Factoring x^2n + x^n + 1. To Math nerds like me...?

Question

1) I have been working with these kind of polynomials

x^2n + x^n + 1

and I realised that they are a multiple of x^2 + x + 1 but for the cases when n is a multiple of 3.

Adding and substracting x^n, I get:

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

And, if n is an even number, n/2 is a natural number, so, I can factore the expression this way:

[x^n + 1 - x^(n/2)][x^n + 1 + x^(n/2)]

But I cant do this if n is an odd number, since I wouldnt get polynomials factors. But, dividing the original polynomial by x^2 + x + 1, I have noticed that the residual is 0 if n is not a multiple of 3. Any suggestions?

2) In these cases:

x^2n - x^n + 1, adding and substracting 3x^n, I get:

x^2n + 2 x^n + 1 - 3x^n = (x^n +1)^2 - 3x^n

And, if n is an even number, n/2 is a natural number, so, I can factore the expression this way:

[x^n + 1 - V3 x^(n/2)][x^n + 1 + V3 x^(n/2)]

Id like suggestions too for n an odd number

Thanks.

To Cattbarf: I dont see which the mistake is, that you are pointing. You did exactly the same than I did. Did you read the question?

Answer 1

Instead of breaking the problem into odd and even cases
right away, it's easier to consider it in it's full generality.

Let n be a natural number; then x^(3n) - 1 =( x^3)^n - 1,
which of course has the factor x^3 - 1 =(x-1)(x^2+x+1), and
also x^(3n) - 1 = (x^n - 1)(x^(2n) + x^n + 1). Therefore,
(x-1)(x^2+x+1) divides (x^n-1)(x^(2n) + x^n + 1). Since (x-1) divides (x^n -1) we can look at : (x^n-1)/(x-1) is relatively
prime to (x^2 +x+1) whenever 3 does not divide n and so
(x^2+x+1) divides (x^(2n)+x^n+1) whenever 3 does not
divide n but fails to do so otherwise.

Answer 2

I can prove the first part and I'll state the result
for cases when n is a multiple of 3. I admit I don't
have a full proof yet, but I'll work on it!
Let's look at x^2n+x^n+1 when n is not a multiple of 3
and let's let w1 and w2 be the roots of x²+x+1=0, i.e.,
the complex cube roots of unity.
Note that w1²=w2 and w2²=w1.
Now look at x^2n+x^n+1, plugging in w1
into this expression, we get
(w1)^2n + (w1)^n + 1
Let's assume n is not a multiple of 3.
If n = 2(mod 3) this expression is just
w1 + w1²+1 = 0.
If n = 1(mod 3) we get
w2 + w2² + 1 = 0, which is also 0.
If we plug in w2 the same sort
of calculation reveals that we also get 0 in all cases.
So every root of x²+x+1 = 0 is also a root
of x^2n+x^n+1 = 0. So every factor(over
the complex numbers) of x²+x+1=0 is
a factor of x^2n+x^n+1 = 0 and thus
x²+x+1 is a factor of x^2n+x^n+1=0.
For example let's look at
x^10 + x^5+1 with n = 5.
w1^10 + w1^5+1 = w1+w1²+1 = 0
w2^10 + w2^5 + 1 = w2 + w2² +1 = 0,
so x²+x+1 is a factor of x^10+x^5+1.

Results for n a multiple of 3.

If n is a power of 3 x^2n + x^n + 1 is irreducible.
For example, x^6+x^3+1 is irreducible.
Now suppose n is not a power of 3.
If n is a multiple of 3 but not a multiple of 9,
x^2n+x^n+1 is divisible by x^6+x^3+1.
If n is a multiple of 9 but not a multiple of 27,
x^2n+x^n+1 is divisible by x^18 + x^9 +1,
etc.
I'll try to find proofs for these.
The point of all this is not that n is even or odd
but whether n is of the form 3m or not.

Answer 3

Your adding and subtracting step is incorrect.
(x^n +1)^2 = x^2n +2x^n + 1
You need go no further.

Answer 4

the respond is: (4a - one million)(4a-one million) you may desire to apply the trick with the aid of skill of multiplying sixteen and one million and making it a^2-8a+sixteen. whilst it aspects to (a-4)(a-4), than you upload the unique A making it (16a-4(16a-4) which as quickly as you element each and every with the aid of skill of dividing with the aid of skill of four, comes out to (4a-one million)(4a-one million).