Find all real constants c for which there exist strictly increasing sequence a of positive integers such that (a2n−1+a2n)/an=c for all positive intеgers n.
Problem
Source: Bosnia Herzegovina TST 2017, problem 3
Tags: algebra
RagvaloD
22.07.2017 12:49
Obvious, that c is rational and >2. Let c=mk, m>2k and (m,k)=1 a1+a2=ca1→a2=(c−1)a1 a3+a4=ca2=c(c−1)a1 a5+a6+a7+a8=c(a3+a4)=c2(c−1)a1 ... a2n+1+...a2n+1=c(a2n−1+1+...+a2n)=...=cn(c−1)a1=mn(m−k)a1kn So kn|a1 for every n so k=1 and c is integer ≥3
Vietnamisalwaysinmyheart
25.07.2017 17:54
I got all the answer is c≥4, and is a natural number. Here it goes:
The first thing to show is c≥4. This is not too hard. Using the old trick:
Consider (Δn) is a positive integer sequence satisfy: Δn=an+1−an, for convenient, denote:
Δ0=0.
Since: a2n−1+a2n=can and: a2n+1+a2n+2=can+1. Subtract and replace by Δ, we have:
cΔn=Δ2n−1+2Δ2n+Δ2n+1
Hence, the minium of Δ2n−1,Δ2n,Δ2n+1, must not exceed c4Δn, i.e, we can find i1, such that: Δi1≤c4Δn. Continue this progress, we find that, there exist a sequence: (ik) such that: Δik≤(c4)k.Δn.
Now, if: c<4, then the above obviously converges to 0, which is impossible!
For c=4, we can use the sequence: 1,3,5,7,... (i.e odd natural numbers sequence)
Consider c>4. I will prove that: c must be an integer.
Assume c is not integer, then c=pq with q>2 and: p,q are two relative primes number.
The idea is to use infinite descendant: pqΔn=Δ2n−1+2Δ2n+Δ2n+1, since the right-hand side is natural, we must have: q|Δn forall n. Continue, we must have: qk|Δn forall k, which can't be true with q≥2
Thus, c is an integer greater that 4. The last part, I will construct a sequence satisfy:
a1=1,a2=c,a3=c(c−1)2−1,a4=c(c−1)2+1, easily check that: a1<a2<a3<a4 under the condition.
Now, if an has been constructed, there are two cases: can is odd, takes: a2n−1=can−12,a2n=can+12, and if can is even, takes: a2n−1=can2−1,a2n=can2+1
We only need to check the condition that: (an) is strictly increasing. I will prove by induction.
Of course: a2n−1<a2n forall n, by our construction.
Suppose at some n, we have: a2n−1≤a2n−2
Then, since: a2n−1≥can2−1, a2n−2≤can−12+1, we have:
c(an−an−1)≥4, but this is of course wrong since: c≥5, an−an−1≥1
Thus, we have the answer: All positive integer greater or equal to 4
jerker
05.10.2018 04:14
RagvaloD wrote: Obvious, that c is rational and >2. Let c=mk, m>2k and (m,k)=1 a1+a2=ca1→a2=(c−1)a1 a3+a4=ca2=c(c−1)a1 a5+a6+a7+a8=c(a3+a4)=c2(c−1)a1 ... a2n+1+...a2n+1=c(a2n−1+1+...+a2n)=...=cn(c−1)a1=mn(m−k)a1kn So kn|a1 for every n so k=1 and c is integer ≥3 it is just not enough...