the sequence {an:n=1,2,3,....} is defined by: a1=20 ,a2=30, and a(n+2)=3a(n+1) - a(n), n>=1 find all natural

Question

The sequence {a(n):n=1,2,3,....} is defined by: a(1)=20 ,
a(2)=30, and a(n+2)=3a(n+1) - a(n), n>=1 find all natural numbers (n) sush that 1+5a(n)a(n+1) is perfect square

Answer 1

I have not been able to solve this either, but if it helps, I found an explicit formula for a_n: The auxilliary equation for
a_(n+2) - 3a_(n+1) + a_n = 0 is u^2 - 3u + 1 = 0. The solutions are (3 +/- sqrt(5))/2. Thus,

(1) a_n = Ac^n + Bd^n, for some constants A and B,

where c = (3 + sqrt(5))/2 and d = (3 - sqrt(5))/2. To facilitate computations, notice that c^2 = 3c - 1 and d^2 = 3d - 1 (becase they are solutions of the aux. eqn.); also cd = 1,
c + d = 3, and c - d = sqrt(5).

We can find A and B by using the initial values a_1 = 20 and a_2 = 30. We have to solve the system

Ac + Bd = 20, and
Ac^2 + Bd^2 = 30.

We find A = 10d and B = 10c. Substituting into (1), we get

a_n = 10c^(n-1) + 10d^(n - 1).

That's as far as I could go.

m_math, kyle, Ben: Here's what I've been trying. We have noted that from the eqn
1 + 5(a_n)*(a_(n+1)) = x^2, since both a_n and a_(n+1) are multiples of 10, it follows that x^2 is 1 (mod 500). I found that the only integers whose squares are 1 (mod 500) are 1, 249, 251, 499 (mod 500). Of course, Ben has already found that 251^2 is a solution of this problem. I've been trying to use these other forms (i.e., 500K + 1, 500k + 249, 500k + 251, 500k + 499) to find solutions. My lack of progress has been ego-shattering.

Answer 2

a(n+2) = 3a(n+1) - a(n)

1 + 5a(n)a(n+1) = 1, 4, 9, 16, 25, etc.

First off,
a(n)a(n+1) ≥ 600 (since a(1)=20 and a(2) = 30 and the sequence is monotonic increasing)

Which implies that
1 + 5a(n)a(n+1) ≥ 1 + 5(600)
1 + 5a(n)a(n+1) ≥ 3001 > 54^2

So 55^2 is the first one that would even work

Next restriction...
a(n) and a(n+1) will always end in a 0, regardless.
So a(n)a(n+1) will also end in a 0, regardless.
Let p be the perfect square
(p - 1) / 5 has to be an integer that is divisible by 10

So p definitely has to end in a 1 or 6

So any perfect square of a number that ends in 1, 4, 6 or 9 will work.

(p - 1) / 5 = 10m, m an integer

If sqrt(p) ends in a 1, the second to last digit must be a 0.
If sqrt(p) ends in a 4, (p - 1) / 5 will never be divisible by 10
If sqrt(p) ends in a 6, (p - 1) / 5 will never be divisible by 10
If sqrt(p) ends in a 9, the second to last digit must be a 9.

So sqrt(p) must end in a 01 or 99

Examples: 99, 101, 199, 201, etc.


Now, looking at the a(n)a(n+1) term:
[3a(n+1) - a(n)][3a(n+2) - a(n+1)] = (p - 1) / 5
= 9a(n+1)a(n+2) - 3a(n)a(n+2) - 3a(n+1)^2 + a(n)a(n+1)
= 9a(n+1)[3a(n+1) - a(n)] - 3a(n)[3a(n+1) - a(n)] - 3a(n+1)^2 + a(n)a(n+1)
= 27a(n+1)^2 - 9a(n)a(n+1) - 9a(n)a(n+1) + 3a(n)^2 - 3a(n+1)^2 + a(n)a(n+1)
= 24a(n+1)^2 - 17a(n)a(n+1) + 3a(n)^2
= (8a(n+1) - 3a(n)) (3a(n+1) - a(n)) = (p - 1) / 5
= (9a(n+1) - a(n+1) - 3a(n)) ((3a(n+1) - a(n))
= (9a(n+1) - 3a(n) - a(n+1)) ((3a(n+1) - a(n))
= (3(3a(n+1) - a(n)) - a(n+1)) ((3a(n+1) - a(n))
= (3a(n+2) - a(n+1))(3a(n+1) - a(n))
= and yeah I just went around in a big circle.

Someone take my work and make something from it.

I'm stumped, but I had to try.


edit:
Tony, here is the equation that I found with an internet search:
a(n) = 10 * ((3-sqrt(5))^n + (3+sqrt(5))^n)*(1/2)^n
Source: http://www.research.att.com/~njas/sequences/?q=2+3+7+18+47&language=english&go=Search

Answer 3

Nicely done with the closed-form Tony. I was trying to get that with the hopes of substituting into the formula for p (the perfect square), but haven't done that sort of thing before. In response to kyle's edit: the formula you've given is assuming we start at n=0, but we're starting with n=1. Therefore Tony's is correct, as is yours if you substitute n-1 for n in your exponents.

I've attempted substituting the closed form into p as mentioned above:
p = 1 + 500 * (c^(n-1)+d^(n-1)) * (c^n + d^n)
= 1 + 500 * (c^(2n-1) + c^(n-1)d^(n) + c^(n)d^(n-1) + d^(2n-1))

Remembering Tony's helpful notes on the properties of c and d (or messily expanding things and then recollecting):
= 1 + 500 * (c^(2n-2)*c + c^(n-1)*d^(n-1)*d + c^(n-1)*d^(n-1)*c + d^(2n-2)*d)
= 1 + 500 * ((3c-1)^(n-1)*c + d + c + (3d-1)^(n-1)*d)
= 1 + 500 * (c*(3c-1)^(n-1) + 3 + d*(3d-1)^(n-1)).

Hmm, this isn't working out too well I don't believe. However, let's go ahead and return the first and last term in the parentheses back to their original form:
p= 1 + 500*(c^(2n-1) + d^(2n-1) + 3)
= 1501 + 500*(c^(2n-1) + d^(2n-1)).

But here I too am stuck. I'll continue working on it, but I thought I would post up my little contribution (hopefully) to a possible solution.

Oh, and I used my calculator to go through values of n from 1 to just shy of 50. Only for n=3 is p a perfect square, 251^2.

Answer 4

scythian is on the mark here: the useful integers are carefully built making use of Peano's axioms, and a million+a million is the on the spot successor of a million, it extremely is defined as 2. yet i could completely disagree with scythian's argument that Godel's Incompleteness Theorems are arguable; the only mathematicians that heavily experience this way are fringe logicians and maniacal set theorists. Godel's theorems unfold out alot of recent venues in arithmetic, and a few could say "freed" us from the exceedingly much particular rigorous death to which we've been headed. actually, Hilbert in 1900 asked "ought to somebody please set up a equipment of axioms it extremely is thoroughly consistent and serves as a foundation for all math?", to which Godel replied, numerous years later, "no, no possible; any axiomatic equipment describing the integers could have particular unprovable statements, and a few that are consistent while dealt with the two as real and as fake." To summarize: a million+a million is two via fact it incredibly is defined that way, axiomatically, and as a result won't be in a position to be shown under the typical equipment of Peano's axioms. Steve EDIT - Above, once I say "unprovable statements", I recommend statements dealt with as real, yet no longer proved as such (no longer including axioms). once I talk some assertion being "consistent while dealt with the two as real and as fake", I recommend autonomous of the present axiomatic framework; it incredibly is such as announcing the framework can not tutor its own consistency.