1. The problem statement, all variables and given/known data Prove that for all n ≥ 1, there exists a polynomial f(x) ∈ ℚ[x] with deg(f(x)) = n such that f(x) is irreducible in ℚ[x]. 2. Relevant equations In mathematics, a rational number is any number that can be expressed as the quotient or fractionp/q of two integers, a numeratorp and a non-zero denominatorq f(x) is called irreducible if it is not reduicble. f(x) is reducible if there exists g(x)*h(x) such that f(x) = g(x)*h(x) and deg(g(x)) < deg(f(x)), deg(h(x)) < deg(f(x)). http://mathworld.wolfram.com/IrreduciblePolynomial.html 3. The attempt at a solution Is it possible to use the quadratic discriminant here to show when no roots exist? b^2 - 4ac ≥ 0 So we say whenever b^2-4ac < 0 then f(x) is irreducible. And it would seem clear that you could always make a=c and a^2 > b^2.... Ouch. Nvm, this doesn't work for higher degree terms actually. In that case, hm. Could we work with cases, possibly induction? Case n = 1. Then by definition f(x) is irreducible. Induction step: n+1 can be irreducible. Then no, this is not working. Is the division algorithm necessary? Theorem 7 Let f be a polynomial with rational coefficients in lowest terms so that the numerators of the coefficients are relatively prime (i.e., have no common prime factor). Let d be the least common multiple (lcm) of the denominators of the coefficients of f. Let g = df. Then g has integer coef- ficients. Furthermore, f is irreducible over the rationals if and only if g is irreducible over the integers Am I going in the right direction by observing the division algorithm?