Proof: x^n + 1 is Valid for Odd n".

  • Thread starter ascheras
  • Start date
  • Tags
    Proof
In summary, to prove that the statement is valid only when n is an odd integer, we can use a simple 3 line proof. We know that n is an integer from the given notation. By checking the sign of the leading order term on the right-hand side, we can see that even values of n do not satisfy the statement. Therefore, n must be odd for the statement to be valid. Additionally, we can also prove that (2^n)+1 is prime only when n=(2^r) for some r.
  • #1
ascheras
14
0
Prove the following is valid only when n is an odd integer.

x^n + 1= (x+1)(x^n - X^n-2 + ... + (x^2 - x + 1).

It's an easy 3 line proof.
 
Physics news on Phys.org
  • #2
The notation given already implies that n is an integer. From there, it's easy to check that even values of n result in the wrong sign for the leading order term on RHS.
 
  • #3
(x+1) a factor implies x = -1 is a root, but it ain't. (one liner)
 
  • #4
The way that is meant is (X^(2n+1)+1) =(X+1)(X^(2n)-X^(2n-1)+X^(2n-2)-+-..+1).
 
  • #5
robert, how can you tell he meant the converse of what he said? what he said was to prove his statement false for even n, not to prove it true for odd n. that at least is how i translate the word "only".
 
  • #6
mathwonk said:
robert, how can you tell he meant the converse of what he said? what he said was to prove his statement false for even n, not to prove it true for odd n. that at least is how i translate the word "only".

Yes, that is true. Even so statement is not correct for odd n since

(X^3+1) not equal to (X+1)(X^3-X+1) for all X.
 
Last edited:
  • #7
Let n= (2^r)*q, where q has no prime divisors. Then you can write (2^n) +1= (y^q) +1, where y=(2^r). Then (2^n) +1= (y+1)(y^(q-1)- y^(q-2) +...+ (y^2) - y+1). Here, y+1= (2^r)+1 >1 and there are two factors. This (2^n)+1 cannot be prime if the other factor is also > 1. That happens unless y^(q-1) +...+ (y^2)- y+1 reduces to 1, i.e. q=1. Therefore (2^n)+1 is prime, which implies that q=1 and n=(2^r) for some r as claimed.
 

FAQ: Proof: x^n + 1 is Valid for Odd n".

What is the proof for x^n + 1 being valid for odd n?

The proof for x^n + 1 being valid for odd n is based on mathematical induction. It involves showing that the statement is true for n=1, and then assuming it is true for some arbitrary odd number k. By using algebraic manipulation and the properties of odd numbers, it can be shown that the statement is also true for k+2. This completes the proof by induction.

Why is it necessary to use mathematical induction for this proof?

Mathematical induction is necessary for this proof because it allows for a general statement to be proven true for an infinite number of cases. In this case, we are proving that x^n + 1 is valid for all odd values of n, which is an infinite number of cases. Mathematical induction provides a systematic way to prove this statement for all possible values of n.

Can this proof be extended to even values of n?

No, this proof cannot be extended to even values of n. This is because the statement x^n + 1 is only valid for odd values of n. When n is an even number, the statement becomes x^n + 1 = x^even + 1 = (x^2)^k + 1, which is not true for all values of x. Therefore, this proof is only valid for odd values of n.

What are the practical applications of this proof?

This proof has several practical applications in mathematics and computer science. It is used in fields such as number theory, algebra, and cryptography. It is also used in computer algorithms that involve exponents and finite fields. Additionally, this proof can be used to simplify complex expressions and solve equations involving odd exponents.

Can this proof be applied to other similar statements?

Yes, this proof can be applied to other similar statements involving odd exponents. For example, it can be extended to the statement x^n - 1 is valid for odd n, or x^n + 2 is valid for odd n. The same principles of mathematical induction can be used to prove these statements for all odd values of n.

Similar threads

Replies
13
Views
2K
Replies
25
Views
2K
Replies
2
Views
2K
Replies
11
Views
2K
Replies
17
Views
2K
Back
Top