# On a generalisation of a problem on factorisation

How can we check whether a polynomial of higher degree is factorisable or not? In this article, Siddhartha Sankar Chattopadhyay, shares his idea, in At Right Angles (AtRiA) magazine.

In high school mathematics, we come across the topic of factorisation of polynomials with integer or rational coefficients.

Usually, we only consider factorisation of polynomials of low degrees. Checking whether or not a polynomial of higher degree such as 3727X 11 + 3 is factorisable turns out to be a difficult problem.

Although there are some methods available in higher mathematics to deal with such problems, the tools and techniques of high school mathematics seem to be of little use for addressing such problems.

In this article, we discuss the factorisation of a particular infinite family of polynomials of arbitrary degree.

### Statement of the problem

In , it is shown that for any integer n 1 and distinct integers a1, a2,. .. , an, the polynomial

is not factorisable. It is interesting to note that if we consider a variant of this polynomial, namely

then the polynomial is sometimes factorisable.
For instance we have:

However, keeping +1 in place of −1, we can show that a large number of such polynomials are not factorisable. More precisely, we prove the following theorem.

Theorem 1. Let n 1 be an integer and let a1, a2,. . ., an be distinct odd integers. Then the polynomial

is not factorisable over the integers.

Remark 1. The presence of the factor (X 2) in Theorem 1 is crucial, otherwise the following serves as a counter-example:

is factorisable over the integers. Let f(X)= g(X)h(X) for some two polynomials and with integer coefficients. We note that

Consequently, we have

Since and are polynomials with integer coefficients, g(2), h(2), g(a1), h(a1), . .., g(an), h(an) are all integers. Therefore, from (4),

Let P(X)= g(X) − h(X). Then is a polynomial with integer coefficients, and P(2)= P(ai)= 0 for all

i ∈ {1,. .. , n}. Now,

This means that P(X) is a polynomial of degree < n + 1, having at least n + 1 distinct zeros (namely, 2, a1,. .. , an). Hence P(X) is identically zero, which implies that g(X)= h(X).

Therefore, the relation f(X)= g(X)h(X) becomes f(X)= (g(X))2.

Let g(0)= k. Then from the equation f(X)= (g(X))2, we get f(0)= (g(0))2 = k2. That is,

Since the ai are all odd integers, is odd. Hence k21 = (k 1)(k + 1) is divisible by 8 (as it is the product of two consecutive even integers). This implies that

is divisible by 8, which is not possible as all the aiare odd.

This contradiction shows that the stated factorisation is not possible.

This completes the proof of Theorem 1.

### Reference

 A. Engel, Problem-Solving Strategies, Springer.