## KingGeorge 3 years ago [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 $$n\in\mathbb{N}$$

1. bmp

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

2. Ishaan94

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?

3. Ishaan94

For n=0.

4. bmp

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.

5. KingGeorge

I would suggest induction as bmp is attempting.

6. KingGeorge

@bmp, check your value for $$a_7$$

7. bmp

Ah, got it, I think.

8. bmp

Thanks @KingGeorge :-)

9. bmp

Fibonacci it is, then :-)

10. Ishaan94

Fibonacci How?

11. bmp

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

12. Ishaan94

But how are you going to prove it?

13. bmp

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.

14. bmp

to prove*

15. Ishaan94

$a_{n+3} = 2a_{n+2} + 2a_{n+1} -a_n$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.

16. bmp

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 :-(

17. KingGeorge

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

18. KingGeorge

Another hint that may be helpful: Use strong induction.

19. bmp

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.

20. KingGeorge

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

21. m_charron2

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?

22. bmp

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

23. m_charron2

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?

24. KingGeorge

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.

25. bmp

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

26. m_charron2

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...

27. KingGeorge

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.

28. KingGeorge

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

29. m_charron2

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

30. KingGeorge

Good night.