Let **P** be the perimeter

**(1)**

Can you explain the following:

The generators and 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 are given by

where

The corresponding generators are given by

, and

So,

The generators of PPT with consecutive legs are given by the consecutive Pell numbers and , where

The lengths and of the corresponding legs are given by

and

The hypotenuse:

we have the identity

conversely, suppose is a solution the Pell equation

Then

, and

So

and

then

that is,

and are consecutive integral lengths.

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.

I have adapted this to read:

(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)

I have adapted this to read:

[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.

Yes, indeed