Show that $x^4 + 8$ is irreducible over Z

Solution 1:

Since $\ 9^4\!+8\ $ is prime, it is irreducible by Cohn's irreducibility test. Or, with smaller prime, consider $\,f(2x) = 8(2x^4\!+1).\,$ Then $2x^4\!+1$ is irreducible by Cohn, since $\,2\cdot 3^4+1 = 163\,$ is prime (in fact, a very famous prime, the largest Heegner number, which explains why Euler's famous prime producing polynomial $\,n^2-n+41\,$ yields distinct primes for $\,n = 1,2,\ldots,40).$

Remark $\ $ Cohn's criterion can be viewed as an example of the general idea that the factorizations of a polynomial are constrained by the factorizations of the values that it takes. If one pushes this idea to the hilt one obtains a simple algorithm for polynomial factorization using factorization of its integer values and (Lagrange) interpolation. The ideas behind this algorithm are due in part to Bernoulli, Schubert, Kronecker. The algorithm is of more theoretical value than practical, since nowadays much more efficient algorithms are known.

There are also other closely related results. In $1918$ Stackel published the following simple:

Theorem If $\rm\, f(x)\,$ is a composite integer coefficient polynomial then $\rm\, f(n)\, $ is composite for all $\rm\,|n| > B,\, $ for some bound $\rm\,B.\,$ In fact $\rm\, f(n)\, $ has at most $\rm\, 2d\, $ prime values, where $\rm\, d = {\rm deg}(f)$.

The simple proof can be found online in Mott & Rose [3], p. 8. I highly recommend this delightful and stimulating $27$ page paper which discusses prime-producing polynomials and related topics.

Contrapositively, $\rm\, f(x)\, $ is prime (irreducible) if it assumes a prime value for large enough $\rm\, |x|\, $. As an example, Polya-Szego popularized A. Cohn's irreduciblity test, which states that $\rm\, f(x) \in \mathbb Z[x]\,$ is prime if $\rm\, f(b)\, $ yields a prime in radix $\rm\,b\,$ representation (so necessarily $\rm\,0 \le f_i < b).$

For example $\rm\,f(x) = x^4 + 6\, x^2 + 1 \pmod p\,$ factors for all primes $\rm\,p,\,$ yet $\rm\,f(x)\,$ is prime since $\rm\,f(8) = 10601\rm$ octal $= 4481$ is prime. Cohn's test fails if, in radix $\rm\,b,\,$ negative digits are allowed, e.g. $\rm\,f(x)\, =\, x^3 - 9 x^2 + x-9\, =\, (x-9)\,(x^2 + 1)\,$ but $\rm\,f(10) = 101\,$ is prime.

Conversely Bouniakowski conjectured $(1857)$ that prime $\rm\, f(x)\, $ assume infinitely many prime values (excluding cases where all the values of $\rm\,f\,$ have fixed common divisors, e.g. $\rm\, 2\: |\: x(x+1)+2\, ).$ However, except for linear polynomials (Dirichlet's theorem), this conjecture has never been proved for any polynomial of degree $> 1.$

Note that a result yielding the existence of one prime value extends to existence of infinitely many prime values, for any class of polynomials closed under shifts, viz. if $\rm\:f(n_1)\:$ is prime, then $\rm\:g(x) = f(x+ n_1\!+1)\:$ is prime for some $\rm\:x = n_2\in\Bbb N,\:$ etc.

For further detailed discussion of Bouniakowski's conjecture and related results, including heuristic and probabilistic arguments, see Chapter 6 of Ribenboim's The New Book of Prime Number Records.

[1] Bill Dubuque, sci.math 2002-11-12, On prime producing polynomials.

[2] Murty, Ram. Prime numbers and irreducible polynomials.
Amer. Math. Monthly, Vol. 109 (2002), no. 5, 452-458.

[3] Mott, Joe L.; Rose, Kermit. Prime producing cubic polynomials.
Ideal theoretic methods in commutative algebra, 281-317.
Lecture Notes in Pure and Appl. Math., 220, Dekker, New York, 2001.

Solution 2:

Proffering a different route. The polynomial $$ x^4+8\equiv x^4-2\pmod5. $$ Therefore it suffices to show that $x^4-2$ is irreducible as a polynomial in $\Bbb{Z}_5[x]$. To see that we observe that $2$ is of order four modulo $5$. Therefore the zeros of this polynomial (in some extension field of $\Bbb{Z}_5$) are primitive sixteenth roots of unity. The smallest exponent $m>0$ with the property $16\mid 5^m-1$ is $m=4$. This means that the zeros of $x^4-2$ all belong to a quartic extension of the prime field. The claim follows.