For which integers \(n\) is \[ x^4+n \] reducible in \(\mathbb Z[x]\)?
Solution
1. If\[ n=-m^2 \]for some \(m\in\ZZ\), then \(x^4+n\) is reducible in \(\ZZ[x]\).
One has \[ x^4-m^2 = (x^2-m)(x^2+m). \] Both factors have positive degree, so the polynomial is reducible.
2. If\[ n=4m^4 \]for some \(m\in\ZZ\), then \(x^4+n\) is reducible in \(\ZZ[x]\).
Direct multiplication gives \[ \begin{aligned} &(x^2+2mx+2m^2)(x^2-2mx+2m^2)\\ &\qquad= (x^2+2m^2)^2-(2mx)^2\\ &\qquad= x^4+4m^4. \end{aligned} \] Thus \(x^4+n\) factors nontrivially over \(\ZZ\).
3. Conversely, suppose \(x^4+n\) is reducible in \(\ZZ[x]\). Itsfactors may be chosen monic.
The polynomial \(x^4+n\) is monic. In any factorization into positive-degree integer polynomials, the product of the leading coefficients is \(1\). Hence both leading coefficients are units \(\pm1\); changing both signs if necessary gives monic factors.
4. If \(x^4+n\) has a linear factor, then\[ n=-m^2 \]for some \(m\in\ZZ\).
A monic linear factor has the form \(x-r\) with \(r\in\ZZ\), so \[ r^4+n=0. \] Thus \[ n=-r^4=-(r^2)^2. \] Taking \(m=r^2\) gives the stated form.
5. If \(x^4+n\) is reducible and has no linear factor, then it factorsas two monic quadratics\[ x^4+n = (x^2+ax+b)(x^2+cx+d) \]with \(a,b,c,d\in\ZZ\).
By step 3 the factors may be chosen monic. Their positive degrees sum to \(4\). Since there is no factor of degree \(1\), the only possible degree split is \(2+2\), giving the displayed form.
6. In the factorization of step 5,\[ c=-a \]and\[ a(d-b)=0, \qquad b+d-a^2=0, \qquad bd=n. \]
Expanding after comparing the coefficient of \(x^3\) gives \[ c=-a \] and hence \[ \begin{aligned} (x^2+ax+b)(x^2-ax+d) &= x^4 + (b+d-a^2)x^2\\ &\qquad + a(d-b)x + bd. \end{aligned} \] The coefficients of \(x^2\) and \(x\) in \(x^4+n\) are zero, and its constant term is \(n\), yielding the three equations.
7. If \(a=0\) in step 6, then\[ n=-m^2 \]for some \(m\in\ZZ\).
When \(a=0\), the equation \[ b+d-a^2=0 \] becomes \[ d=-b. \] Therefore \[ n=bd=-b^2. \] Taking \(m=b\) gives the required form.
8. If \(a\ne0\) in step 6, then\[ n=4m^4 \]for some \(m\in\ZZ\).
Since \[ a(d-b)=0 \] and \(a\ne0\), one has \[ d=b. \] Then \[ b+d-a^2=0 \] becomes \[ 2b=a^2. \] Thus \(a^2\) is even, so \(a\) is even. Write \[ a=2m. \] Then \[ b=\frac{a^2}{2}=2m^2, \] and hence \[ n=bd=b^2=4m^4. \]
9. Therefore\[ \boxed{ x^4+n\text{ is reducible in }\ZZ[x] \iff n=-m^2\text{ or }n=4m^4 \text{ for some }m\in\ZZ. } \]
Steps <1>1–<1>2 prove that every integer in either displayed family gives a reducible polynomial. Conversely, if the polynomial is reducible, step 4 handles the linear-factor case, while steps <1>5–<1>8 handle the only remaining factorization type.
10. Q.E.D.
Step 9 is the complete classification.