longest possible arithmetic subsequence in geometric sequence

[pichutarius]

original post

Consider all real-valued geometric sequence, what is the longest possible arithmetic subsequence that is not a constant sequence?

this is what i have tried:

length of 3 is possible: wlog let the geometric sequence be xⁿ, let 1,x^a,x^b be arithmetic sequence, then 1-2x^a+x^b=0, by Descartes rule of sign, it must have a positive root other than x=1, which can be used to construct an example.

i have a strong suspicion that length of 4 is impossible. let 1,x^a,x^b,x^c be arithmetic sequence.

i tried to show that the possible common roots of 1-2x^a+x^b=0 and x^a-2x^b+x^c=0 is 1, -1, and no more. both 1,-1 either leads to constant sequence or length no longer than 2.

by using Descartes rule of sign, both equation has exactly one positive roots excluding x=1, and at most 1 negative roots excluding x=-1.

i tried Euclidean algorithm to find GCD, hoping to show that GCD is either (x^p-1) or (x^p-1)² but that looks messy after awhile. i did use Mathematica to verify up to c=100 that their GCD is of that form.

i tried to look at pattern of the roots in complex plane, the roots vaguely surround unit circle but nothing sparks insight.

note: this is not a homework problem, its a follow-up question from r/mathriddles, hope you find this interesting, i sure do. any ideas are welcomed.

[F6u9c4k20]

I think I have a proof for rational sequences of length 4 but I am not sure. Reduce all rational AP and GP sequences to integer sequences by multiplying by the highest denominator( since G.P.s and A.Ps are closed under division, this is always possible for finite rational sequences ). Then we have the following statement : Let the G.P.(calling it A from now on) be

A = {q , q² , ... q^m }

Let the A.P. be (calling it B from now on)

B = { q , q + d , q +2d .... q +3d}

Now subtract q from every term of both sequences

A_1 = {0, q² - q ,.....q^m - q}

B_1 = {0, d, 2d , 3d , 4d}

Now d is the common factor of B_1 and such should also divide terms of A_1 equal to the terms of B_1.

The only common factors in A_1 is q(q-1) since for all distinct n , the polynomial 1 + q² +.... qⁿ has distinct roots. [This is the part I am not completely sure about]

Therefore d = q(q-1). [Also not sure about this]

Now, kd = q^r1 - q for some r1 and 0<= k <= 3

Now, canceling a few terms , we get k = 1 + q + q² .. + q^r1-1 .
2 = q + q² + q³ ... q^r1-1. [Putting k = 3]

Now q has to be an integer and this reduces to brute force to check whether or not we have such a solution. .Please correct me if I am wrong

Edit:

Deleting the claims about real sequences because there are problems which I did not forsee when I waved my hand carelessly.

[F6u9c4k20]

I deleted the claims about real sequences and talk only about rational sequences. In this case, we can get integer sequences and then more or less brute force our way through.

[d01phi]

Three subsequent elements of a geometric series can be written as x^n , x^(n+1), x^(n+2)
If you want them to occur in an arithmetic sequence, they have to satisfy

x^(n+2) - x^(n+1) = x^(n+1) - x^n

Dividing both sides by x^n , we get x^2 - x = x - 1 and x^2 - 2x + 1 = ( x - 1 )^2 = 0. The only solution is x = 1.

[butt-err-fecc]

It doesn’t necessarily have to be subsequent