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I developed an alternative to $(m^2-n^2), 2mn, (m^2+n^2)$ for the generation of Pythagorean triples by gleaning the results of 8 million spreadsheet formulas. Almost every time I show any part of my $formula$ in this venue, I seem to get downvoted. I can show how I developed it but that would just be a distraction. Here is my no-frills theorem. The answer I need addresses: "Is my proof flawed?"

Theorem: There is a Pythagorean triple for every pair of natural numbers (n,k). $$\forall n,k \in \mathbb{N}, \exists A,B,C\in \mathbb{N}:A^2+B^2=C^2 \iff A=(2n-1)^2+2(2n-1)k$$

Proof: Let $$A=(2n-1)^2+2(2n-1)k$$

Solving $A^2+B^2=C^2$ for $B$ and $C$, respectively, and substituting $A$, we find that $$B=2(2n-1)k+2 k^2$$ $$C=(2n-1)^2+2(2n-1)k+2k^2$$ We can then see that $$A^2=(2n-1)^4+4(2n-1)^3 k+4(2n-1)^2 k^2$$ $$B^2=4(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$ $$C^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$ $$A^2+B^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4=C^2$$

$\therefore \forall n,k \in \mathbb{N},\exists A,B,C\in \mathbb{N}:A^2+B^2=C^2 \iff A=(2n-1)^2+2(2n-1)k\text{ } \blacksquare$

These generate all triples where GCD(A,B,C) is the square of an odd number. $\mathbf {\text{This includes all primitives}}$ and excludes all non-odd-square multiples of primitives. In the following sample of sets of triplets ($Set_1, Set_2, Set_3, \text{ and }Set_{25}$), we can also see that $\mathbf {(C-B) \text{ is the }n^{th} \text{odd square}}$. In the example: $C_1-B_1=1^2, C_2-B_2=3^2, C_3-B_3=5^2\text{ and }C_{25}-B_{25}=49^2=2401$.

$$\begin{array}{c|c|c|c|c|} \text{$Set_n$}& \text{$Triplet_1$} & \text{$Triplet_2$} & \text{$Triplet_3$} & \text{$Triplet_4$}\\ \hline \text{$Set_1$} & 3,4,5 & 5,12,13& 7,24,25& 9,40,41\\ \hline \text{$Set_2$} & 15,8,17 & 21,20,29 &27,36,45 &33,56,65\\ \hline \text{$Set_3$} & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 \\ \hline \text{$Set_{25}$} &2499,100,2501 &2597,204,2605 &2695,312,2713 &2793,424,2825\\ \hline \end{array}$$

poetasis
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    As stated, the theorem makes little sense. Something about the quantifiers is greatly at odds (thereby making the theorem trivially true). Probably the theorem does not express what you want to express, and consequently your proof argument can hardly be evaluated. As stated, the argument is certainly logically flawed as you nowhere show an equivalence. – Hagen von Eitzen Apr 28 '19 at 17:51
  • ... but at its core, your $A, B, C$ are just as in the standard formulas, with $m\leftarrow 2n-1+k$, $n\leftarrow k$ (so you can only cover the cases where $A$ is odd) – Hagen von Eitzen Apr 28 '19 at 17:59
  • @Hagen von Eitzen The theorem in my paper is more extensive but, here, I stripped out everything except that which says $\text{There is a Pythagorean triple for every pair of natural numbers iff }A=(2n-1)^2+2(2n-1)k$ – poetasis Apr 28 '19 at 18:01
  • As stated, the theorem is satisfied for any $n,k$ by taking $A,B,C$ such that $A^2+B^2 \ne C^2$ and $A \ne (2n-1)^2+2(2n-1)k$. Then the double implication $A^2+B^2=C^2 \iff A=(2n-1)^2+2(2n-1)k$ is true, because both parts are false. – Misha Lavrov Apr 28 '19 at 18:05
  • @ Misha Lavrov I have another theorem in my paper proving that $A=(2n-1)^2+2(2n-1)k +x$ will not produce a valid triple unless $x$ is a multiple of $2(2n-1)$ because then $B$ would not be an integer. This means there are no missing values of $A$ for GCD(A,B,C)=(2n-1), which means there are no missing primitives in my sets. – poetasis Apr 28 '19 at 18:19
  • I'm not discussing the theorem in your paper because I don't see it. I'm saying that the theorem in your question doesn't actually make any nontrivial statement about the existence of Pythagorean triples. – Misha Lavrov Apr 28 '19 at 19:08
  • @Misha Lavrov The question is: Is there a flaw in this proof that there is a Pythagorean triple for every pair of natural numbers? The standard formula will not, for example, produce a Pythagorean triple for $m=n=1$; instead, it will produce $0,2,2$, a trivial 1-dimensional triple. – poetasis Apr 28 '19 at 19:50

1 Answers1

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Your formula $$ (2n-1)^2 + 2(n-1)k, \quad 2(2n-1)k + 2k^2, \quad (2n-1)^2 + 2(2n-1)k + 2k^2 $$ is equivalent to the standard formula $$ m^2-n^2, \quad 2mn, \quad m^2+n^2 $$ if we set $(m,n) = (2n+k-1, k)$.

Your claim that the formula produces a Pythagorean triple for every pair of let's say positive integers $(n,k)$ is true; your proof has some problems.

One is that you can't "solve for $B$, $C$" given $A$; there are sometimes many Pythagorean triples sharing a side length. To get a true statement, you have to provide the formula in terms of $n,k$ for at least two of $A,B,C$; you might as well provide the formula for all three.

But yes, you've correctly shown that if $A,B,C$ are defined as above, then $A^2+B^2=C^2$.

Misha Lavrov
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  • I know the standard formula generates the same triples as mine but I believe mine does it in a more predictable way. It also allows me so solve for $k$ and find triples with matching sides, perimeters, or areas if they exist. Mine also works for every pair of natural numbers – poetasis Apr 28 '19 at 18:08
  • Anything you can do with one formula, you can do with the other. (Well, except for the triples generated by the standard formula in cases where $m$ and $n$ are both even or both odd.) – Misha Lavrov Apr 28 '19 at 18:09
  • I am reconsidering your answer. It was not immediately apparent that it was helpful because it seemed to dismiss my work off-hand and not address the validity of the proof. For matching sides, I should have used this post where I have addressed multiples. The part of the proof I need to address is to add an intersection $GCD(A,B,C)=(2m-1), m \in \mathbb{N}$ and showing there are no missing values of $A$ as I have done in my paper. That is needed for the $\iff$ part of the proof. – poetasis May 01 '19 at 17:06