Factors of the form N^2 - 3N +1

In summary, the conversation discusses a relation for the form F_{n} = n^{2}-3n+1 and its factors. The relation states that if p_n is a whole factor of F_n and p_{n}^{'} is the quotient, then the following relation holds: [tex]\frac{F_{n}}{p} * \frac{F_{(n+p)}}{p} = F_{(n+p_{n}^{'})}. An example is given with p = 1. It is also mentioned that the notation used in the conversation may not be standard.
  • #1
ramsey2879
841
3
I noted the following relation for the form [tex] F_{n} = n^{2}-3n+1[/tex]
let [tex]p_n[/tex] be any whole factor of [tex] F_n[/tex] and [tex]p_{n}^{'}[/tex] be the quotient. The following relation then holds

[tex]\frac{F_{n}}{p} * \frac{F_{(n+p)}}{p} = F_{(n+p_{n}^{'})}

A trivial example would be to let p = 1. Then
[tex] F_{n} * F_{(n+1)} = F_{(n+F_{n})}

Is this something that is of interest?
 
Physics news on Phys.org
  • #2
You could at least put closing /tex's in. It is common to let quotients be q's and i notice that the subscript n's vanish in your (non-closed off) latex, should they?
 
  • #3
I think you're saying something equivalent to: f(n)=n^2-3*n+1, and if f(n)=a*b for some n, then f(n+a)/a=f(n+b)/b. This is actually true for any monic quadradic polynomial with integer coefficients.
 

What is the general formula for factors of the form N^2 - 3N + 1?

The general formula for factors of the form N^2 - 3N + 1 is (N-1)(N-2).

How do I find the factors of a specific number in the form N^2 - 3N + 1?

To find the factors of a specific number in the form N^2 - 3N + 1, you can use the general formula (N-1)(N-2). Plug in the value of N to find the factors.

What is the relationship between the factors of N^2 - 3N + 1 and the factors of N?

The factors of N^2 - 3N + 1 are related to the factors of N through the general formula (N-1)(N-2). This means that if N has factors a and b, then N^2 - 3N + 1 will have factors (a-1)(b-2).

Are there any special cases for factors of the form N^2 - 3N + 1?

Yes, there are two special cases for factors of the form N^2 - 3N + 1. When N = 1, the only factor is 1. When N = 2, the factors are 1 and 3.

Can I use the factors of N^2 - 3N + 1 to find the prime factors of N?

No, the factors of N^2 - 3N + 1 are not necessarily prime factors of N. However, they can be used to find the prime factors of N by factoring them further.

Similar threads

Replies
2
Views
712
Replies
3
Views
494
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Linear and Abstract Algebra
Replies
17
Views
1K
  • Mechanical Engineering
Replies
9
Views
1K
  • General Math
Replies
4
Views
2K
  • Linear and Abstract Algebra
Replies
8
Views
897
  • Differential Geometry
Replies
13
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
904
  • Linear and Abstract Algebra
Replies
8
Views
785
Back
Top