A recurrence relation of the Pell and Negative Pell Equation.

Question

In solving for isosceles almost and nearly Pythagorean triples, I have observed that the set of triples:

$$ S_n = \{ (1,1,1), (2,2,3) ,(5,5,7), (12, 12, 17), (29, 29, 41) \dots \}$$

are iso-APTs for odd $n$ and iso-NPTs for even $n$. Moreover, I've observed that the next terms are given by: $$(x+y, x+y, x+2y)$$ where $(x,y)$ is a solution to $x^2-2y^2 = \pm 1$

How can I show that if I have $(x,y)$ as a solution to either signs of $x^2-2y^2 = \pm 1$, then $(x+2y, x+y)$ is a solution to the Pell equation of the opposite sign?

Yes. $(x+2y)^2-2(x+y)^2=x^2+4xy+4y^2-2x^2-4xy-2y^2=-(x^2-2y^2)$ — J. W. Tanner, Mar 11 '24 at 00:33

score 1 · Answer 1 · answered Mar 11 '24 at 00:15

1

Note the two variable "Special Algebraic Identity"

$$ 0 = (a + 2b)^2 - 2(a + b)^2 + a^2 - 2 b^2 \tag1 $$

which is a special case of the four variable identity

$$ 0 = (ad-bc)^2 - ab(c-d)^2 - (a-b)(ad^2-bc^2). \tag2 $$

In equation $(1)$ replace $a$ with $x$ and $b$ with $y$ to get

$$ (x+2y)^2 - 2(x+y)^2 = -(x^2 - 2y^2). \tag3 $$

answered Mar 11 '24 at 00:15

Somos

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This is exactly what I needed. Thank you. – numbergobbler Mar 11 '24 at 02:03
@numbergobbler Thank you for the opportunity to add two new identities to my "Special Algebraic Identity" collection. – Somos Mar 11 '24 at 02:39
I am curious on whether this can be applied to other Pell-like equations.
I have applied this on the case where $x^2 - 2y^2 = n$ by the way.
– numbergobbler Mar 11 '24 at 03:03
@numbergobbler The Wikipedia aritcle Pell's equation gives much information about the long history of solutions. Does it answer your question? – Somos Mar 11 '24 at 03:15

A recurrence relation of the Pell and Negative Pell Equation.

1 Answers1