# Thread: Nuber Theory and Sequences

1. ## 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))

2. ## Re: Nuber Theory and Sequences

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?

3. ## Re: Nuber Theory and Sequences

of course, in fact I'm now still trying to solve it.

4. ## Re: Nuber Theory and Sequences

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.

5. ## Re: Nuber Theory and Sequences

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.

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.

6. ## Re: Nuber Theory and Sequences

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).

7. ## Re: Nuber Theory and Sequences

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)
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).

8. ## Re: Nuber Theory and Sequences

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.

9. ## Re: Nuber Theory and Sequences

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.

10. ## Re: Nuber Theory and Sequences

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.

11. ## 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.

12. ## Re: Nuber Theory and Sequences

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.