## Primitive Pythagorean triples (a, b=a+1, c)

Let P be the perimeter

(1)

Can you explain the following:

The generators   $m$   and   $n$     of the first few triangles are the Pell numbers
2,   5,   12,   29, …,   13860,   and   1,   2,   5, …,   5741,   respectively.

The lengths of their hypotenuses are the Pell numbers,   5,   29,   169,   …

A000129
1,   2,   5,   12,   29,   70,   169,   408,   985,   2378,   5741,   13860,   33461,   80782,   195025,   470832,   1136689,   2744210,   6625109,   15994428,   38613965,   93222358,   225058681,   543339720,   1311738121,   3166815962,   7645370045,   18457556052,   44560482149,   107578520350,   259717522849,   …

The solutions of   $u^2 \; - \; 2 \, v^2 \; = \; \pm 1$   are given by
$( \,u_{k}, \; v_{k} \,) = ( \,Q_{k}, \; Q_{k} \,)$
where   $k \geq 1$

The corresponding generators   $m_{k}, \; n_{k}$   are given by
$m_{k} \; - \; n_{k} \; = \; u_{k} \; = \; Q_{k}$,   and
$n_{k} \; = \; v_{k} \; = \; P_{k}$

So,   $m_{k} \; = \; n_{k} \; + \; Q_{k} \; = \; P_{k} \; + \; Q_{k} \; = \; P_{k+1}$

The generators of PPT with consecutive legs are given by the consecutive Pell numbers   $P_{k+1}$   and   $P_{k}$,   where   $k \geq 1$

The lengths   $x_{k}$   and   $y_{k}$   of the corresponding legs are given by

$x_{k} \; = \; ( \,m_{k} \,)^2 \; - \; (n_{k})^2$
$= \; ( \,P_{k+1} \,)^2 - ( \,P_{k} \,)^2$
$= \; ( \,P_{k+1} + P_{k} \,) \,( \,P_{k+1} - P_{k} \,)$
$= \; Q_{k+1} \, Q_{k}$

and

$y_{k} \; = \; 2 \, m_{k} \, n_{k} \; = \; 2 \, P_{k+1} \, P_{k}$

The hypotenuse:

$z_{k} \; = \; ( \,m_{k} \,)^2 \; + \; ( \,n_{k} \,)^2$
$= \; ( \,P_{k+1} \,)^2 \; + \; ( \,P_{k} \,)^2$
$= \; P_{2 \,k + 1}$

we have the identity

$( \,Q_{k} \, Q_{k+1} \,)^2 \; + \; ( \,2 \, P_{k} \, P_{k+1} \,)^2 \; = \; ( \,P_{2 \,k+1} \,)^2$

conversely, suppose   $( \,u_{k}, \; v_{k} \,) = ( \,Q_{k}, \; P_{k} \,)$   is a solution the Pell equation   $u^2 \; - \; 2 \, v^2 \; = \; \pm 1$

$( \,Q_{k} \,)^2 \; - \; 2 \, ( \,P_{k} \,)^2 \; = \; \pm \, 1$

Then

$m_{k} \; - \; n_{k} \; = \; Q_{k}$
$n_{k} \; = \; P_{k}$
$m_{k} \; = n_{k} + Q_{k} \; = \; P_{k} + Q_{k} \; = \; P_{k+1}$,    and
$m_{k} + n_{k} \; = \; P_{k+1} + P_{k} \; = \; Q_{k+1}$

So

$x_{k} \; = \; (m_{k})^2 \; - \; (n_{k})^2 \; = \; Q_{k} \; Q_{k+1}$

and    $y_{k} \; = \; 2 m_{k} \; n_{k} \; = \; 2 \, P_{k} \; P_{k+1}$

then    $x_{k} - y_{k} \; = \; Q_{k} \; Q_{k+1} - 2 \, P_{k} \; P_{k+1} = (-1)^{k}$

that is,    $| \, x_{k} - y_{k} \, | \; = \; 1$

$x_{k}$   and   $y_{k}$   are consecutive integral lengths.

math grad - Interest: Number theory
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### 2 Responses to Primitive Pythagorean triples (a, b=a+1, c)

1. John McMahon says:

Pytharogas’s Theorem: a^2 + b^2 = c^2 yields Primitive Pythagorean Triangles with consecutive legs, when:

a = (Pn + 1)^2 – (Pn)^2

b = 2 (Pn) (Pn + 1)

c = (Pn + 1)^2 + (Pn)^2

where (Pn), (Pn + 1), (P2n), (P2n + 1) and (P2n + 2) are all Pell numbers, and n is ≥ 0.

I am going to use three well-known Pell Identities to show that the perimeter of a consecutive-legs Primitive Pythagorean Triangle, calculated as per the equations above, is (P2n + 2).

I am going to use the letter d to designate the perimeter, in order to avoid confusion between P, the perimeter, and P(n), the notation for Pellian numbers.

Adding a, b and c gives the perimeter;

d = (Pn + 1)^2 – (Pn)^2 + 2 (Pn) (Pn + 1) + (Pn + 1)^2 + (Pn)^2

= 2 (Pn + 1)^2 + 2 (Pn) (Pn + 1)

= 2 (Pn + 1) [ P(n) + P(n + 1) ]

The first identity I would like to use, is from Thomas Koshy, Pell and Pell-Lucas Numbers with Applications, 7.7 Fundamental Pell Identities, (1) on page 122:

(1) (Pn) + (Pn – 1) = (Qn)

where (Qn), (Qn + 1), (Q2n), (Q2n + 1) and (Q2n + 2) are all Pell-Lucas numbers, and n is ≥ 0; see A001333.

(Pn + 1) + (Pn) = (Qn + 1)

So that the perimeter,

d = 2 (Pn + 1) [ (Pn) + (Pn + 1) ]

now becomes;

d = 2 (Pn + 1) (Qn + 1)

The second identity is also from Thomas Koshy, Pell and Pell-Lucas Numbers with Applications, 7.7 Fundamental Pell Identities, (28) on page 123:

(28) (P2n) = 2 (Pn) (Qn)

[P2(n + 1) ] = 2 (Pn + 1) (Qn + 1)

or

(P2n + 2) = 2 (Pn + 1) (Qn + 1)

The third identity is also from Thomas Koshy, Pell and Pell-Lucas Numbers with Applications, 7.7 Fundamental Pell Identities, (18) on page 122:

(18) (Pn + 1)^2 + (Pn)^2 = P(2n + 1)

This is the Hyptenuse, c, above.

This identity is also to be found in A Primer on the Pell Sequence and Related Sequence | Marjorie Bicknell at (8) on Page 2.