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

 
 
Let P be the perimeter

a, a+1, c 1
 
(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,   …
 
PELL PPT 1

 

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.

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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.

    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.

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