4
Question
The sequence (an)n≥0 is defined by a0=a1=1 and an+1=14an−an−1 for all n≥1. Prove that the number 2an−1 is a square for all n≥0.
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Solution
We have 2a0−1=1,2a1−1=1,2a2−1=52,2a3−1=192,2a4−1=712.
We observe that if we define b0=−1,b1=1,b2=5,b3=19,b4=71,
then bn+1=4bn−bn−1
for 1≤n≤3.
We will prove inductively that 2an−1=b2n, where b0=−1 and bn+1=4bn−bn−1
For all n≥1.
Suppose this is true for 1,2,...,n and observe that
2an+1−1=14(2an−1)−(2an−1−1)+12=14b2n−b2n−1+12
=16b2n−8bnbn−1+b2n−1−2b2n+8bnbn−1−2b2n−1+12
=(4bn−bn−1)2−2(b2n+b2n−1−4bnbn−1−6)
=b2n+1−2(b2n+b2n−1−4bnbn−1−6).
Hence it suffices to prove that b2n+b2n−1−4bnbn−1−6=0.
This follows also by induction. It is true for n=1 andn=2.
Suppose it holds for n and observe that b2n+1+b2n−4bn+1bn−6=(4bn−bn−1)2+b2n−4(4bn−bn−1)bn−6
as desired.
We observe that if we define b0=−1,b1=1,b2=5,b3=19,b4=71,
then bn+1=4bn−bn−1
for 1≤n≤3.
We will prove inductively that 2an−1=b2n, where b0=−1 and bn+1=4bn−bn−1
For all n≥1.
Suppose this is true for 1,2,...,n and observe that
2an+1−1=14(2an−1)−(2an−1−1)+12=14b2n−b2n−1+12
=16b2n−8bnbn−1+b2n−1−2b2n+8bnbn−1−2b2n−1+12
=(4bn−bn−1)2−2(b2n+b2n−1−4bnbn−1−6)
=b2n+1−2(b2n+b2n−1−4bnbn−1−6).
Hence it suffices to prove that b2n+b2n−1−4bnbn−1−6=0.
This follows also by induction. It is true for n=1 andn=2.
Suppose it holds for n and observe that b2n+1+b2n−4bn+1bn−6=(4bn−bn−1)2+b2n−4(4bn−bn−1)bn−6
as desired.
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