = z3 has integer solutions -> xyz = 0

(1) Let’s assume that we have solutions x,y,z to the above equation.

(2) We can assume that x,y,z are coprime. [See here for the proof]

(3) First, we observe that there must exist p,q such that (see here for proof):

(a) gcd(p,q)=1

(b) p,q have opposite parities (one is odd; one is even)

(c) p,q are positive.

(d) 2p*(p2 + 3q2) is a cube.

(4) Second, we know that gcd(2p,p2+3q2) is either 1 or 3. (see here for proof).

(5) If gcd(2p,p2+3q2)=1, then there must be a smaller solution to Fermat’s Last Theorem n=3. (see here for proof).

(6) Likewise, if gcd(2p,p2+3q2)=3, then there must be a smaller solution to Fermat’s Last Theorem n=3. (see here for proof).

(7) But then there is necessarily a smaller solution and we could use the same argument on this smaller solution to show the existence of an even smaller solution. We have thus shown a condition of infinite descent.

QED

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Well, that proof has links to fill in gaps, minor proofs.

So, here’s the link to the above proof:

http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-proof-for-n3.html

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The middle sentence of my comment of 15 March 2014 is not correct, apologies. I can summarise my views as follows. In the binomial difference expansion of (p +q)^n – (p- q)^n , 2 is always a common factor of all the binomial terms, regardless of whether or not p equals q.

If p equals q , then the binomial difference expansion can be condensed to 2(2^[n-1]. q^n). This coefficient of q^n will only be 2^[n-1] provided p equals q. Taking the real nth root, we then

have 2^(1/n) [2^{1 -(1/n)} .q] equalling q +q, but only if p equals q.

If p does not equal q , the coefficient of q will not be 2^{1 -(1/n)}, so that the irrationality of 2^(1/n) will not be corrected but will persist. This proves Fermat’s Last Theorem, but any counter example needs to correct the irrationality of 2^(1/n).

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Your views are clear and consistent. The issue is: why should we be especially convinced that the remaining terms of the binomial difference cannot reconcile the 2 (contain n – 1 2’s as factors in their sum)? How are you so sure that what’s in the parentheses below cannot contain exactly n – 1 2’s as factors?

2( nC1 pn – 1 q + nC3 pn – 3 q3 + . . . + nCn-1 p1 q n – 1 )

or (if n is odd)

2( nC1 pn – 1 q + nC3 pn – 3 q3 + . . . + q n )

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Ah, the subscripts and supers did not come out in the post!

In ‘summation’, how do we know that the sum of the odd terms of the binomial difference for the exponent n does not contain exactly n – 1 2’s as factors?

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In reply to your comments of 13 April 2014, I have altered my equations slightly by assuming that q can be replaced by p/r, with r being rational and greater than or equal to 1, but is not otherwise an integer. This avoids p and q having common factors.

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Further to my comments of 4 May 2014, my replacement of q by p/r seems to solve the remaining problems. 1/(r^n)will be a common factor of all the binomial terms. If r^n exceeds 1, then the original 2^(1 -1/n) will decline and will cease to correct the irrationality of 2^(1/n).

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My comment of 8 May 2014 contains one confused error on my part, the word increase should replace the word decline. Apologies, I hope this clarifies.

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Again, you really must flesh out your statements to at least outline their proofs. But if at this point you are tinkering substantially with a possible proof, history would suggest it will neither fit in a margin nor a couple of tightly spaces pages.

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. . . spaced . . .

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For the integer nth root for n greater than 2,

[(p +q)^n – (p-q)^n]^(1/n) = 2^(1/n). [ (n*1)p^(n-1)q +(n*3) p^(n-3) q^3 +(n*5) q^5…]^1/n. For q substitute p/r on the right hand side of the equation. = 2^(1/n). (p/r) [(n*1) r^(n-1) +(n*3) r^(n-3) + (n*5) r^(n-5) …..]^(1/n) which will be irrational unless r=1 and p=q, demonstrated as follows. If r=1, then 2^(1/n)p [(n*1) + (n*3 +(n*5)…]^(1/n) will be 2(1/n)p [2^(n-1)]^(1/n) which is rational when r =1. If r is greater than 1, and p is greater than q, then

2^(1/n). (p/r) [(n*1) r^n-1) +(n*3) r^(n-3) +(n*5) r^(n-5) …]^(1/n) is less than 2^(1/n) p 2^(1-[1/n]) and will be irrational since

(1/r) [(n*1)r^(n-1) +(n*3)r^(n-3) +(n*5) r^n-5) …}^1/n reduces 2^(1 -{1/n}) which consequently will not correct the irrationality of 2^(1/n), with r greater than 1. This is the factorial explanation and proof of Fermat’s Last Theorem. From Peter L. Griffiths. Am I wrong?

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What do you mean here?

(1/r) [(n*1)r^(n-1) +(n*3)r^(n-3) +(n*5) r^n-5) …}^1/n reduces 2^(1 -{1/n})

I would run what you have so far by a professor who specializes in combinatorics or number theory.

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While it’s definitely true that (1/r) [(n*1)r^(n-1) +(n*3)r^(n-3) +(n*5) r^n-5) …}^1/n must be less than 2^(1 -{1/n}), [(n*1)r^(n-1) +(n*3)r^(n-3) +(n*5) r^n-5) …}^1/n must be more than

2^(1 -{1/n}).

On the face of it, it appears you must now prove that [(n*1)r^(n-1) +(n*3)r^(n-3) +(n*5) r^n-5) … ] cannot contain n – 1 2’s as factors.

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Reference your comment 10.27 27 July, if r equals 1, then (1/r)[(n*1)r^(n-1)+(n*3)r^(n-3) +n*5r^(n-5) …]^(1/n) will equal 2^(1 -{1/n}) provided r equals 1. If r is greater than 1, then 2^(1 -{1/n}) will be reduced and so will cease to correct the irrationality of 2^(1/n). This (1/r) is conveniently a common factor of the binomial expansion. I am not sure that my first lines are clear so I will repeat. (1/r)[(n*1)r^(n -1) +(n*3)r^(n-3) +(n*5)r^(n-5)…]^(1/n) will equal 2^(1-{1/n}) provided r equals 1.

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Yes, my last comment assumes r is greater than 1. But that is where we started, with just p and q, and there was a need to rule out the possibility of n – 1 2’s as factors from an expression.

Now, when r is greater than 1, there appears to remain an expression from which we again must rule out that same possibility.

And yes, the expression you are talking about being less than 2^(1 – 1/n) indeed is less than that when r is greater than 1, but only when multiplied by 1/r. Without that factor it is greater than 2^(1 – 1/n) when r is greater than 1, and you must rule out the possibility it contains n – 1 2’s as factors. Because if that happens, division by r will not result in irrationality.

I repeat: On the face of it, it appears you must now prove that [(n*1)r^(n-1) +(n*3)r^(n-3) +(n*5) r^n-5) … ] cannot contain n – 1 2′s as factors.

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Sorry, last comment (first paragraph changed) should have read:

Yes, my last comment assumes r is greater than 1. When r = 1 we get the rather trivial case of p = q and thus taking the nth root of (2p)^n. So, I was focusing only on r being greater than 1.

Before you introduced the r, when we were working with just p and q, and there was a need to rule out the possibility of n – 1 2′s as factors from an expression.

Now, when r is greater than 1, there appears to remain an expression from which we again must rule out that same possibility.

And yes, the expression you are talking about being less than 2^(1 – 1/n) indeed is less than that when r is greater than 1, but only when multiplied by 1/r. Without that factor it is greater than 2^(1 – 1/n) when r is greater than 1, and you must rule out the possibility it contains n – 1 2′s as factors. Because if that happens, division by r will not result in irrationality.

I repeat: On the face of it, it appears you must now prove that [(n*1)r^(n-1) +(n*3)r^(n-3) +(n*5) r^n-5) … ] cannot contain n – 1 2′s as factors.

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Please explain what you mean by n-12’s.

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We know the difference in question is even, has that common factor of 2. If the expression in question has exactly n – 1

two’s as factors, then the nth root of the difference will not necessarily be irrational. You would be taking the nth root of 2^n.

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Of course, we know from Fermat’s Last Theorem that the nth root of the difference must be irrational. But we have to prove it.

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If 2^(1 -[1/n])is reduced then it will not correct the irrationality of

2^(1/n) which will persist when r is greater than 1.

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Not necessarily. Suppose it is reduced by a fraction. It will still correct the irrationality, but the result will be 2 multiplied by that fraction.

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Is your fraction inside the power or outside the power? If the fraction is inside the power then the irrationality of 2^(1/n) will persist.

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Precisely! It’s outside. In fact, in our most recent expression, prior to taking the nth root, it was 1/r^n (inside the radical in other words), or 1/r after taking the root (or pulled outside of the radical).

If you can isolate those powers of 2 inside the radical, prove you have less than 2^(n – 1) and prove it must be whole number, then you have proved irrationality. Or you must prove there must be fewer than n – 1 factors of 2 inside the radical some other way.

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bbbI can challenge you to give an example of a decline of 2^(1 -{1/n}) which when multiplied by 2^(1/n) leaves a rational product, that is which corrects the irrationality of 2^(1/n). The rational product of 2 and indeed any rational product only arises when 2^(1/n) is multiplied by the full 2^(1 -{1/n}) that is when r equals 1.

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A decline by 1/2 or 1/3 outside the radical would do it. Or a decline by 1/2^n or 1/3^n inside the radical, which is the same thing.

You could also challenge me to come up with a counterexample to FLT. But what would that prove, literally?

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The decline of 1/2 0r1/3 which you mention outside the radical would still have to be multiplied by 2^(1/n) producing irrationals. The decline by 1/2^n or 1/3^n which you mention inside the radical would not affect the irrationality. In spite of my disagreements I think there is probably a neater way of explaining this argument. Of course if you are right then you will have your counterexample to FLT.

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1/2[ 2^(1 -{1/n}) ] times 2^(1/n) equals 1

1/3[ 2^(1 -{1/n}) ] times 2^(1/n) equals 2/3

Find a counterexample to FLT and you’ll be in the papers. The absence of a counterexample is not the same as proof, it just heightens and maintains a conjecture, and motivates a proof.

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Demonstrating that a counterexample cannot exist is proof, of course.

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Further to my comment of 28 July 2014, r is rational being p/q, so that if r is greater than 1, the irrational 2^(1 -[1/n]) which contains 1/r will be reduced by a rational amount leaving another irrational amount, confirming FLT.

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Well, aren’t there multiple reductions of 1/r raised to increasing powers WITHIN the radical?

Also, won’t there be a multiplication by the irrational 2^(1/n)?

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The sum of the powers of 1/r should still be rational. The multiplication by irrational 2^(1/n) should still preserve the irrationality unless 2^(1/n) is multiplied by 2^(1 -[1/n]). The irrationality needs to be preserved to confirm FLT.

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But we’re talking about powers of 1/r multiplying different quantities all WITHIN a radical.

And, right, 2^(1/n) won’t bring about irrationality unless it multiplies 2^(1 – 1/n), yet that’s what must be proven, that there cannot exist (n – 1) 2’s inside the radical.

Of course, employing FLT, we know what must be irrational. I thought the goal here was to do the opposite. Figure out what must be irrational in order to prove FLT.

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I stick to my comment of 28 July 2014. It is 2^(1/n) which determines whether the equation is irrational. The reduction of 2^(1 -[1/n]) from r being greater than 1, may be either to a rational or irrational number, neither of which will correct the irrationality of 2^(1/n), which will continue unaffected. I think you have to bear in mind that an irrational number can be corrected but is otherwise unique.

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I have come across the possibility of a counter example if p is very high compared with q, leading to irrational 2^(1/n) being corrected by 1/[2^(1/n)].

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If r is slightly greater than 1, and n is an integer greater than 2, then (r +1)^n -(r-1)^n will be greater than 2^n. If we substitute p/q for r we have ([p/q] +1)^n -([p/q] -1)^n will be greater than 2^n. This can be expressed as (p +q)^n -(p -q)^n will be greater than q^n X 2^n . The nth root of this will be integer 2q, which ( unless p equals q) will always be less than

[(p +q)^n – (p -q)^n]^(1/n). The above may not be sufficient for a complete proof.

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When r=1 and p=q, there are only 2 terms

[(n*1) + (n*3) …]^(1/n) = 2^(1-1/n). However if the 3 term assumption is restored,

[(n*1)(1/r) + (n*3) 1/r^3……]^(1/n) will be less than 2^(1-1/n), with r greater than 1, and

p greater than q, and will be less than 2 when multiplied by 2^(1/n). There can only be equality with 2 terms not the 3 terms assumed in Fermat’s Last Theorem.

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The Wiles proof of Fermat’s Last Theorem lacks any explanation of the absence of the power n=2 being the Pythagorean Triples situation from the conclusions. This could mean that the whole proof is wrong particularly since it is only thought to be understood by a small clique of mathematicians. The situation when power 1/n=1/2 is I think fairly clearly brought out in my comment of 28 December 2015.

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My full proof Fermat’s Last Theorem in less than 300 words is contained in the February 2017 issue of M500

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