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Working with Pythagorean triples, I found that Pell numbers (1,2,5,12,29,70...) provide needed $m,n$ input values for Euclid's formula $\quad(A=m^2-n^2\quad B=2mn\quad C=m^2+n^2)\quad$ when seeking $\quad (B-A=\pm1).\quad$ For example $F(2,1)=(3,4,5)\quad F(5,2)=(21,20,29)\quad F(12,5)=(119,120,169)...\quad. $

I found a formula to generate them in The Online Encyclopedia of Integer Sequences. I also found that I could generate these needed pairs sequentially with a formula found by solving $(B-A)$ for $x$. I generalized the formula and now it generates both positive and negative Pell numbers in sequential order of increasing value. Any input value $n$ that does not yield an integer is not a Pell number. Any input Pell number yields the next (or prior) Pell number depending on the input sign.

$$p=\pm\bigg(n+\sqrt{2n^2+(-1)^{n}}\bigg)\quad\text{where $\pm$ is the sign of the input Pell number. Note:}\frac{n}{|n|}\text{ does not work for }\frac{0}{|0|}$$ For example $$p_{-12}=-\bigg(-12+\sqrt{288^2+(-1)^{-12}}\bigg)=(-1)(-12+17)=-5$$ $$p_{-5}=-\bigg(-5+\sqrt{50+(-1)^{-5}}\bigg)=(-1)(-5+7)=-2$$ $$p_{-2}=-\bigg(-2+\sqrt{8+(-1)^{-2}}\bigg)=(-1)(-2+3)=-1$$ $$p_{-1}=-\bigg(-1+\sqrt{2+(-1)^{-1}}\bigg)=(-1)(-1+1)=0$$ $$p_{0}=\bigg(0+\sqrt{0+(-1)^{0}}\bigg)=(0+1)=1$$ $$p_{1}=\bigg(1+\sqrt{2+(-1)^{1}}\bigg)=(1+1)=2$$ $$p_{2}=\bigg(2+\sqrt{8+(-1)^{2}}\bigg)=(2+3)=5$$ $$p_{5}=\bigg(5+\sqrt{50+(-1)^{5}}\bigg)=(5+7)=12$$

Has anyone seen this formula before? Is it useful or anything else I can't think to ask?

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    $\begingroup$ If we fix the difference $B-A$, we can determine the solutions by solving a corresponding pell-like equation. Shall I try to work this out in an answer ? $\endgroup$
    – Peter
    May 30, 2020 at 11:27
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    $\begingroup$ This would also mean that $2n^2+(-1)^n$ is a perfect square when $n$ is a Pell number, which is pretty interesting IMO. Great question! $\endgroup$ May 30, 2020 at 11:30
  • $\begingroup$ I have $3$ answer(s) to (B-A). My question is about the originality of my formula. $\endgroup$
    – poetasis
    May 30, 2020 at 11:37
  • $\begingroup$ I found this, where they talk about the equation I commented above and how it's related to the numerator and denominator of the closest rational approximation of $\sqrt{2}$ (which is exactly what the Pell numbers are) books.google.fi/… $\endgroup$ May 30, 2020 at 11:56
  • $\begingroup$ @Peter I have generalized this formula even more to find solutions for any $B-A=x$. FYI, for primitives, x is limited to prime numbers $(p)$ where $p=\pm1\mod8$ raised to any non-negative power. Under $100, x \in \{1,7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97\}$ $\endgroup$
    – poetasis
    May 30, 2020 at 12:29

1 Answer 1

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Starting with the recurrence relation $p_ {n + 1} = 2 p_ {n} + p_ {n - 1} $ (where $p_n$ is the $n$'th Pell number) immediately gives $$p_ {n + 1} - p_ {n} = p_ {n} + p_ {n - 1} $$ and $$ (p_ {n + 1} - p_ {n})^2 - p_n^2 = (p_n + p_ {n - 1})^2 - p_n^2 = (p_ {n + 1}) (p_ {n - 1}) \tag{1}$$

using the identity $(p_{n+1})(p_{n-1})=p_{n}^2+(-1)^n$ derived in A. F. Horadam, “Pell identities”, The Fibonacci Quarterly, 9 (3), (1971); https://www.fq.math.ca/9-3.html (Eq. 30)

we can obtain from (1) $$ (p_ {n + 1} - p_ {n})^2 - p_n^2=p_{n}^2+(-1)^n$$ which on rearranging becomes your formula

$$p_{n + 1}= p_ {n}+\sqrt{2p_{n}^2+(-1)^n}\tag{2}$$

https://mathworld.wolfram.com/PellNumber.html almost gives the equivalent identity on taking the square root of both sides of (11)

$$\frac{Q_n}{2}=p_{n + 1}-p_ {n}=\sqrt{2p_{n}^2+(-1)^n}$$

where $Q_n$ are the Pell-Lucas numbers

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  • $\begingroup$ Thanks, you've shown me that my discovery is not unique. $\endgroup$
    – poetasis
    May 30, 2020 at 17:47

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