# Nuber Theory and Sequences

• Sep 10th 2013, 05:28 AM
Orpheus
Nuber Theory and Sequences
Hello MHF, Here's a problem that I couldn't solve, could anyone help me.

- Here's the things i already figured out :
U(1) = U(2) = 1
U(n+1) = U(n) + U(n-1) \ n is greater than 2
U(n+p) = U(n) * U(P-1) + U(n+1) * U(p)

Show that GCD(U(n+p),U(n)) = GCD(U(n), U(p))
• Sep 10th 2013, 06:49 AM
Plato
Re: Nuber Theory and Sequences
Quote:

Originally Posted by Orpheus
Hello MHF, Here's a problem that I couldn't solve, could anyone help me.
- Here's the things i already figured out :
U(1) = U(2) = 1
U(n+1) = U(n) + U(n-1) \ n is greater than 2
U(n+p) = U(n) * U(P-1) + U(n+1) * U(p)
Show that GCD(U(n+p),U(n)) = GCD(U(n), U(p))

That is a nice question. Do you plan to show any effort or discussion on it?
• Sep 10th 2013, 06:56 AM
Orpheus
Re: Nuber Theory and Sequences
of course, in fact I'm now still trying to solve it.
• Sep 10th 2013, 06:57 AM
Orpheus
Re: Nuber Theory and Sequences
Quote:

Originally Posted by Plato
That is a nice question. Do you plan to show any effort or discussion on it?

I think i should show that U(n+p) = U(n) * K + U(p) but i couldn't do it.
• Sep 10th 2013, 10:10 AM
emakarov
Re: Nuber Theory and Sequences
Quote:

Originally Posted by Orpheus
U(1) = U(2) = 1
U(n+1) = U(n) + U(n-1) \ n is greater than 2
U(n+p) = U(n) * U(P-1) + U(n+1) * U(p)

Is the third line a part of the definition of U, just like the first two lines? If not, then it should be indicated.

Quote:

Originally Posted by Orpheus
I think i should show that U(n+p) = U(n) * K + U(p) but i couldn't do it.

There is no constant K that works for all n and p. For example, for p = 2, K has to be 1 for n = 1, and K has to be 2 for n = 2.
• Sep 10th 2013, 10:46 AM
Orpheus
Re: Nuber Theory and Sequences
Quote:

Originally Posted by emakarov
Is the third line a part of the definition of U, just like the first two lines? If not, then it should be indicated.

There is no constant K that works for all n and p. For example, for p = 2, K has to be 1 for n = 1, and K has to be 2 for n = 2.

- Yes the third part is a part of the definition, and is true for all n and p that are greater than 2
- K is not a constant, i mean i could show that U(p) is the remainder of the division of U(n+p) on U(n).
• Sep 10th 2013, 11:29 AM
emakarov
Re: Nuber Theory and Sequences
Quote:

Originally Posted by Orpheus
U(1) = U(2) = 1
U(n+1) = U(n) + U(n-1) \ n is greater than 2
U(n+p) = U(n) * U(P-1) + U(n+1) * U(p)

Quote:

Originally Posted by Orpheus
- Yes the third part is a part of the definition, and is true for all n and p that are greater than 2

That's not the response I was hoping for. Look at the first two lines. The first line defines U at 1 and 2. The second line defines U at 3 when n = 2. Indeed, the right-hand side refers to U(2) and U(1), which are already defined. Similarly, it defined U at 4 when n = 3, and in general it defines U at every n ≥ 2. So, what can the third line add? At best it agrees with what the first two lines say; at worst it contradicts them. But it is not a part of the definition; it is a separate claim. It turns out that this claim is true and is given as a lemma without a proof to be used in this problem.

Not knowing the definitions involved in a statement is worse than not being able to prove that statement, but here are further hints anyway. Replace U(n+p) with U(n) * U(p-1) + U(n+1) * U(p) in GCD(U(n+p), U(n)) and use the following facts.

Proposition 1. GCD(kn + m , n) = GCD(m, n) for k ∈ ℤ.
Proposition 2. GCD(U(n+1), U(n)) = 1. Proof: By induction on n.
Proposition 3. If GCD(m, n) = 1, then GCD(km, n) = GCD(k, n).
• Sep 10th 2013, 11:38 AM
Orpheus
Re: Nuber Theory and Sequences
Quote:

Originally Posted by emakarov
That's not the response I was hoping for. Look at the first two lines. The first line defines U at 1 and 2. The second line defines U at 3 when n = 2. Indeed, the right-hand side refers to U(2) and U(1), which are already defined. Similarly, it defined U at 4 when n = 3, and in general it defines U at every n ≥ 2. So, what can the third line add? At best it agrees with what the first two lines say; at worst it contradicts them. But it is not a part of the definition; it is a separate claim. It turns out that this claim is true and is given as a lemma without a proof to be used in this problem.

Not knowing the definitions involved in a statement is worse than not being able to prove that statement, but here are further hints anyway. Replace U(n+p) with U(n) * U(p-1) + U(n+1) * U(p) in GCD(U(n+p), U(n)) and use the following facts.

Proposition 1. GCD(kn + m , n) = GCD(m, n) for k ∈ ℤ.
Proposition 2. GCD(U(n+1), U(n)) = 1. Proof: By induction on n.
Proposition 3. If GCD(m, n) = 1, then GCD(km, n) = GCD(k, n).

the 3rd line is the previous question in this exercice and I already solved it, I tought it may be used in this question that's why i put there.
as for GCD(U(n+1), U(n)) = 1 I proved it with induction.
• Sep 10th 2013, 11:43 AM
emakarov
Re: Nuber Theory and Sequences
Quote:

Originally Posted by Orpheus
the 3rd line is the previous question in this exercice and I already solved it, I tought it may be used in this question that's why i put there.
as for GCD(U(n+1), U(n)) = 1 I proved it with induction.

Good, then proving GCD(U(n+p), U(n)) = GCD(U(n), U(p)) should not be hard.
• Sep 10th 2013, 11:48 AM
Orpheus
Re: Nuber Theory and Sequences
Quote:

Originally Posted by emakarov
Good, then proving GCD(U(n+p), U(n)) = GCD(U(n), U(p)) should not be hard.

I couldn't do it, because i tought the only way is by showing that U(p) is the remainder of U(n+p) divised by U(n)
and I don't see how can i use those propositions.
• Sep 10th 2013, 11:57 AM
emakarov
Re: Nuber Theory and Sequences
GCD(U(n+p), U(n))
= GCD(U(n) * U(p-1) + U(n+1) * U(p), U(n))
= GCD(U(n+1) * U(p), U(n)) by Proposition 1
= GCD(U(p), U(n)) by Propositions 2 and 3.

The sequence U(n) is called Fibonacci numbers. A similar fact is discussed here.
• Sep 10th 2013, 12:05 PM
Orpheus
Re: Nuber Theory and Sequences
Quote:

Originally Posted by emakarov
GCD(U(n+p), U(n))
= GCD(U(n) * U(p-1) + U(n+1) * U(p), U(n))
= GCD(U(n+1) * U(p), U(n)) by Proposition 1
= GCD(U(p), U(n)) by Propositions 2 and 3.

The sequence U(n) is called Fibonacci numbers. A similar fact is discussed here.

OHH I see thank you, I tought that i couldn't use a product of two numbers for the first proposition that was dumb of me.
anyway thank you.