Why we can't find a lot more general pattern between pythagorean triples?

Question

Although there are patterns between primitive Pythagorean triples but why we can't find any pattern between all of them(i.e. all existing Pythagorean triplets)or why there is not a very generalisation of this concept?Why it's random? Any specific reason?

Answer 1

There is definitely a classified pattern to all Pythagorean triples. However, the primitive triples are the exciting ones to classify.

Once we have established a classification of all the primitive Pythagorean triples (which is a cool result), the classification of all Pythagorean triples follows automatically and almost trivially. We simply multiply each primitive triple by all possible natural numbers to get all possible triples. Thus authors rarely even bother to mention this broader classification.

Note that for a Pythagorean triple $(a, b, c)$, $\gcd(a, b, c) = 1$ is equivalent to $a, b, c$ being pairwise coprime (and, more generally, $\gcd(a, b, c) = d$ is equivalent to $\gcd(a, b) = \gcd(b, c) = \gcd(a, c) = d$). This is because their defining relationship $a^2+b^2=c^2$ forces any factor shared by two of the numbers to also be shared by the third.

Answer 2

Perhaps you are wondering if there is a pattern in their generation? Euclid's formula here shown as $F(m,k)$ $ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$ generates primitives triangles if $(m,k)$ are mutually prime and of opposite parity. All triples may be generated by multiplying each term by a constant but those are not "interesting" because who cares about $(6,8,10), (9,12,15), etc?$ one problem with Euclid's formula is that it generates trivial triples such as $(0,2,2)$ if $m=k$ and non-primitives if$(m,n)$ are the same parity, i.e. both odd or both even. Viewing the results of this formula "can" give one the impression that there are no "simple" patterns in Pythagorean triples.

I developed a formula $G(n,k)$ that generates only the subset where $GCD(A,B,C)=(2x-1)^2, x\in\mathbb{N}$.

\begin{equation} A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2 \end{equation} This formula makes some patterns easier to "see" (as seen in this sample table). \begin{array}{c|c|c|c|c|c|c|} n & k=1 & k=2 & k=3 & k=4 & k=5 & k=6 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 & 13,84,85 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 & 45,108,117 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 & 85,132,157 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 & 133,156,205 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 & 189,180,261 \\ \hline \end{array}

Note that the difference between $A$-values in any set is always the same $A_x+2(2n-1)=A_{x+1}$. Note that only $Set_1$ or when $k=1$ are all triples primitive and only in $Set_1$ can both $A$ and $C$ be prime.

Note that $(C-B)=(2n-1)^2$ for all triples generated.

Note that $C-A=2k^2$ for all members of any column.

Note how (for primitives) $|B=A|= X$ where $X$ is a prime number $p$ raised to any non-negative power $n$ and $p\equiv \pm 1 \mod 8$. Under $100$, $X\in \{1,7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97\}$

Note how $G(n,k)=F(2m-1+k,k) $ if you separately compare the formulas' triples generated.

Of course there are many more relationships to be found here and here and elsewhere. The two formulas shown appear to be the most useful in "finding triples" but that is another topic altogether.