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

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.

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.

Your adding and subtracting step is incorrect.

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

You need go no further.

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