All members of geometrical progression (bn)n≥1 are members of some arithmetical progression. It is known that b1 is an integer. Prove that all members of this geometrical progression are integers. (progression is infinite)
Problem
Source: Moldova TST 2020
Tags: progressions, Arithmetic Progression, geometric progression, number theory
07.03.2020 22:45
*b_n has to be infinite I think
07.03.2020 23:44
Hamel wrote: *b_n has to be infinite I think Yes, its infinite
08.03.2020 03:20
Set the geom progression to be (brn)n≥0 with b=b1∈Z, then there are a,c such that brn−ca∈Z∀n. Define ck:=ba(1−rk), then ⟹ckrn=brn−ca−brn+k−ca∈Z∀n∈N∪{0},k∈N.Thus r=ckrn+1ckrn∈Q, so ck∈Q∀k. Fix k, and suppose νq(r)<0 for some prime q. Then νq(ckrn)→−∞ as n→∞, contradiction to ckrn always being an integer. Thus νq(r)≥0∀q prime, so r∈Z. YUH.
19.08.2020 21:38
stroller wrote: Fix k, and suppose νq(r)<0 for some prime q. Then νq(ckrn)→−∞ as n→∞, contradiction to ckrn always being an integer. Thus νq(r)≥0∀q prime, so r∈Z. Can you (or someone, who understands) explain theorem/technique/trick and the notation behind νq(r). Thanks in advance!
21.08.2020 11:24
Asking for help, but i helped myself, so redacted