A pattern within the differences of pythagorean triples?

Question

Looking at some of the more common pythagorean triples, I noticed a trend that there are usually two numbers which differ by only one or two, with more much smaller or larger number in comparison. For example:

$$5,12,13$$ $$8,15,17$$ $$7,24,25$$ $$20,21,29$$ $$12,35,37$$ $$9,40,41$$ and so on...

I was wondering whether there is some deeper algebraic or geometric reason for this, or whether it is just a coincidence with the numbers I have chosen?

Answer 1

This comes from the fact that, when $n$ is odd$$\left(n,\frac{n^2-1}2,\frac{n^2+1}2\right)$$is a pythagorean triple and, when $n$ is even,$$\left(n,\left(\frac n2\right)^2-1,\left(\frac n2\right)^2+1\right)$$is a pythagorean triple too. In the first case, the second and the third numbers differ by $1$ and, in the second case, they differ by $2$.

Answer 2

All pythagorean triples are given by

$$a = m^2 - n^2 ,\ \, b = 2mn ,\ \, c = m^2 + n^2 $$

for any pair $m>n>0$.

In order to have a small $a$ we can set $m=n+k$ with $k$ small and therefore

• $b=2n^2+2nk$
• $c=2n^2+2nk+k^2$

therefore

$$b-c=k^2$$

which tends to be relatively "small".

Answer 3

If you want for example that $$c=b+1$$

you choose $a$ such that

$$a^2=2b+1=(2k+1)^2$$

or $$b=2k^2+2k$$ for $k=3$ you get $$b=24 , a=7 \text{ and } c=25$$

Answer 4

A primitive Pythagorean triple is of the form $$m^2-n^2, 2mn, m^2+n^2$$ where $m\gt n$, $m,n$ have no common prime factor and one of $m,n$ is even.

If $n=1$ you automatically have two numbers which differ by $2$. If $m=n+r$ then you have $2n^2+2nr$ and $2n^2+2nr+r^2$ so if $m-n=1$ you automatically have sides which differ by $1$.

Now look at the first few possible pairs for $m,n$

We have $2,1$ giving $3,4,5$ Then $3,2$ giving $5,12,13$ And $4,1$ with $15, 8, 17$ And $4,3$ with $7, 24, 25$ And $5,2$ with $21, 20, 29$ etc

This last is not one of the ones I caught, but the closeness here is essentially because the numbers $m$ and $n$ are small.

As numbers get larger, cases like $10,7$ giving $51, 140, 149$;

or $10,3$ leading to $91, 60, 109$ become more typical.

Answer 5

The well-known Euclid's formula is $A=m^2-n^2\quad B=2mn\quad C=m^2+n^2$ and generates trivial triples, all primitives, doubles and square multiples of primitives but in a seemingly random pattern. A variation replaces $(m,n)$ with $(2n-1+k,k)$, produces no trivial triples, and generates only the subset where GCD(A,B,C) is an odd square:

$$A=(2n-1)^2+2(2n-1)k\quad B=2(2n-1)k+2k^2\quad C=(2n-1)^2+2(2n-1)k+2k^2$$ This formula generates a distinct pattern of sets that can be seen in the sample below where the increment between values of $A$ in each set is $2(2n-1)$ and $C-B=(2n-1)^2$.

$$\begin{array}{c|c|c|c|c|} n & Triple_1 & Triple_2 & Triple_3 & Triple_4 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41\\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65\\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137\\ \hline \end{array}$$

Setting $n=1$ reduces formula to $$A=2k+1\quad B=2k(k+1)\quad C=2k(k+1)+1$$ and generates only $Set_1$ where $C-B=1$.

Setting $k-1$ reduces the formula to $$A=4n^2-1\quad B=4n\quad C=4n^2+1$$ and generates only the first member of each set where $C-A=2$.

Another unrelated formula generates only the rare triples where $B-A=1$ such as $(3,4,5), (20,21,29), (119,120,169), (696,697,985)$. That formula begins with $A_1,B_1,C_1=3,4,5$ and then builds each triple based on its predecessor: $$A_{n+1}=3A_n+2C_n+1\quad B_{n+1}=3A_n+2C_n+2\quad C_{n+1}=4A_n+3C_n+2$$

There are an infinite number of these but they get scarcer with altitude; Excel can only generate 19 or them before resulting in trailing zeros. The $19^{th}$ is $(211929657785303,211929657785304,299713796309065)$.