# Do All Primitive Triples Belong to the Plato, Pythagoras and Fermat Families?

After running into this terminology on the "Formulas for generating Pythagorean triples" Wikipedia page, I was curious whether all triples fit into these categories.

The article states:

Plato: c - b = 1, Pythagoras: c - a = 2, Fermat: |a - b| = 1

Another page lists these as the first primitive triples where c<100:

(3, 4, 5)       (5, 12, 13)     (8, 15, 17)     (7, 24, 25)
(20, 21, 29)    (12, 35, 37)    (9, 40, 41)     (28, 45, 53)
(11, 60, 61)    (16, 63, 65)    (33, 56, 65)    (48, 55, 73)
(13, 84, 85)    (36, 77, 85)    (39, 80, 89)    (65, 72, 97)


However, I noticed that some sets like the (36, 77, 85) triple have a very large difference between each integer. Does this mean that not all pythagorean triples are part of one of those families? And if so, is there a formula for generating all the pythagorean triples?

• I have edited my answer to an old question of yours to include what I have found about the families you mentioned. I hope it helps. May 1, 2019 at 19:10

The views of these people only apply to a small number of triples and sometimes depend on which of $$A or B$$ is odd or even. Below are original functions gleaned from 8 million spreadsheet formulas.

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 see a great variation in the differences between $$A,B,C$$. We can also see that $$\mathbf {(C-B) \text{ is the }n^{th} \text{odd square}}$$. In the example: $$C_1-B_1=1^2, C_2-B_2=3^2, C_3-B_3=5^2\text{ and }C_{25}-B_{25}=49^2=2401$$.

$$\begin{array}{c|c|c|c|c|} \text{Set_n}& \text{Triplet_1} & \text{Triplet_2} & \text{Triplet_3} & \text{Triplet_4}\\ \hline \text{Set_1} & 3,4,5 & 5,12,13& 7,24,25& 9,40,41\\ \hline \text{Set_2} & 15,8,17 & 21,20,29 &27,36,45 &33,56,65\\ \hline \text{Set_3} & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 \\ \hline \text{Set_{25}} &2499,100,2501 &2597,204,2605 &2695,312,2713 &2793,424,2825\\ \hline \end{array}$$ Note: These triples can be generated by a variation of Euclid's formula where $$A=(2m-1+n)^2-n^2\quad B=2(2m-1+n)n\quad C=(2m-1+n)^2+n^2\quad$$ but we will use a formula developed from observations of these sets because it is easier to see that $$n$$ is a set number and $$k$$ is a member number.

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

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 $$B=2(2n-1)k+2 k^2$$and$$C=(2n-1)^2+2(2n-1)k+2k^2$$ We can then show that $$A^2=(2n-1)^4+4(2n-1)^3 k+4(2n-1)^2 k^2$$ $$B^2=4(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$ $$C^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$ $$A^2+B^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4=C^2$$

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

The Plato family $$C-B=1$$ applies only to members of $$Set_1$$ as we can see in the sample above. The Pythagoras family $$C-A=2$$ applies only to the first members of all sets. I haven't heard of the Fermat family $$| A-B |=1$$ and it seems to apply at random as in these examples of $$f(n,k)=A,B,C$$. $$f(1,1)=3,4,5$$ $$f(2,2)=21,20,29$$ $$f(4,5)=119,120,169$$ $$f(9,12)=697,696,985$$ $$f(21,29)=4059,4060,5741$$ $$f(50,70)=23661,23660,33461$$ I stopped checking at $$Set_{50}$$ and I don't see a pattern, do you? Do note that the first two families are nothing more than the members of $$Set_1$$: f(1,k) or the first members of all sets: f(n,1). I hope I have shown you there is much more to explore than the families mentioned.

If you wanted to explore other configurations, I have another set of equations for side $$A$$ even:$$A=2n^2+2{N}n+4kn-4n\text{, }B=2n(2k-1)+(2k-1)^2\text{, }C=2n^2+2n(2k-1)+(2k-1)^2$$ and it generates the same triplets rotated $$90$$ degrees in the example with $$A$$ and $$B$$ reversed.

Update: I have recently inferred, but not proven, that triples where $$GCD(A,B,C)\ne 2$$ and $$GCD(A,B,C)\ne x^2,x\in\mathbb{N}$$ cannot be generated by Euclid's formula $$(A=m^2-n^2\quad B=2mn\quad C=M^2+n^2)$$ alone without a multiplier, as shown here.

General solution. Let $$m\gt n$$ be any pair of integers. To get P. triples, let $$a=m^2-n^2,\ b=2mn,\ c=m^2+n^2$$

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

Above is a complete discussion.

• Does this mean that not all Pythagorean triples are part of one of those families? How about the triple [3,4,5], it satisfies all of the requirements (Plato: c - b = 1, Pythagoras: c - a = 2, Fermat: |a - b| = 1) so does that mean that it is part of all of them or none of the families? Sep 30, 2018 at 7:52
• The formula (Euclid's) gives all triplets. As far as belonging to specific families, it is obvious from the list you presented that some do not. I don't understand what you are looking for. Sep 30, 2018 at 22:35
• I don't understand what is troubling you. Some triplets don't belong to any of the families, some belong to one or more. Looking at your table, it is obvious. (3,4,5) belongs to all three. Oct 1, 2018 at 19:48

Yes,

For $$~a~$$ is an even number then $$~b= \frac{1}{2i}(a^2-i^2)~$$ , where $$~i~$$ is an positive even integer such that $$~i<[a/2]~$$, $$~b~$$ belong to $$~N~$$ , and $$~c= b+i~$$ e.g., $$~a=36~$$

For $$~i=~$$ even positive number $$~2n^2; ~~~n~$$ belong to $$~N~$$

$$~~~~~~= 2,8<[36/2]$$

$$i=2,~~~~ (36,323,325)~$$

$$i=8, ~~~~(36,77,85)~$$

From this it is clear that for even number a, we can find more than one primitive triples, it may have Plato, pythagorean , and Fermat family triples....this is the generating formula for even primitive triples....

For odd generating primitive Pythagorean Triples, I will send it later on.... Regards, Shailesh