Question

[SOLVED] Define a sequence of real numbers by \(a_0=0\), \(a_1=1\), \(a_2=1\) and \[\large a_{n+3}=2a_{n+2}+2a_{n+1}-a_n\]Show that \(a_n\) is a perfect square for every \(n\in\mathbb{N}\)

Answer 1

The recurrence relation seems pretty messy, so I will try to solve it inductively, T_T

Answer 2

How about if we reduce every term into the \(ka_{n+2} + ma_{n+1} + na_n\)?
The term \(a_n\) won't matter as its Zero. We can express terms as just \(k + m\). Maybe?

Answer 3

For n=0.

Answer 4

I tried to find a pattern in the roots, but couldn't. a3 = 2*2, a4 = 3*3, a5 = 5*5, a6 = 8*8, a7 = 12*12. Will try something else now.

Answer 5

I would suggest induction as bmp is attempting.

Answer 6

@bmp, check your value for \(a_7\)

Answer 7

Ah, got it, I think.

Answer 8

Thanks @KingGeorge :-)

Answer 9

Fibonacci it is, then :-)

Answer 10

Fibonacci How?

Answer 11

I think every
\[a_{n} = Fib(n)*Fib(n)\]that is a perfect square for every n in N.

Answer 12

But how are you going to prove it?

Answer 13

I am trying to refine my proof, but I think that assuming that that is true(by induction), it's not too hard to proof that it is correct.

Answer 14

to prove*

Answer 15

\[a_{n+3} = 2a_{n+2} + 2a_{n+1} -a_n\]Lets assume \(a_{n+3} = q^2\).
\[a_{n+4} = 2a_{n+3} + 2a_{n+2} -a_{n+1} \implies a_{n+4} = 2q^2 + 2a_{n+2}-a_{n+1}\]Now what? How do we prove \(a_{n+4}\) is perfect square as well.

Answer 16

I think that the proof would go more or less like this:
Assume:\[a_{k} = Fib(k)*Fib(k)\]Clearly, that holds for k <= 2. So, multiply it by -1, we get:\[-a_{k} = -(Fib(k)*Fib(k))\]Add 2*(ak+2 ak+1) in both sides, we get:\[2(a_{k+2} + a_{k+1}) - a_{k} = 2(a_{k+2} + a_{k+1}) - Fib(k)Fib(k) \rightarrow a_{k+3} = a_{k+3}\]What I just did I am pretty sure is a fallacy, but the idea is that we have to prove that every {ak} has the form Fib^2(k). I am still working on it, was never good with proofs :-(

Answer 17

I'm also pretty sure what you just did was a fallacy. However, induction is the correct path to go. Keep trying.

Answer 18

Another hint that may be helpful:

Use strong induction.

Answer 19

I see. It makes sense, since I will need a(1) ... a(k) to be true to prove that a(k+1) is true. This is a nice problem @KingGeorge :-) Thanks for it.

Answer 20

You are also correct in that it is the sequence of squares of the Fibonacci numbers. You still need to prove it though. :)

Answer 21

I think I may have gotten it?
a5 = 2a4 + 2a3 - a2 = 2 F(4) ^2 + 2 F(3) ^2 - F(2) ^2
= F(4)^2 + 2F(3)^2 +(F(3)+F(2))^2 - F(2)^2
= F(4)^2 + 2F(3)^2 +F(3)^2+2F(3)F(2)
= F(4)^2 + 2F(3)(F(3)+F(2)) + F(3)^2
= (F(4) + F(3))^2
= (F(5)^2)
and this would be true always?
tadam?

Answer 22

Yeah, I just now got to it. I think that would be the way, if you can prove for a(1)...a(k).

Answer 23

well, I guess that, if a4 = F(4)^2 and a5 = F(5)^2, then a6 = F(6)^2, so a7 = F(7)^2, so a8 = F(8)^2, etc? Am I allowed to do this?

Answer 24

Try using n, n+1, n+2, n+3 instead of 4, 5, 6, 7. Your proof above is very very close to correct. You basically just have to use a variable instead of numbers.

Answer 25

@m_charron2 Kudos for beating me on the proof, mate :-). Well done.

Answer 26

oh, couldn't've done it if you hadn't told us about the fibonacci here, so Kudos to you as well
I can't find how to start replacing the numbers with variables without assuming things. I can't prove that a_(n+3) = F(n+3)^2 without first knowing that a_(n+2) = F(n+2)^2, a_(n+1) = F(n+1)^2 and a_n = F(n)^2... I'm lost in a mental paradox of my own making I think...

Answer 27

You're first method of showing it's true for \(a_5\) is so close it might as well be correct. Just substitute \[\large a_5 =a_{n+1}\]\[\large a_4 =a_{n}\]\[\large a_3 =a_{n-1}\]\[\large a_2 =a_{n-2}\]And similar substitutions for the Fibonacci numbers. Then go through the same steps. You will then have proved it for the general case.

Answer 28

For those interested, here's the solution I got when I was challenged with this problem.

Answer 29

on this note, off to sleep for me, that drained me lol. Excellent problem KingGeorge!

Answer 30

Good night.