Is n a power of 3 if and only if x^(2n) + x^n + 1 is irreducible?

Let below f(x) = x²ⁿ + xⁿ + 1, where n is a natural number.

If n is not a power of 3, it turns out that f(x) is reducible over the rationals.

Indeed since (x² + x + 1)(x - 1) = x³ - 1 by modulo x² + x + 1 we have

x² ≡ -x - 1 (mod x² + x + 1) and x³ ≡ 1 (mod x² + x + 1)

/these congruences are by modulo principal ideal, generated by the polynomial x² + x + 1/

1) Let n ≡ 1 (mod 3), or n = 3k + 1 for some k - natural; then f(x) is divisible by x² + x + 1 (see Rita's examples n = 4 and n = 7). Proof:

f(x) = x²ⁿ + xⁿ + 1 = x^(2(3k + 1)) + x^(3k + 1) + 1 ≡

≡ (x³)^(2k) * x² + (x³)^k * x + 1 ≡ 1*x² + 1*x + 1 ≡ 0 (mod x² + x + 1)

2) Let n ≡ 2 (mod 3), or n = 3k + 2 for some k - natural; then f(x) is divisible again by x² + x + 1 (Rita's examples n = 2, n = 5 and n = 8).

Proof: f(x) = x²ⁿ + xⁿ + 1 = x^(2(3k + 2)) + x^(3k + 2) + 1 ≡

≡ (x³)^(2k) * x⁴ + (x³)^k * x² + 1 ≡ 1*1*x + 1*x² + 1 ≡ 0 (mod x² + x + 1)

3) Let n ≡ 0 (mod 3), but n isn't a power of 3; or n = 3^k * m for some k - natural and m > 1, m not divisible by 3; then f(x) is divisible by

x^(2(3^k)) + x^(3^k) + 1. See Rita's example n = 6; also:

x³⁰ + x¹⁵ + 1 = (x⁶ + x³ + 1)(x²⁴ - x²¹ + x¹⁵ - x¹² + x⁹ - x³ + 1);

x³⁶ + x¹⁸ + 1 = (x¹⁸ + x⁹ + 1)(x¹⁸ - x⁹ + 1);

x⁹⁰ + x⁴⁵ + 1 = (x¹⁸ + x⁹ + 1)(x⁷² - x⁶³ + x⁴⁵ - x³⁶ + x²⁷ - x⁹ + 1)

/Hope I haven't mistyped anywhere - a check is highly recommended!/

Proof: A simple change of variable x^(3^k) = y brings the things to a congruence by modulo y² + y + 1, cases 1) and 2) above, since m is not divisible by 3.

From 1) and 2) follows that f(x) is reducible over the rationals if n is not power of 3, because the proven divisibility implies the coefficients of the divisor and quotient are also rational.

Now it remains to prove the irreducibility if n is a power of 3. I shall think it over, Y!A has no obligation to close the question in 4 days or to send it to voting as before (also I wonder why Y!A didn't email me about this question as usual), so please keep it open for a while (many of my long ago answered questions are still open, probably the askers have lost interest). This incomplete answer may also motivate some of our contributors to resolve it completely.

* * * * *

Next attempt: f(x) reminds the standard approach to prove the irreducibility of the cyclotomic polynomial - the substitution x = t + 1. If we follow that approach for example to x⁶ + x³ + 1, we shall have:

(t + 1)⁶ + (t + 1)³ + 1 = t⁶ + 6t⁵ + 15t⁴ + 21t³ + 18t² + 9t + 3

Now the prime number 3 does not divide the leading coefficient 1; 3 divides all other coefficients, but 9 = 3² does not divide the last coefficient - according Eisenstein's Criterion it is irreducible. Probably it works in case n = 3^k - again the leading coefficient is 1, the last one is again 3 and 9 does not divide it, the only problem remains to prove that the middle coefficient C[2*3^k, 3^k] + 1 is divisible by 3. All other coefficients are sum of two binomials, divisible by 3.

Attempts continue...

* * * * * * *

4) Let below n is a power of 3 (n = 3^k). The standard approach to use the popular Eisenstein's Criterion to prove the irreducibility of the cyclotomic polynomial by substitution x = t + 1 can help here too.

http://en.wikipedia.org/wiki/Eisenstein%27s_criter...

Here is the proof: (t + 1)²ⁿ + (t + 1)ⁿ + 1 =

= ∑[i=0 to 2n] C[2n, i] t^i + ∑[i=0 to n] C[n, i] t^i + 1 =

= t²ⁿ + C[2n, 1] t²ⁿ⁻¹ + C[2n, 2] t²ⁿ⁻² + ... + C[2n, n-1] tⁿ⁺¹ +

+ (C[2n, n] + 1) tⁿ + (C[2n, n-1] + C[n,1]) tⁿ⁻¹ + ...

... + (C[2n, 1] + C[n, n-1]) t + 1 + 1 + 1

All C[2n, i] = (2n)(2n - 1)...(2n - i + 1)/i!, i = 1,2,...n-1 are divisible by 3 (they are integers and 3 goes in each numerator in higher power than in the corresponding denominator); the same is true about C[n, i], i = 1,2,...n-1, finally

C[2n, n] = ∑[i=0 to n] C[n, i]² =

= 1 + (C[3^k, 1])² + (C[3^k, 2]² + ... + (C[3^k, 1])² + 1 ≡ 1 + 1 (mod 3),

hence the middle coefficient C[2n, n] + 1 ≡ 1 + 1 + 1 ≡ 0 (mod 3)

Now the prime number 3 does not divide the leading coefficient 1; 3 divides all other coefficients, but 9 = 3² does not divide the last coefficient 3, what proves the required irreducibility.

I would welcome a shorter or an easier solution, if any.

Irreducible over the rationals?

By rational root theorem, only 1 or -1 can be roots, and with 3 terms all equal to ±1, they obviously can't be.