[seqfan] Re: Sum of Pell numbers and companion Pell numbers

[Robert Dougherty-Bliss]

Dear Max,

Thanks for the clarification! Forgive me, but I am still confused about the rank
of apparition part of the argument. Why do either m = q(p) / 2 or m = q(p) /
4 satisfy the identity (1 + sqrt(2))^(2m) (-1)^m == -1 (mod p)? I understand
that m = (p - 1) / 2 or m = (p - 1) / 4 would work, but I don't understand why
we can replace p - 1 with q(p), or why this would be the smallest such m.

It seems like you know where the zeros of a(n) mod p will be, but I can't
figure out how to find them explicitly.

Thanks again for your help.

Robert

On Sat, Apr 29, 2023 at 8:59 AM Max Alekseyev <maxale at gmail.com> wrote:
>
> It all follows from the explicit formula:
> a(n) = (1 + sqrt(2))^n + (1 - sqrt(2))^n.
>
> For an odd prime p, we have
> (1 +- sqrt(2))^(p-1) == 1 (mod p) whenever 2 is a quadratic residue modulo
> p, ie. p == 1,7 (mod 8);
> and
> (1 +- sqrt(2))^(p+1) == 1 (mod p) whenever 2 is a non-quadratic residue
> modulo p, ie. p == 3,5 (mod 8).
> It follows that the period length q(p) of a() modulo p divides p-1 or p+1,
> respectively.
>
> Then, a(m) == 0 (mod p) is equivalent to
> (1 + sqrt(2))/(1-sqrt(2))^m = (1 + sqrt(2))^(2m) * (-1)^m == -1 (mod p).
> If m is the smallest, ie. m = r(p) is the rank of apparition, then r(p) =
> q(p)/2 when it's odd; or r(p) = q(p)/4 when it's even.
> In the former case, there is only one zero in the period modulo p, ie. if
> a(m) == 0 (mod p) and q(p)/2 is odd, then m == q(p)/2 (mod q(p)) and thus
> q(p)/2 divides m.
> In the latter case, there are two zeros in the period modulo p, ie. if a(m)
> == 0 (mod p) and q(p)/4 is even, then m == q(p)/4 (mod q(p)/2) and thus
> q(p)/4 divides m.
>
> Let me know if anything is still unclear.
>
> Regards,
> Max
>
> On Sat, Apr 29, 2023 at 2:51 AM Robert Dougherty-Bliss <
> robert.w.bliss at gmail.com> wrote:
>
> > Dear Max,
> >
> > Thank you for the proof! I am excited about it, because this is
> > exactly what I suspected was true. However, I'm having some trouble
> > following the details of your argument.
> >
> > > Let p be the smallest prime dividing m and r(p) the rank of apparition of
> > > a() modulo p, then the period of a() modulo p is either 2*r(p) or 4*r(p),
> > > and it divides p-1 or p+1.
> >
> > Why is the period either 2 * r(p) or 4 * r(p)? And why does it divide
> > p - 1 or p + 1?
> >
> > > That is, r(p) <= (p+1)/2 < p and r(p) divides m, which together with
> > > minimality of p implies r(p)=1.
> >
> > Why does r(p) divide m?
> >
> > I know some similar facts / arguments about Fibonacci numbers, but I
> > don't see how the ideas apply here.
> >
> > Robert
> >
> > > On Thu, Apr 27, 2023 at 11:05 PM Robert Dougherty-Bliss <
> > > robert.w.bliss at gmail.com> wrote:
> > >
> > > > Dear Sequence Fans,
> > > >
> > > > Consider the sequence a(n) which satisfies the recurrence
> > > >      a(n) = 2 * a(n - 1) + a(n - 2)
> > > > with initial conditions a(0) = a(1) = 2. (This is A2203, the companion
> > > > Pell numbers.)
> > > >
> > > > The sequence of positive n such that a(n) = 2 (mod n) begins
> > > > 1, 2, 3, 4, 5, 7, 8, 11, 13, 16, 17, 19, 23, 24, 29.
> > > > Except for the leading 1 and 2, this looks like A270342, the sequence
> > > > of n such that n divides the sum of the first n Pell numbers, but
> > > > there is no comment to this effect. Can anyone prove this?
> > > >
> > > > The sum of the first n Pell numbers S(n) turns out to equal (a(n) - 2)
> > > > / 4, so everything in A270342 is in this sequence. For the other way,
> > > > I have managed to prove that if a(n) = 2 (mod n) and n is odd or
> > > > divisible by 4, then n is in A270341, but I cannot figure out the
> > > > "divisible by just 2" case.
> > > >
> > > > Robert
> > > >
> > > > --
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> > > >
> > >
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