Note: This article is based extensively on the content of 《数学奥林匹克小丛书》, supplemented by my own insights and reorganized to cover additional topics not addressed in the original text. Therefore, it should be regarded as a newly compiled set of notes rather than a completely original work.
Basics
Definition
The definition of “factorize $A$ over $B$” is: To write the polynomial $A$ as a product of several factors where each factor has coefficients in field $B$, and every factor in this expression is irreducible over the specified field.
($A$ is irreducible over $B$ means that “$A$ cannot be written as a product of at least two nonconstant polynomials with coefficients in $B$”.)
Factorization depends on the field over which you are working. You can factorize over the rational, the real, the complex, or other fields, depending on your goals.
Here are some examples of factorizations:
$$
\begin{aligned}
&24 =2^3 \times 3 \\
&a^2-5b^2 =(a+\sqrt{5}b)(a-\sqrt{5}b) \\
&64a^4+1 =(8a^2-4a+1)(8a^2+4a+1) \\
&a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\
\end{aligned}
$$
If you are wondering how such factorizations are found, do not worry. This article aims to guide you through methods and reasoning that will help you understand and perform these steps on your own.
Fundamentals
Before we moving on to the following parts, I think it’s necessary to introduce the basic concepts and the fundamentals of factorizations first…
Consider this factorization for polynomial over the real numbers field:
$$x^3+2x^2+3x+6$$
Student A:
$$
\begin{aligned}
&x^3+2x^2+3x+6 \\
=\ & (x^3+3x)+ (2x^2+6) \\
=\ & x(x^2+3)+ 2(x^2+3) \\
=\ & (x+2)(x^2+3) \\
\end{aligned}
$$
Student B:
$$
\begin{aligned}
&x^3+2x^2+3x+6 \\
=\ & (x^3+2x^2)+ (3x+6) \\
=\ & x^2(x+2)+ 3(x+2) \\
=\ & (x+2)(x^2+3) \\
\end{aligned}
$$
We can see that student A and B both get the same result, though they are using different approaching methods to factorize.
But here is the question, how can we be sure that even if we use different approaching methods to factorize, we are still going to get the same factorization result?
In other words, is the factorization unique over the real numbers field?
The answer is yes, the factorization over the real numbers field is unique.
And here is a brief informal reasoning which we won’t discuss the details further, because these mathematical concepts are way more beyond the contents that we are focusing on currently:
Based on the fundamental theorem of algebra and the complex conjugate root theorem, we are able to conclude that:
(Assume $P(x)$ and $f(x)$ to be any polynomial of $1$ variable $x$ with real coefficients)
1. $\forall \ P(x)$ can always be factorized to the product form of several $f(x)$ of 1st degree and 2nd degree.
2. The factorization of $\forall \ P(x)$ over the real numbers field is unique.
It is essential to understand the two fundamental properties of factorization as we will encounter various factorization problems. Thus, I would like to claim these two conclusions at this point. Please keep them in mind to aid your comprehension of factorization.
Formulas
$$
\begin{aligned}
& a^2-b^2=(a+b)(a-b) \\
& a^3+b^3=(a+b)\left(a^2-a b+b^2\right) \\
& a^3-b^3=(a-b)\left(a^2+a b+b^2\right) \\
& a^2+2 a b+b^2=(a+b)^2 \\
& a^2-2 a b+b^2=(a-b)^2 \\
& a^3+3 a^2 b+3 a b^2+b^3=(a+b)^3 \\
& a^3-3 a^2 b+3 a b^2-b^3=(a-b)^3 \\
&a^2+b^2+c^2+2 ab+2 bc+2 ca =(a+b+c)^2\\
&a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2} b+a^{n-3} b^2+\cdots+a b^{n-2}+b^{n-1}\right), \quad n \in \mathbb{Z^+}\\
&a^n+b^n=(a+b)\left(a^{n-1}-a^{n-2} b+a^{n-3} b^2-\cdots-a b^{n-2}+b^{n-1}\right), \quad n \in (2\mathbb{Z^+}-1)\\
\end{aligned}
$$
The listed factorization formulas above are so important that you should master. And all these formulas are easy to prove except the last 2, which requires you to apply the polynomial remainder theorem (which we will discuss later).
Strategies
Note: In the following exposition, I will attempt to simulate the thought process of the problem setters, exploring how they might craft and refine problems to push contestants beyond standard solution templates and inspire deeper reasoning.
Method of undetermined coefficients
MUC is an approach that does the factorization reversely through assuming the factors expressions and then expand forwardly.
Problem: Please factorize
$$Ax^2+Bx+C, \quad A,B,C \in \mathbb{Q} $$
over the integer numbers field.
First, let’s be clear that when we are talking about the factorization of a polynomial, it means we are trying to find its polynomial factors. So a polynomial won’t have some factors like $(ax^{0.5}+b)$ or $(ax^{-1}+b)$ according to the definition of the polynomial.
If $Ax^2+Bx+C$ can be factored over the integers, it must have 2 factors polynomial of 1st degree in the form of $ax+b$, since it is a 2nd degree polynomial. If it cannot be factored in this way, then it can only have factors of 2nd or higher degree, which would imply that the polynomial cannot be factored over the integer numbers field or even over the real numbers field.
Now, assume $\exists \ a,b,c,d \in \mathbb{Q}, \quad c,d \ne 0$ such that:
$$
\begin{aligned}
(ax+b)(cx+d)&= Ax^2+Bx+C \\
acx^2+ (ad+bc)x+bd &= Ax^2+Bx+C \\
\end{aligned}
$$
Then we have:
$$
\begin{cases}
&ac &= A \\
&ad+bc&=B \\
&bd&=C \\
\end{cases}
\label{eq:function}
\implies
\begin{cases}
&a &= \frac{A}{c} & (1)\\
&ad+bc&=B & (2)\\
&b&=\frac{C}{d} & (3)\\
\end{cases}
$$
Substitute $(1), (3)$ into $(2)$ get:
$$
\begin{aligned}
\frac{A}{c}d+\frac{C}{d}c&=B \\
Ad^2-Bcd+Cc^2&=0 \\
\end{aligned}
$$
Now the question has changed to factorize $Ad^2-Bcd+Cc^2$ over the integer numbers field, but it is even more complicated than the original problem $Ax^2+Bx+C$ …
So that indicates, maybe we should change the way we think.
Let’s solve the following problem first.
Problem: Please factorize
$$x^2+x-6$$
over the integer numbers field.
Assume $\exists \ a,b,c,d \in \mathbb{Q}$ such that:
$$
\begin{aligned}
(ax+b)(cx+d)&= x^2+x-6 \\
acx^2+ (ad+bc)x+bd &= x^2+x-6 \\
\end{aligned}
$$
Then we have:
$$
\begin{cases}
&ac &= 1 \\
&ad+bc&=1 \\
&bd&=-6 \\
\end{cases}
$$
Since attempting to solve the entire system of equations would be difficult, it may be more effective to use a different approach. One possible method is to find integer solutions by substituting numbers into the equations system first… and we luckily got:
$$
\begin{cases}
a &= 1 \\
b&=3 \\
c &= 1 \\
d&=-2 \\
\end{cases}
$$
Therefore $x^2+x-6=(ax+b)(cx+d)=(x+3)(x-2)$
Now, let’s think about how to make this problem harder by doing some tiny modifications:
$$x^2+x-6 \ \to\ \dfrac{x^2+x-6}{3} = \dfrac{x^2}{3}+\dfrac{x}{3}-2$$
Here it is! Then let’s try to solve them.
Problem: Please factorize
$$\dfrac{x^2}{3}+\dfrac{x}{3}-2$$
over the integer numbers field.
Are you going to apply the previous solving method? Yes, you can, but this time, you’ll definitely need to deal with fractions (because integer $\times$ integer $\ne$ fraction… etc). Since the polynomial in the problem contains fractions, it may be more convenient to eliminate them at first. Therefore, transform:
$$\dfrac{x^2}{3}+\dfrac{x}{3}-2 = \dfrac{1}{3}(x^2+x-6)$$
Then we can focus on solving $x^2+x-6$ first, then times the result with $\dfrac{1}{3}$ . So $\dfrac{x^2}{3}+\dfrac{x}{3}-2=\dfrac{1}{3}(x+3)(x-2)$, accordingly the answer is $\dfrac{x^2}{3}+\dfrac{x}{3}-2$ cannot be factorized over the integer numbers field because its factors contain non-integer coefficients.
Similarly…
$$x^2+x-6 \ \to\ 7(x^2+x-6) = 7x^2+7x-42 $$
Problem: Please factorize
$$7x^2+7x-42$$
over the integer numbers field.
Using the same approach, we can get the answer: $7x^2+7x-42= 7(x+3)(x-2)$
How about this…
$$x^2+x-6= (x+3)(x-2) \ \to\ (x+3y)(x-2y)=x^2+xy-6y^2 $$
Problem: Please factorize
$$x^2+xy-6y^2$$
over the integer numbers field.
We can still apply the MUC for this problem, but please make sure that you set up $(ax+by)(cx+dy)$ instead of some others factors like $(ax+b)(cx+dy^2)$, because that product cannot produce the term of $xy$. Furthermore, it is important to note that any factors polynomials of a polynomial cannot have a degree higher than the degree of the polynomial itself (you can try to prove it using the MUC, it is quite easy).
So the result is: $x^2+xy-6y^2= (x+3y)(x-2y)$
Keep modifying the problem…
$$x^2+x-6 = (x+3)(x-2) \ \to\ (x+2y+3)(x-5y-2) = x^2 -3xy -10y^2 + x -19y -6 $$
Problem: Please factorize
$$x^2 -3xy -10y^2 + x -19y -6$$
over the integer numbers field.
For this problem, maybe you will not know how to set up the factors with undetermined coefficients unless you saw the process how I derive it. Then what should you do to develop your “MUC” ability effectively? One word, practicing.
For $x^2 -3xy -10y^2 + x -19y -6$, we can notice that the there are terms of 2nd , 1st and 0 degree, so we should set up the factors polynomials with the max degree of 1st in the form of $(ax+by+c)(dx+ey+f)$ , though you can also try to set up the factors polynomials in the form of $(ax+c)(by+d)$ , this would tell “no solution” instead, and it’s not because $x^2 -3xy -10y^2 + x -19y -6$ cannot be factorized over the integer numbers fields, it’s due to your wrong factors form setting at the beginning.
Therefore, by processing the right factors form setting, we can get $x^2 -3xy -10y^2 + x -19y -6 = (x+2y+3)(x-5y-2)$ through testing the integers substitutions finitely.
$$x^2+x-6 \ \to\ x^2+x-6-1 = x^2+x-7 $$
Problem: Please factorize
$$x^2+x-7$$
over the integer numbers field.
Similarly, we have:
$$
\begin{cases}
&ac &= 1 \\
&ad+bc&=1 \\
&bd&=-7 \\
\end{cases}
$$
However, through setting up the right form of factors and testing the integers substitutions finitely, we discovered that the problem can only have the answer: $x^2+x-7=x^2+x-7$, because the equations system fails to satisfy all the integer substitution, in other words, this equations system has no integer solutions. However, it can be factorized over the real numbers field:
$$
x^2+x-7=-\dfrac{1}{4}(-2 x+\sqrt{21}-1)(2 x+\sqrt{21}+1)
$$
Method of substitutions
MS is a powerful technique for factorizing polynomials with high degrees. By using MS, we can simplify the original polynomial by reducing its degree or creating the symmetric situation. This approach can greatly simplify the process of factoring complicated polynomials.
Now, let me set up a problem first.
$$x^2+5x-6 \ \to\ (x^2)^2+5(x^2)-6 = x^4+5x^2-6$$
Problem: Please factorize
$$x^4+5x^2-6$$
over the integer numbers field.
Assume $y=x^2$, then
$$x^4+5x^2-6=y^2+5y-6=(y+6)(y-1)=(x^2+6)(x^2-1)=(x^2+6)(x+1)(x-1)$$
Similarly…
$$x^2+5x-6 \ \to\ (x^2+5x-6)^2+5(x^2+5x-6)-6$$
Problem: Please factorize
$$(x^2+5x-6)^2+5(x^2+5x-6)-6$$
over the integer numbers field.
Assume $y=x^2+5x-6$, then
\begin{align*}
(x^2+5x-6)^2+5(x^2+5x-6)-6
&=y^2+5y-6=(y+6)(y-1) \\
&= (x^2+5x-6+6)(x^2+5x-6-1) s\\
&= x(x+5)(x^2+5x-7)
\end{align*}
What if I expend the question:
$$(x^2+5x-6)^2+5(x^2+5x-6)-6 = x^4+10 x^3+18 x^2-35 x$$
Since in this form, the problem won’t provide you any hint to set up the substitution $y=x^2+5x-6$ directly, you need to discover that hidden property yourself instead, but it is very hard to notice that \((x^2+5x-6)^2+5(x^2+5x-6)-6 = x^4+10 x^3+18 x^2-35 x\) reversely… So, is there a better method except MS to solve this problem? Luckily, yes, and we will discuss later.
How about this…
$$x^2+5x-6 \ \to\ x^2+5x(x^2+5x-6)-6(x^2+5x-6)^2$$
Problem: Please factorize
$$x^2+5x(x^2+5x-6)-6(x^2+5x-6)^2$$
over the integer numbers field.
Assume $y=x^2+5x-6$, then
$$x^2+5x(x^2+5x-6)-6(x^2+5x-6)^2=x^2+5xy-6y^2=(x+6y)(x-y)$$
Problem: Please prove that the product of (any 4 consecutive positive integers) plus 1 can become a perfect square number.
$\forall \ x \in \mathbb{Z^+}$, we have:
$$x(x+1)(x+2)(x+3)+1=\Big( x(x+3) \Big)\Big( (x+1)(x+2) \Big) +1 =(x^2+3x)(x^2+3x+2) +1 $$
Assume $y=x^2+3x+1$, then we derive the answer
\begin{align*}
(x^2+3x)(x^2+3x+2)+1
&=(x^2+3x+1-1)(x^2+3x+1+1)+1\\
&=(y-1)(y+1)+1\\
&=y^2-1+1=y^2=(x^2+3x+1)^2
\end{align*}
It’s time to explain my motivation:
Our primary goal is to factorize $x(x+1)(x+2)(x+3)+1$ and satisfy
$$x(x+1)(x+2)(x+3)+1=(?)^2$$
And according to the 2 fundamental properties of factorizations:
1. $\forall \ P(x)$ can always be factorized to the product form of several $f(x)$ of 1st degree and with 2nd degree.
2. The factorization of $\forall \ P(x)$ over the real numbers field is unique.
We know that the polynomial $x(x+1)(x+2)(x+3)+1$ must be factorable over the integer field since this problem is solvable, and also using different approaches won’t affect the result of the factorization over the integer field of this polynomial.
However, expending the polynomial directly may be not a good approach, unless you’ve discovered that there is no other way. So then I have:
$$x(x+1)(x+2)(x+3)+1=\Big( x(x+3) \Big)\Big( (x+1)(x+2) \Big) +1 =(x^2+3x)(x^2+3x+2) +1$$
But why didn’t I factorize $x(x+1)(x+2)(x+3)$ in this way:
$$x(x+1)(x+2)(x+3)+1=\Big( x(x+1) \Big)\Big( (x+2)(x+3) \Big) +1 =(x^2+x)(x^2+5x+6) +1$$
Because $(x^2+3x)(x^2+3x+2)+1$ indicates the hidden identity of $x^2+3x$, and then I could apply the substitution easily to replace $x^2+3x$, but for $(x^2+x)(x^2+5x+6) +1$, you cannot do so.
Then why did I set up $y=x^2+3x+1$ instead of $y=x^2+3x$, though both of them contain $x^2+3x$ ? It is because the former could reveal the hidden symmetry property of $(x^2+3x)(x^2+3x+2)+1$ and then I can naturally apply the formula $(y-1)(y+1)=y^2-1$ and derive the perfect square after adding 1, but the latter cannot.
Problem: Please factorize
$$x^4+(x-8)^4$$
over the integer numbers field.
If you know the binomial theorem, you can expand the polynomial $x^4+(x-8)^4=2 x^4-32 x^3+384 x^2-2048 x+4096$ and then do the factorization. However, it is not recommended, want to know why? Here is the answer:
Assume $y=(x-4)$ and $z=y^2$, then we rewrite the express as
$$(y+4)^4+(y-4)^4=2y^4+192y^2+512=2(y^2)^2+192(y^2)+512=2z^2+192z+512$$
When comparing $2z^2+192z+512$ and $2x^4-32x^3+384x^2-2048x+4096$, it is evident that factoring $2z^2+192z+512$ is a much simpler task than factoring the latter polynomial. The reason is because the substitution of $y=(x-4)$ what we were using could reveal the symmetry property of the polynomial and this property can provide a significant advantage in problem-solving, for this problem, it helped us eliminate the odd degree terms and accordingly decrease the complexity of the factorization.
Now, how should we factorize $2z^2+192z+512$ ? Applying MUC? Yes, it is a valid approach, but not efficient enough for this case, we will discuss it later.
Polynomial Remainder Theorem and Rational Root Theorem
Assume $P(x)$ is a polynomial and $\forall \, r \in \mathbb{R}, \quad \exists! \, Q(x), R$ such that $P(x)=Q(x)(x-r)+R$, where $Q(x)$ is a polynomial and $R \in \mathbb{R}$, accordingly we have $P(r)=Q(r)\times 0+R=R$ . If $P(r)=0$, then $R=0$, which $P(x)=Q(x)(x-r)$ . In other words, if $P(r)=0$ , then $(x-r)$ must be one of its factors.
Problem: Please factorize
$$x^3-2x^2-5x+6$$
over the integer numbers field.
Through substituting some small integers, we found that $P(1)=0$ , therefore $(x-1)$ must be one of its factors. Then applying the polynomial division and MUC:
$$x^3-2x^2-5x+6=(x-1)(x^2-x-6)=(x-1)(x-3)(x+2)$$
Problem: Please factorize
$$2z^2+192z+512$$
over the real numbers field.
$2z^2+192z+512=2(z^2+96z+256)$, for this problem, applying the MUC is definitely not a good idea since the coefficients are too large. Besides, substituting random real numbers to test is also not an efficient approach… Therefore, we can try to solve the equation $P(z)=z^2+96z+256=0$ directly to find the $r$ that let $P(r)=0$ through applying the quadratic formula $z=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}=-48\pm 32\sqrt{2}$ , so the solution is:
$$2z^2+192z+512=2(x+48-32\sqrt{2})(x+48+32\sqrt{2})$$
It’s time to introduce a corollary of PRT — RRT (Rational Root Theorem) (it is not hard to prove):
For a polynomial equation $P(x)=\displaystyle \sum_{i=0}^{n}\left ( a_{i}x^{i}\right )=0$ with $a_{i}\in \mathbb{Z}$, $n\ge 0$ and $a_{n} \ne 0$ , then every rational solution $x=\dfrac{p}{q}$ which this equation satisfies: $p$ is the factor of $a_{0}$ and $q$ is the factor of $a_{n}$.
So the steps to apply this corollary are:
- Write down all the factors of $a_{0}$ and $a_{n}$ respectively.
- Use these factors ($p \mid a_{0}$ and $q \mid a_{n}$) to form all the possible rationals $\left \{\dfrac{p}{q}\right \}$.
- Test these candidates by directly substituting them into the polynomial $P(x)$ to check if they satisfy $P(\dfrac{p}{q}) = 0$ with PRT.
Problem: Please factorize
$$x^4+10 x^3+18 x^2-35x$$
over the integer numbers field.
$x^4+10 x^3+18 x^2-35x = x(x^3+10 x^2+18 x-35)$, for $x^3+10 x^2+18 x-35$, as $a_{0}=-35$, $a_{3}=1$, then all the possible rationals are $\left \{ -35, -7, -5, -1, 1, 5, 7, 35 \right \}$. By applying MS, $x=-5$ is the only rational root for the polynomial equation $P(x)=0$, which proves that $x+5$ is the only rational factor of $x^3+10 x^2+18 x-35$ by PRT. Then by using polynomial division, $x^3+10 x^2+18 x-35 = (x+5)(x^2+5x-7)$, since $x+5$ is the only rational factor of it, therefore $x^2+5x-7$ is irreducible over the rational numbers field. So the factorization result is $x^4+10 x^3+18 x^2-35x = x(x+5)(x^2+5x-7)$.
Clearly, RRT is a powerful tool for finding rational roots and factoring polynomials over the integers (or rationals). However, they do not help us factor polynomials that lack linear rational or integer factors, because no rational root doesn’t necessarily implies irreducibility.
Recall that any polynomial $P(x)$ with real coefficients can be written (uniquely, up to constant factors) as a product of one or more 1st-degree or 2nd-degree polynomials.
- So for polynomials of degree $2$ or $3$ with rational coefficients, if RRT finds no rational root, then the polynomial is irreducible over $\mathbb{Q}$. This is because any nontrivial factorization of a quadratic or cubic must include a linear factor $(x -r)$, which would imply the existence of a rational root $r$.
- For polynomials of degree $4$ or higher, having no rational root does not guarantee irreducibility over $\mathbb{Q}$. A higher-degree polynomial may factor into two or more irreducible polynomials each of degree $\ge 2$. For instance, $ x^4 + 2x^2 + 1 = (x^2 + 1)^2 $ has no rational (or real) roots, yet it is clearly reducible over $\mathbb{Q}$.
Cyclotomic Polynomials and Primitive Roots
For the previous example, $x^4 + 2x^2 + 1 = (x^2 + 1)^2$ is indeed a valid factorization over $\mathbb{Z}$, yet RRT does not reveal any rational root. Thus, to handle more advanced factorizations, especially when a polynomial’s roots lie on the complex unit circle, we turn to cyclotomic polynomials and primitive roots. These concepts allow us to incorporate the PRT in an extended setting: instead of substituting a rational number to test if $(x -r)$ divides the polynomial, we substitute a complex number (a root of unity) to see if its minimal polynomial (a cyclotomic polynomial) divides the polynomial in question.
A complex number \(\omega\) is called an \(n\)-th root of unity if
$$
\omega^n = 1
$$
They are precisely the solutions to the equation \(x^n -1 = 0\), and can be represented by
$$
\omega_k = e^{\tfrac{2\pi i k}{n}},
\quad
k = 0,1,\dots,n-1
$$
Geometrically, these \(n\)-th roots of unity lie on the unit circle at equal angular intervals of \(\dfrac{2\pi}{n}\).
An \(n\)-th root of unity \(\omega\) is called primitive if \(\omega^n = 1\) and, for every divisor \(d < n\), \(\omega^d \neq 1\). Equivalently, the order of \(\omega\) is exactly \(n\).
Now let us see how the PRT can still be used when the suspected root is complex. Suppose \(P(x)\in \mathbb{Z}[x]\) and we guess that a non-rational complex number \(\omega\) might be a root of $P$. By the PRT,
$$
P(\omega) = 0
\implies
(x -\omega) \mid P(x) \quad \text{in the relevant extension field}
$$
Even though \(\omega\) is not rational, we can substitute \(x=\omega\) into \(P(x)\). If \(P(\alpha)=0\), the polynomial \(P\) must have \(\omega\)’s minimal polynomial (over \(\mathbb{Q}\)) as a factor. When \(\omega\) is a primitive $n$-th root of unity, its minimal polynomial is precisely a \(n\)-th cyclotomic polynomial $\Phi_n(x)$. (Wait, what is a \(n\)-th cyclotomic polynomial and its proof? Don’t worry, I’ll show you later.) Hence, by substituting \(\omega\) into \(P(x)\), we effectively confirm that
$$
\Phi_n(x) \mid P(x)
\quad
\text{if \(\omega\) is a primitive \(n\)-th root of unity}
$$
And the reason is because if $\alpha$ is a primitive root and $(x-\omega) \mid P(x)$, then all unique primitive roots $\omega_1, \omega_2, \dots, \omega_{\varphi(n)}$ (you will know what is $\varphi(n)$ later) must have
$$
(x-\omega_1)(x-\omega_2)\cdots(x-\omega_{\varphi(n)}) \mid P(x)
$$
Thus, an interesting fact about cyclotomic polynomial just revealed,
$$
\Phi_n(x) = (x-\omega_1)(x-\omega_2)\cdots(x-\omega_{\varphi(n)})
$$
The \(n\)-th cyclotomic polynomial, denoted by \(\Phi_n(x)\), is defined as the unique monic polynomial in \(\mathbb{Z}[x]\) whose roots are exactly the primitive \(n\)-th roots of unity. Symbolically,
$$
\Phi_n(x)
=
\prod_{\substack{1 \le d \le n \\ \gcd(d,n)=1}}
\Bigl(x -e^{\tfrac{2\pi i d}{n}}\Bigr)
$$
A very important fact is that each cyclotomic polynomial \(\Phi_n(x)\) is irreducible over \(\mathbb{Q}\), and that’s why it is precisely a minimal polynomial stated previously. A common way to prove irreducibility uses Gauss’s lemma, which we won’t discuss here. Moreover, a classical result in number theory shows that the total number of the unique primitive \(n\)-th roots of unity is \(\varphi(n)\), where \(\varphi\) denotes Euler’s totient function:
$$
\deg(\Phi_n) = \varphi(n)
$$
A key theorem called Cyclotomic Polynomial Factorization Theorem states that
$$
x^n -1 = \prod_{d \,\mid\, n} \Phi_d(x)
$$
where the product is over all positive divisors \(d\) of \(n\). Because each factor \(\Phi_d(x)\) is irreducible in \(\mathbb{Q}[x]\), this gives a complete factorization of \(x^n -1\) over the integers.
Several small cyclotomic polynomials appear frequently in factorization problems:
$$
\Phi_1(x) = x -1,
\quad
\Phi_2(x) = x + 1,
\quad
\Phi_3(x) = x^2 + x + 1,
\quad
\Phi_4(x) = x^2 + 1,
\quad
\Phi_6(x) = x^2 -x + 1,
\quad\dots
$$
Using these, one can obtain some factorizations like
$$
x^6 -1
=
\Phi_1(x)\,\Phi_2(x)\,\Phi_3(x)\,\Phi_6(x)
=
(x-1)(x+1)(x^2 + x + 1)(x^2 -x + 1)
$$
Problem: Please factorize
$$
P(x) = x^5 + x^4 + x^2 + x + 2
$$
By the RRT, no rational root emerges from checking divisors of the constant term. We next consider \(\omega\), a primitive cube root of unity satisfying \(\omega^3 = 1\) and \(\omega \neq 1\). By the definition of primitive root, \(\omega\) also satisfies \(\omega^2 + \omega + 1 = 0\). Using the PRT perspective in complex field, we substitute \(x = \omega\):
$$
\omega^5 + \omega^4 + \omega^2 + \omega + 2
=
\omega^2 + \omega + \omega^2 + \omega + 2
=
2(\omega^2 + \omega) + 2
=
2(\,\omega^2 + \omega + 1\,)
=
2 \cdot 0
=
0
$$
Hence, \(\omega\) is a root of \(P(x)\). By the properties of the primitive root \(\omega\), \(\Phi_3(x) = x^2 + x + 1\) (again, the minimal polynomial of \(\omega\) over \(\mathbb{Q}\)) must divide \(P(x)\). Indeed, a polynomial long division confirms:
$$
x^5 + x^4 + x^2 + x + 2
=
(x^2 + x + 1)\,(x^3 + x^2 + 1)
$$
and both factors are irreducible over \(\mathbb{Q}\) by RRT. Thus, we have obtained the complete factorization of \(P(x)\) in \(\mathbb{Z}[x]\).
Cyclotomic polynomials and primitive \(n\)-th roots of unity form a cornerstone of advanced polynomial factorization. Although the RRT succeeds in uncovering rational linear factors, many polynomials require identifying complex roots of unity. The PRT, when interpreted in the relevant extension field, tells us that if \(\alpha\) is a root of a polynomial \(P\), then the minimal polynomial of \(\alpha\) divides \(P(x)\). For the roots of unity, this minimal polynomial is precisely a cyclotomic polynomial. Consequently,
$$
\Phi_n(x)
\mid
P(x)
\quad
\text{if \(P(\omega)=0\) for a primitive \(n\)-th root of unity \(\omega\)}
$$
In practice, testing certain cyclotomic substitutions and applying polynomial long division can yield nontrivial factorizations that remain inaccessible via rational-root checks alone. The fundamental identity
$$
x^n -1 = \prod_{d \,\mid\, n} \Phi_d(x)
$$
provides the prototype for such factorizations, and memorizing small \(\Phi_n(x)\) values is often a substantial aid in problem-solving contexts.
Eisenstein’s Criterion and Modular Methods
Having explored the RRT and cyclotomic polynomials, we now turn to another cornerstone of polynomial factorization techniques: Eisenstein’s criterion and modular arithmetic checks. These methods are especially helpful when rational or obvious cyclotomic factors fail to show up, yet we still seek to prove irreducibility or discover a hidden factorization.
Eisenstein’s Criterion provides a powerful and quick way to prove a polynomial is irreducible over $\mathbb{Q}$, relying on prime divisibility patterns in the coefficients. More precisely, Eisenstein’s Criterion says:
Let $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 \in \mathbb{Z}[x]$. If there exists a prime $p$ such that:
- $p \mid a_k$ for all $k<n$ ($p$ divides every coefficient except possibly the leading one)
- $p \nmid a_n$ (the leading coefficient is not divisible by $p$)
- $p^2 \nmid a_0$ (the constant term is not divisible by $p^2$)
then $P(x)$ is irreducible over $\mathbb{Q}$.
In plain language, this means if all coefficients but the leading one are multiples of some prime $p$, yet the constant term is “not too divisible” by $p$ (that is, only by one power of $p$ but not $p^2$), then $P(x)$ cannot be factored into polynomials of smaller degree with rational coefficients.
But, failure to meet these conditions (for every prime) does not imply reducibility. Many irreducible polynomials do not satisfy Eisenstein’s divisibility conditions. Why? Because Eisenstein’s Criterion is a sufficient condition for irreducibility, not a necessary one. If a polynomial $P(x)$ meets Eisenstein’s divisibility pattern for some prime $p$, it is guaranteed irreducible over $\mathbb{Q}$. Hence, it is a one-way test: if it works, you get an instant proof of irreducibility; if not, it gives no conclusion.
Here is a typical example:
Problem: Please factorize
$$
P(x) = x^2 + 1
$$
over the rational numbers field.
Obviously it is irreducible by using quadratic formula. Yet, there is no prime $p$ dividing $1$ and also meeting the “$p^2 \nmid 1$” pattern in the sense required by Eisenstein. No single $p$ divides all coefficients except the leading one, so Eisenstein’s Criterion gives no verdict. Still, $x^2+1$ is irreducible in $\mathbb{Q}[x]$.
Sometimes no prime $p$ directly fits $P(x)$’s coefficients, however, the polynomial
$$
P(x + c)
$$
(for some small integer shift $c$) might meet Eisenstein’s conditions. This trick can salvage many cases where direct Eisenstein fails. An example is, again:
Problem: Please factorize
$$
P(x) = x^2 + 1
$$
over the rational numbers field.
Set $x = y + 1$. Then $x^2 + 1 = (y+1)^2 + 1 = y^2 + 2y + 2$, hence, $p=2$ satisfies all the Eisenstein conditions for $P(y)$, so it is irreducible over $\mathbb{Q}$.
Problem: Please factorize
$$
x^5 + 5x^3 + 10
$$
over the rational numbers field.
As $n=5$ and pick $p=5$, then we have $p \mid a_{k}$ for any $k<n$, and $p \nmid n$, with $p \nmid a^2_{0}$, which satisfies the Eisenstein’s Criterion, so this polynomial is irreducible over $\mathbb{Q}$.
When Eisenstein’s Criterion does not apply, one can still appeal to modular techniques (often reducing coefficients modulo a small prime $p$) to gather hints about whether a polynomial
$$
P(x) = a_n x^n + \cdots + a_0 \quad \in \quad \mathbb{Z}[x]
$$
is irreducible or not. The strategy is to reduce its coefficients modulo a prime $p$. We get $$ \overline{P}(x) = (a_n \bmod p)\,x^n + \dots +(a_0 \bmod p) \quad \in \quad \mathbb{F}_p[x] $$
The overall idea is:
- Pick a small prime $p$ (often $2,3,5,\dots$)
- Reduce the coefficients of $P(x)$ modulo $p$, obtaining $\overline{P}(x) \in (\mathbb{Z}/p\mathbb{Z})[x]$
- Check if $\overline{P}(x)$ factors in $\mathbb{F}_p[x]$
- If $\overline{P}(x)$ does factor nontrivially over $\mathbb{F}_p$, that suggests (not guarantee) $P(x)$ might also factor over $\mathbb{Q}$. You can try lifting (via Hensel’s lemma or systematic guessing) to recover the factorization over $\mathbb{Z}$.
- If $\overline{P}(x)$ remains irreducible modulo $p$ for several distinct primes $p$, you start suspecting $P(x)$ is irreducible over $\mathbb{Q}$. (Though keep in mind: a single prime $p$ giving irreducibility does not absolutely prove $P(x)$ irreducible in $\mathbb{Q}[x]$, but multiple prime checks strongly reinforce it.)
But the question is, why does this help?
If $P(x)$ factors over $\mathbb{Q}$, say
$$
P(x) = f(x)\,g(x),
\quad
f(x),\,g(x) \in \mathbb{Q}[x]
$$
then by multiplying through by a suitable integer to clear denominators, we get a factorization in $\mathbb{Z}[x]$. Reducing that factorization modulo $p$,
$$
P(x)\bmod p = \bigl(f(x)\bmod p\bigr)\,\bigl(g(x)\bmod p\bigr)
\quad \text{in } \mathbb{F}_p[x].
$$
Hence, if $P(x)$ is reducible in $\mathbb{Q}[x]$, it will usually (not guarantee) appear reducible mod $p$ as well.
So when doing such reduction, the factorization “transfers” (reduces) to a factorization in $\mathbb{F}_p[x]$ except in rare cases of so-called accidental factorization or accidental irreducibility. Checking multiple primes reduces the chance of such accidents.
From my perspective, these modular checks are less straightforward or less commonly the “go-to” technique than methods which were already discussed in previous chapters. Therefore, we will not delve into the details of how a factorization found in $\mathbb{F}_p[x]$ can be systematically “lifted” to $\mathbb{Z}[x]$. It suffices to note that if such a modular factorization exists (and meets certain technical conditions), then in principle one can recover a genuine integer-coefficient factorization for $P(x)$ (due to Hensel’s Lemma).
Thus, while modular methods can indeed be powerful (especially in computational settings or more advanced number theory), they are, in practice, not always the most efficient or elementary path to factorization in a contest or instructional context, particularly when simpler tools are already available…
Symmetric Polynomials
When dealing with factorization problems in multiple variables, one special class of polynomials stands out: symmetric polynomials. A polynomial $f(x_1, x_2, \dots, x_n)$ of $n$ variables is said to be symmetric if, for every permutation $\sigma$ of the set $\{1, 2, \dots, n\}$, we have
$$
P \left (x_{\sigma(1)}, x_{\sigma(2)}, \dots, x_{\sigma(n)} \right ) = P(x_1, x_2, \dots, x_n)
$$
In other words, swapping any two variables does not change the value of a symmetric polynomial. For example, $x^3+y^3$, $x^2+y^2+z^2-5xyz$, and $x^2y^2-x-y$ are symmetric polynomials, but $x^2+y^3$ is not one of them, because after swapping $x$ and $y$, the polynomial becomes $y^2+x^3 \ne x^2+y^3$.
A central result in the theory of symmetric polynomials is the Fundamental Theorem of Symmetric Polynomials, it states that every symmetric polynomial can be expressed in terms of the elementary symmetric polynomials. The elementary symmetric polynomials are defined as follows:
\begin{align*}
e_1(x_1, x_2, \dots, x_n) &= x_1 + x_2 + \cdots + x_n \\[6pt]
e_2(x_1, x_2, \dots, x_n) &= \sum_{1 \leq i < j \leq n} x_i x_j \\[6pt]
e_3(x_1, x_2, \dots, x_n) &= \sum_{1 \leq i < j < k \leq n} x_i x_j x_k \\[6pt]
&\vdots \\[6pt]
e_n(x_1, x_2, \dots, x_n) &= x_1 x_2 \cdots x_n
\end{align*}
In general, for $1 \leq k \leq n$,
$$
e_k(x_1, x_2, \dots, x_n) = \sum_{1 \leq i_1 < i_2 < \cdots < i_k \leq n} x_{i_1} x_{i_2} \cdots x_{i_k}
$$
These elementary symmetric polynomials form a natural set of “building blocks” for any symmetric polynomial. Moreover, each $e_k$ is homogeneous of degree $k$, since every term in $e_k$ is a product of degree $k$. This homogeneity and the fundamental theorem of symmetric polynomials together imply that if you aim to factor a symmetric polynomial, it is often advantageous to consider factorization patterns that involve these elementary symmetric polynomials. In many problems, this approach simplifies the process dramatically.
When factorizing symmetric polynomials, a natural strategy is to first identify factors using a substitution-based reasoning similar to the PRT, and then leverage the fundamental structure provided by the elementary symmetric polynomials to complete the factorization with MUC.
Let $ P(x_1,x_2,\dots,x_n) $ be a symmetric polynomial in $ n $ variables over a field $\mathbb{K}$. Suppose there exists a polynomial function $ f:\mathbb{K}^n \to \mathbb{K} $ such that
$$
P\bigl(f(x_1,x_2,\dots,x_n),\,x_2,\dots,x_n\bigr) = 0
$$
By the symmetry of $ P $ and the bound variable renaming (formally known as $\alpha$-conversion), it follows that, for each $ i \in \{1,2,\dots,n\} $,
$$
P\bigl(x_i, x_2,\dots,x_{i-1},f(x_i,x_2,\dots,x_{i-1}, x_{1}, x_{i+1},\dots,x_n),x_{i+1},\dots,x_n\bigr) = 0
$$
By PRT, for every variable we have
$$
(x_i -f(x_i,x_1,\dots,x_{i-1},x_{i+1},\dots,x_n)) \mid P(x_1,x_2,\dots,x_n)
$$
Thus we can conclude
$$
(x_1 -f(x_1,x_2,\dots,x_n)) (x_2 -f(x_2,x_1,\dots,x_n)) \cdots (x_n -f(x_n,x_1,\dots,x_{n-1})) \mid P(x_1,x_2,\dots,x_n)
$$
Problem: Please factorize
$$
P(a,b,c) = a^3(b-c) + b^3(c-a) + c^3(a-b)
$$
First, we check what happens if we set $a=b$. Substitute $a=b$:
$$
P(b,b,c) = b^3(b-c) + b^3(c-b) + c^3(b-b) = b^4 -b^4 + c^3(0) = 0
$$
Since $P(b,b,c)=0$, so, again, by following the property of symmetric polynomials and bound variable renaming (formally known as $\alpha$-conversion), we know that $P(b,b,c)=P(a,c,c)=P(a,b,a)=0$, which indicates $(a-b)$, $(b-c)$, $(c-a)$ are also factors of $P$ by using PRT. Therefore, we have:
$$
(a-b) \mid P(a,b,c) \implies (a-b)(b-c)(c-a) \mid P(a,b,c)
$$
So we’ve identified a cubic factor $(a-b)(b-c)(c-a)$ in our homogeneous polynomial $P(a,b,c)$ of degree $4$. This leaves us needing one more factor of degree $1$ to match the total degree of $P(a,b,c)$. Since $P$ is symmetric and homogeneous, the missing factor must itself be a symmetric, homogeneous polynomial of degree $1$.
Returning to our specific polynomial $P(a,b,c)$ in three variables, we need a symmetric, homogeneous polynomial of degree $1$. In three variables, the only such elementary symmetric polynomial of degree $1$ is $e_1(a,b,c) = a + b + c$.
Thus, we propose that the missing factor is $e_1(a,b,c)$ and apply MUC, then we can assume:
$$
P(a,b,c) = k (a-b)(b-c)(c-a)\,e_1(a,b,c)
$$
for some constant $k \in \mathbb{R}$ that we must determine.
To find the constant multiplier $k$, we can substitute convenient numeric values into $P(a,b,c)$ and into our proposed factorization. A convenient choice is to pick some values for $a,b,c$ that simplify computation. Note: These numeric values must satisfy $ a \ne b \wedge b \ne c \wedge c\ne a $ in this case, or that would trigger the PRT and the substitution would be meaningless. So, let us set $(a,b,c)=(0,1,2)$ for simplicity:
\begin{align*}
LHS &= P(0,1,2) = 0^3(1-2) + 1^3(2-0) + 2^3(0-1) = (0)(-1) + (1)(2) + (8)(-1) = 2 -8 = -6 \\
RHS &= k (a-b)(b-c)(c-a)\,e_1(a,b,c) = k (0-1)(1-2)(2-0) (0+1+2) = 6k
\end{align*}
As $LHS = RHS$, so we must have $6k=-6$, giving $k=-1$. Thus:
$$
P(a,b,c) = -(a-b)(b-c)(c-a)(a+b+c)
$$
Problem: Please factorize
$$
P(a,b,c) = a^2(a-b-c)+b^2(b-a-c)+c^2(c-a-b)+2abc
$$
Similarly, let $a=b+c$, then $P(b+c,b,c) = 0$, which implies
$$
(a-(b+c)) \mid P(a,b,c) \implies (a-(b+c))(b-(a+c))(c-(a+b)) \mid P(a,b,c)
$$
By MUC, assume $P(a,b,c) = k(a-(b+c))(b-(a+c))(c-(a+b))$, then we do numeric values substitution to confirm the value of $k$, for simplicity, let $(a,b,c) = (1,0,0)$, then
\begin{align*}
LHS &= 1^2(1-0-0)+0+0+0 = 1 \\
RHS &= k(1-(0+0))(0-(1+0))(0-(1+0)) = k(1)(-1)(-1)=k
\end{align*}
Therefore $LHS = RHS \implies k=1$, which implies
$$
P(a,b,c) = (a-(b+c))(b-(a+c))(c-(a+b))
$$
Problem: Please factorize
$$
P(a,b,c) = (x+y+z)^4-(x^2+y^2+z^2)^2-2(xy+yz+xz)((x+y+z)^2+(x^2+y^2+z^2))
$$
let $a=b$, then $P(b,b,c) = 0$, which implies
$$
(a-b) \mid P(a,b,c) \implies (a-b)(b-c)(c-a) \mid P(a,b,c)
$$
let $a=0$, then $P(0,b,c) = 0$, which implies
$$
(a-0) \mid P(a,b,c) \implies abc \mid P(a,b,c)
$$
So together, we have
$$
abc(a-b)(b-c)(c-a) \mid P(a,b,c)
$$
But $P(a,b,c)$ is a homogeneous symmetric polynomial of degree $4$, since it is divisible by a homogeneous symmetric polynomial of degree $6$, then $P(a,b,c)=0$.
Problem: Please factorize
$$
P(a,b,c) = a^3+b^3+c^3-3abc
$$
let $a=-(b+c)$, then $P(-(b+c),b,c) = 0$, which implies
$$
(a+b+c) \mid P(a,b,c)
$$
Since $P(a,b,c)$ and $(a+b+c)$ are homogeneous, so $\dfrac{P(a,b,c)}{(a+b+c)}$ is homogeneous of degree $2$. By the fundamental theorem of symmetric polynomials and MUC, we can assume
$$
P(a,b,c) = (a+b+c)\left(k_1e_1^2(a,b,c)+k_2e_2(a,b,c)\right)
$$
By doing 2 sets of convenient numeric substitutions, we get
$$
P(a,b,c) = (a+b+c)\left(e_1^2(a,b,c)-3e_2(a,b,c)\right)
$$
In fact, $(e_1^2(a,b,c)-3e_2(a,b,c))$ is irreducible, this can be proven by using MUC with assuming $(e_1^2(a,b,c)-3e_2(a,b,c)) = (\alpha_1 a + \beta_1 b + \gamma_1 c)(\alpha_2 a + \beta_2 b + \gamma_2 c)$ and finding out that no solution exists.
Problem: Please factorize
$$
P(a,b,c) = \left(y^2-z^2\right)(1+x y)(1+x z)+\left(z^2-x^2\right)(1+y z)(1+y x)+\left(x^2-y^2\right)(1+z x)(1+z y)
$$
This problem is a little bit tricky, since it is not homogeneous, so a natural approach is to rewrite it into the sum of several homogeneous polynomials:
\begin{align*}
P(a,b,c) = & \, \left(y^2-z^2\right)(1+x y)(1+x z)+\left(z^2-x^2\right)(1+y z)(1+y x)+\left(x^2-y^2\right)(1+z x)(1+z y) \\
= & \, x y z\left ( x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)\right) +\left ( \left(y^2-z^2\right)+\left(z^2-x^2\right)+\left(x^2-y^2\right)\right) \\
& +\left(x(y+z)\left(y^2-z^2\right)+y(z+x)\left(z^2-x^2\right)+z(x+y)\left(x^2-y^2\right)\right) \\
= & \, x y z\left(x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)\right) \\
& +\left(x(y+z)\left(y^2-z^2\right)+y(z+x)\left(z^2-x^2\right)+z(x+y)\left(x^2-y^2\right)\right)
\end{align*}
Then we can follow our common strategy to conquer these $2$ homogeneous polynomials respectively to finish factorizing the main goal $P(a,b,c)$:
\begin{align*}
P(a,b,c) = & \, x y z\left(x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)\right) \\
& +\left(x(y+z)\left(y^2-z^2\right)+y(z+x)\left(z^2-x^2\right)+z(x+y)\left(x^2-y^2\right)\right) \\
= & \, xyz(x-y)(y-z)(z-x) + (x-y)(y-z)(z-x)(x+y+z) \\
= & \, (xyz+x+y+z)(x-y)(y-z)(z-x)
\end{align*}
Some tips
It is generally recommended to avoid immediately applying formulaic solutions without even thinking when encountering a problem, as taking some time to conduct initial observations can often lead to more effective problem-solving outcomes.
If I have a question like factorize $x^2+x-6$ over the integer numbers field, will you consider applying the quadric formula $x=\dfrac{-b \pm \sqrt{b^2-4 a c}}{2 a}$ first?
No (I mean, you can, but, it’s strongly not recommended to do that). Instead, we can do some simple transformations to generate the result easily:
$$
x^2+x-6=x^2+3x-2x-6=x(x+3)-2(x+3)=(x+3)(x-2)
$$
Or for this problem: Factorize $x^2-28x+195$ over the integer numbers field.
Probably this one is not as simple as the previous one, because it contains larger numbers and is not obvious to observe anything… But, really?
As long as you noticed $195=196-1$ and you are familiar with $196=14^2$, then you are able to solve this problem within 10 seconds, look:
\begin{align*} x^2-28x+195&=x^2-28x+196-1\\&=(x^2-2\times 14x+14^2)-1^2\\&=(x-14)^2-1^2\\&=(x-14-1)(x-14+1)\\&=(x-15)(x-13)\end{align*}
Factorize $x^2-28x+195$ over the integer numbers field.
We can also apply the similar method: $0=8\times2a^2-8\times2a^2$
$$ 64a^4+1 =(8^2a^4 + 8\times2a^2 +1) – 8\times2a^2 = (8a^2+1)^2 – (4a)^2=(8a^2-4a+1)(8a^2+4a+1) $$
Upon reflection, you may be curious about how I was able to recognize the approach needed to solve these problems. The truth is, I wasn’t born with the knowledge. Rather, I have spent countless hours practicing and mastering various problem-solving techniques, which has allowed me to develop my mathematical intuition. What I want to emphasize is that humans have the ability to cultivate and refine their mathematical intuition (expanding your Maths database) through sufficient training and exposure, because this skill is about memorization and recognition (automatically retrieving or invoking a corresponding frame to solve the problem), etc… Just like developing a specific skill of a neural network through billions of training data… But for humans, the mechanized training process is a tedious and non-efficient process comparatively, since we are not as powerful as computers… However, we still need to practice as long as we aspire to participate in the Mathematical Olympiad or Putnam Maths Competition, these designed competitive mathematics for humans…
Have you ever encountered a math problem that left you stumped, unable to even begin solving it? Chances are, you will at some point. But why does this happen?
In competitive mathematics contests, problem setters strive to challenge well-prepared participants who have practiced typical problems extensively. To accomplish this, they must devise “good problems” that are difficult enough to separate the exceptional from the merely competent. These problems often require discovering a clever trick that is not immediately apparent.
For instance, a problem like $64a^4+1=(64a^4 + 16a^2 +1) -16a^2$ may have been considered challenging in the past, but it has become too familiar to many participants. Therefore, problem setters must continue to raise the bar by increasing the difficulty of their problems.
However, it’s important to remember that even the most challenging problems in a Mathematics contest are designed to be solved by humans within a limited time frame, so there’s no need to panic. Rest assured, you won’t encounter a problem as difficult as Goldbach’s conjecture, which is a extremely hard question that is yet to be proven, during the contest.
Now, let’s delve into the Method of Trials and Errors (MTE):
This approach is useful when your maths knowledge database is not sufficient to cover the problem directly, and you cannot call any existing methods from your database to solve the problem effectively. In such cases, try MTE. As we’ve mentioned earlier, even the most challenging problems in a Mathematics contest are designed to be solved by humans within a limited time frame. The difficulty of a problem may stem from the fact that you haven’t discovered the clever trick or technique that could solve it easily. It’s possible that a simple transformation to your approach could lead to a sudden insight or revelation from your observations that makes the solution obvious.
For example, here is a very challenging one for you to solve:
Problem: Please factorize
$$
x^8+x^4+1
$$
over the integer numbers field.
At first sight, you may think applying MS is a good approach by letting $y=x^4$ since the relation of $(x^4)^2 = x^8$ is obvious, but after doing that, you would notice $y^2+y+1$ is not factorizable over the real number field as the discriminant is less than zero ($\Delta = 1-4 = -3 < 0$). Then you would apply RRT, and notice this polynomial doesn’t have any linear rational factor, as its degree is $\ge 4$. You may even start checking whether this polynomial is cyclotomic… But through some simple transformations, intuitively, we can do
\begin{align*}x^8+x^4+1 &= x^8+2x^4+1-x^4 \\&= (x^4+1)^2-(x^2)^2 \\&= (x^4+x^2+1)(x^4-x^2+1) \\&= (x^4+2x^2+1-x^2)(x^4-x^2+1) \\&= \left ((x^2+1)^2-x^2) \right )(x^4-x^2+1)\\&=(x^2-x+1)(x^2+x+1)(x^4-x^2+1)\end{align*}
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