# Cyclic Polynomials

**Cyclic polynomials** are polynomial functions that are invariant under *cyclic permutation* of the arguments. This gives them interesting properties that are useful in factorization and problem solving. These polynomials are closely related to symmetric polynomials as all symmetric polynomials are cyclic (but not vice versa).

#### Contents

## Definition

A polynomial \( f(x_1, x_2, x_3, \ldots, x_{n-1}, x_n ) \) is

cyclicif\[ f(x_1, x_2, x_3, \ldots, x_{n-1}, x_n ) = f(x_i, x_{i+1}, \ldots, x_{n-1}, x_n, x_1,x_2,\ldots,x_{i-1} )\]

for any integer \(i\) in the range \(1<i\leq n\).

What this means is that the polynomials remains the same under *cyclic permutation* or rotation of the variables.

A simple example of a cyclic polynomial is

\[P(x,y,z)=x+y+z.\]

To determine whether a polynomial is cyclic or not, we just need to cycle through the variables like this:

\[P(y,z,x)=y+z+x,\quad P(z,x,y)=z+x+y.\]

Notice that \(P(x,y,z)=P(y,z,x)=P(z,x,y)\). From this, we can know that the polynomial \(P(x,y,z)=x+y+z\) is a

cyclic polynomial.

Here is another cyclic and symmetric polynomial:

A cyclic sum of a polynomial function over several variables could generate a cyclic polynomial. For example,

\[ P(x,y, z) = (x - y)^2 + (y - z)^2 + (z - x)^2. \]

One interesting thing to note is that this cyclic polynomial is also a symmetric polynomial. However, not all cyclic polynomials are symmetric polynomials.

Here's an example for you to try out:

\[P(x,y,z)=x(y+z)-y(z+x)+z(x+y)\]

Is this polynomial a cyclic polynomial?

To determine if a polynomial is cyclic or not, just evaluate \(P(y,z,x)\) and \(P(z,x,y)\):\[\begin{align} P(y,z,x)&=y(z+x)-z(x+y)+x(y+z)\\ P(z,x,y)&=z(x+y)-x(y+z)+y(z+x). \end{align}\]

Notice that \(P(x,y,z)\neq P(y,z,x)\neq P(z,x,y)\). Therefore, we conclude that the polynomial above is

nota cyclic polynomial. \(_\square\)

Sometimes, cyclic polynomials can be long and are troublesome to write out completely. We have a shorter notation for cyclic polynomials:

\[P(x,y,z)=x(y-z)+y(z-x)+z(x-y)=\displaystyle \sum_{\text{cyc}}x(y-z)\]

Cyclic polynomials have an interesting property, which is useful for problem solving:

The addition, subtraction, multiplication, and/or division of two cyclic polynomials always results in a cyclic polynomial.

\[P(x,y,z)=x+y+z,\quad Q(x,y,z)=x^2+y^2+z^2\]

Both \(P(x,y,z)\) and \(Q(x,y,z)\) are cyclic polynomials. Is \((P+Q)(x,y,z)\) also a cyclic polynomial?

We have\[(P+Q)(x,y,z)=x+x^2+y+y^2+z+z^2.\]

Now, we need to determine if \((P+Q)(x,y,z)\) is cyclic or not:

\[\begin{align} (P+Q)(y,z,x)&=y+y^2+z+z^2+x+x^2\\ (P+Q)(z,x,y)&=z+z^2+x+x^2+y+y^2. \end{align}\]

Since \((P+Q)(x,y,z)=(P+Q)(y,z,x)=(P+Q)(z,x,y)\), we know that \((P+Q)(x,y,z)\) is a cyclic polynomial as well. \(_\square\)

## Factorization of Cyclic Polynomials

Let's work through some examples.

Factorize this equation:

\[ f(x,y,z) = x^2 (y-z) + y^2 (z-x) + z^2 (x-y).\qquad (1) \]

We know that \( f(x, y, z) \) is a cyclic polynomial because \( f(x, y, z) = f(y, z, x) = f(z,x,y)\).

We can note that \( f(y, y, z ) = 0 \). By the factor theorem, \( (x-y) \) is a factor of \( f(x, y, z) \).

If \( (x-y) \) is a factor of \( f(x, y, z) \), then it must be true that \( (x-y)(y-z)(z-x) \) is also a factor of \( f(x, y, z) \) because \( f(x, y, z) \) is cyclic.

The degrees of \( f(x, y, z) \) and \( (x-y)(y-z)(z-x) \) are both 3. Thus, the factored result of \( f(x, y, z) \) is equal to

\[ k(x-y)(y-z)(z-x).\qquad (2) \]

If we let \( (x, y, z) = (0, 1, 2) \) and solve the equation \( f(0, 1, 2) = k(0-1)(1-2)(2-0) \), we get

From the initial equation (1),

\[f(0,1,2) = 0^2(1-2) + 1^2(2-0) + 2^2(-1) = -2.\]

And from the second equation (2),

\[f(0,1,2) = k(x-y)(y-z)(z-x) = k(-1)(-1)(2) = 2k.\]

Setting these two equal to each other implies \(k= -1\).

Therefore,

\[x^2 (y-z) + y^2 (z-x) + z^2 (x-y) = -(x-y)(y-z)(z-x).\ _\square\]

Sometimes the factorization requires clever use of the factor theorem.

For all integers \(x,y,z\), prove that \(P(x,y,z) = 8(x+y+z)^3-(x+y)^3-(y+z)^3-(z+x)^3\) is divisible by \(3\).

Let \(x = -\frac{y+z}{2}\), then \(P(x,y,z) = 0\). By the factor theorem, \(2x+y+z\) is a factor of \(P\). Similarly, its cyclic representatives \(2y+z+x\) and \(2z+x+y\) are also factors of \(P\). Notice that the resulting degree matches the initial expression, so we only need to find the constant \(k\) such that

\[k(2x+y+z)(2y+z+x)(2z+x+y) = 8(x+y+z)^3-(x+y)^3-(y+z)^3-(z+x)^3.\]

Substitute \((x,y,z)=(1,0,0)\), then we have \(2k=6\implies k=3\). Thus,

\[8(x+y+z)^3-(x+y)^3-(y+z)^3-(z+x)^3 = 3(2x+y+z)(2y+z+x)(2z+x+y),\]

which is divisible by \(3\). \(_\square\)

Here are some common factors and their cyclic representatives for a cyclic polynomial with 3 variables:

\(x\) | \(xyz\) |

\(x+y\) | \((x+y)(y+z)(z+x)\) |

\(x-y\) | \((x-y)(y-z)(z-x)\) |

\(x+y+z\) | \(x+y+z\) |

\(x-y-z\) | \((x-y-z)(y-z-x)(z-x-y)\) |

\(x+y-z \hspace{20mm} \) | \((x+y-z)(y+z-x)(z+x-y)\) |

However, there are situations when factor theorem doesn't help to complete the factorization. In that case, we may use the following theorem:

The addition, subtraction, multiplication, and/or division of two cyclic polynomials always results in a cyclic polynomial.

This gives us a hint that we can try other non-linear cyclic polynomials as its factor. Look at the example below.

For all integers \(x,y,z\), prove that \(P(x,y,z) = (x+y+z)^5-x^5-y^5-z^5\) is divisible by \(5\).

If \(x=-y\), then \(P(x,y,z)=0\). By the factor theorem, \(x+y\) is a factor of \(P\). Similarly, \(y+z\) and \(z+x\) are also factors of \(P\). This time the resulting degree after we multiply each factor is \(3,\) which is lower than the degree of the initial polynomial \(5\). Thus, we need another polynomial of degree \(2\), say \(k(x^2+y^2+z^2)\) and \(t(xy+yz+zx)\). Thus,

\[(x+y+z)^5-x^5-y^5-z^5 = (x+y)(y+z)(z+x)\big(k(x^2+y^2+z^2)+t(xy+yz+zx)\big).\]

Substitute \((1,1,1)\) and \((0,1,1)\) into \((x,y,z)\), and we obtain

\[\begin{aligned} k+t&=10 \\ 2k+t &= 15.\end{aligned}\]

Solving the system of equations, we get \(k=t=5\). Hence,

\[(x+y+z)^5-x^5-y^5-z^5 = 5(x+y)(y+z)(z+x)\big((x^2+y^2+z^2)+(xy+yz+zx)\big),\]

which is divisible by \(5\). \(_\square\)

Here is the general approach to factorize a cyclic polynomial:

- Verify the polynomial's cyclic property.
- Identify linear factors using the factor theorem. Since the polynomial is cyclic, its factors are also cyclic.
- If you do
**not**have the correct degree of polynomial obtained by the factors (say the original polynomial has a higher degree), note that you get a cyclic polynomial upon multiplying cyclic polynomials, so you must multiply the obtained factors by \( k P \), where \( k \) is a constant and \( P \) is a cyclic polynomial. - If you have the correct degree of polynomial obtained by the factors, substitute random values of \(x, y, z\) into the factors you obtained and also into the original polynomial and determine the relationship (meaning, by what number must you multiply your obtained factors to get the original polynomial?).

The following are some common "tests" to figure out factors of \(f(x, y, z):\)

- If \( f(y, y, z) = 0 \), then a factor is \( (x-y)(y-z)(z-x) \).
- If \( f(0, y, z) = 0 \), then a factor is \( xyz \).
- If \( f( -y-z, y, z ) = 0 \), then a factor is \( x + y + z \).
- If \( f(z-y, y, z) = 0 \), then a factor is \( (x+y-z)(y+z-x)(z+x-y) \).

## Inequalities Of Cyclic Polynomials

## Problem Solving

## Advanced Applications

**Cite as:**Cyclic Polynomials.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/cyclic-polynomials/