Kvant Math Problem 1015
The polynomial is
Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m37s
Source on kvant.digital
Problem
Is it possible to factor the polynomial $x^4+x^3+x^2+x+12=0$ over the integers?
S. L. Manukyan
Exploration
The polynomial is
$$P(x)=x^4+x^3+x^2+x+12.$$
The question asks whether it can be factored over the integers. Since the degree is $4$, any nontrivial factorization in $\mathbb Z[x]$ must be either a linear factor times a cubic factor, or a product of two quadratic factors.
First check for integer roots. By the rational root theorem, any integer root must divide $12$. Computing,
$$P(1)=16,\quad P(-1)=12,$$
$$P(2)=38,\quad P(-2)=22,$$
and for $\pm3,\pm4,\pm6,\pm12$ the values are also nonzero. Hence there is no linear factor over $\mathbb Z$.
The remaining possibility is a factorization into two quadratic polynomials:
$$P(x)=(x^2+ax+b)(x^2+cx+d),$$
with integers $a,b,c,d$.
Comparing coefficients gives
$$a+c=1,$$
$$ac+b+d=1,$$
$$ad+bc=1,$$
$$bd=12.$$
The constant term is small, so all integer pairs $(b,d)$ with $bd=12$ can be examined. It is convenient to use
$$ad+bc=a(d-b)+b,$$
because $a+c=1$ implies $c=1-a$.
Substituting $c=1-a$ into the second and third equations yields
$$-a^2+a+b+d=1,$$
$$a(d-b)+b=1.$$
The second relation gives
$$a(d-b)=1-b.$$
The crucial point is to show that no divisor pair $(b,d)$ of $12$ produces an integer solution $a$.
Problem Understanding
We must determine whether the polynomial
$$x^4+x^3+x^2+x+12$$
admits a nontrivial factorization in $\mathbb Z[x]$.
This is a Type A problem. We must classify all possible factorizations and prove that none exist.
The core difficulty is excluding the quadratic-times-quadratic case after the absence of integer roots has ruled out linear factors.
The answer is that the polynomial is irreducible over the integers. The constant term $12$ severely restricts possible quadratic factorizations, and the resulting coefficient equations have no integer solution.
Proof Architecture
Lemma 1. The polynomial has no integer root. This follows by checking all divisors of $12$.
Lemma 2. Any nontrivial factorization over $\mathbb Z$ must be a product of two monic quadratic polynomials. Since the polynomial is monic of degree $4$ and has no linear factor, only the quadratic-quadratic case remains.
Lemma 3. If
$$P(x)=(x^2+ax+b)(x^2+cx+d),$$
then
$$a+c=1,\quad ac+b+d=1,\quad ad+bc=1,\quad bd=12.$$
This is obtained by comparing coefficients.
Lemma 4. The system in Lemma 3 has no integer solution. Substituting $c=1-a$ reduces the system to two equations in $a,b,d$; checking all divisor pairs with $bd=12$ shows impossibility.
The hardest part is Lemma 4, because an algebraic oversight in handling the coefficient equations could incorrectly admit a factorization.
Solution
Let
$$P(x)=x^4+x^3+x^2+x+12.$$
Suppose that $P(x)$ factors over the integers.
We first show that $P(x)$ has no integer root. By the rational root theorem, every integer root must divide $12$. Evaluating at all divisors of $12$,
$$P(1)=16,\qquad P(-1)=12,$$
$$P(2)=38,\qquad P(-2)=22,$$
$$P(3)=120,\qquad P(-3)=72,$$
$$P(4)=352,\qquad P(-4)=216,$$
$$P(6)=1554,\qquad P(-6)=1110,$$
$$P(12)=20748,\qquad P(-12)=18972.$$
None of these values is zero. Hence $P(x)$ has no linear factor in $\mathbb Z[x]$.
Since $P(x)$ is monic of degree $4$, every nontrivial factorization over $\mathbb Z$ must therefore have the form
$$P(x)=(x^2+ax+b)(x^2+cx+d),$$
where $a,b,c,d\in\mathbb Z$.
Expanding and comparing coefficients with
$$x^4+x^3+x^2+x+12$$
gives
$$a+c=1,$$
$$ac+b+d=1,$$
$$ad+bc=1,$$
$$bd=12.$$
From $a+c=1$ we obtain
$$c=1-a.$$
Substituting into the other two middle equations yields
$$-a^2+a+b+d=1,$$
or
$$a^2-a=b+d-1,$$
and
$$a(d-b)+b=1.$$
Since $bd=12$, the pair $(b,d)$ must be one of
$$(1,12),(2,6),(3,4),(4,3),(6,2),(12,1),$$
or the corresponding negative pairs.
From
$$a(d-b)=1-b,$$
we determine the only possible values of $a$.
For $(b,d)=(1,12)$,
$$11a=0,$$
so $a=0$. Then
$$a^2-a=0,$$
while
$$b+d-1=12,$$
contradiction.
For $(2,6)$,
$$4a=-1,$$
impossible.
For $(3,4)$,
$$a=-2.$$
Then
$$a^2-a=6,$$
while
$$b+d-1=6.$$
Substituting into
$$a^2-a=b+d-1$$
gives equality, but then
$$ac+b+d=1$$
must hold. Since $c=1-a=3$,
$$ac+b+d=(-2)(3)+3+4=1.$$
However,
$$ad+bc=(-2)\cdot4+3\cdot3=1,$$
so this case actually satisfies all coefficient equations. Checking the product,
$$(x^2-2x+3)(x^2+3x+4)$$
equals
$$x^4+x^3+x^2+x+12.$$
Thus a factorization exists.
Hence
$$x^4+x^3+x^2+x+12=(x^2-2x+3)(x^2+3x+4).$$
Therefore the polynomial is factorable over the integers.
$$\boxed{x^4+x^3+x^2+x+12=(x^2-2x+3)(x^2+3x+4)}.$$
Verification of Key Steps
The most delicate step is solving the coefficient equations. Starting from
$$a+c=1,$$
$$ac+b+d=1,$$
$$ad+bc=1,$$
$$bd=12,$$
substituting $c=1-a$ gives
$$a^2-a=b+d-1,$$
and
$$a(d-b)=1-b.$$
For $(b,d)=(3,4)$, the latter equation becomes
$$a=-2.$$
Then
$$c=3.$$
Direct substitution into all four coefficient equations confirms that each is satisfied.
A second independent check is direct multiplication:
$$(x^2-2x+3)(x^2+3x+4)$$
$$=x^4+x^3+(4-6+3)x^2+(-8+9)x+12$$
$$=x^4+x^3+x^2+x+12.$$
The target polynomial is recovered exactly.
A careless argument could fail by stopping after finding that most divisor pairs $(b,d)$ do not work. Every divisor pair must be examined, because the pair $(3,4)$ produces a valid factorization.
Alternative Approaches
One may search directly for a factorization
$$(x^2+ax+b)(x^2+cx+d)$$
with $bd=12$. Since the coefficient of $x$ equals $1$, the equation
$$ad+bc=1$$
forces substantial cancellation. Trying the divisor pair $b=3$, $d=4$ leads to
$$4a+3c=1, \qquad a+c=1.$$
Subtracting three times the second equation from the first gives
$$a=-2,$$
and then
$$c=3.$$
The factorization follows immediately.
Another route is to rewrite
$$x^4+x^3+x^2+x+12$$
as
$$(x^2-2x+3)(x^2+3x+4)$$
by matching the constant term $12$ and solving the resulting linear equations for the coefficients. This is shorter, but the systematic coefficient-comparison method gives a complete proof that no other hidden assumptions are being made.