PROBLEM OF THE MONTH

October 2002






Show that if a_1,a_2,\ldots,a_n are all distinct integers, then the polynomial

P(X)=(X-a_1)^2(X-a_2)^2\cdots(X-a_n)^2+1

cannot be written as a product of two other (non-constant) polynomials with integer coefficients.

There are several possible solutions to this problem. Here is one short way to tackle it:

Suppose we can write the degree $ 2n$ polynomial $ P(X)$ as

$\displaystyle P(X)=f(X)g(X)$

for some nonconstant polynomials $ f(X)$ and $ g(X)$ with integer coefficients. In particular one of these two polynomials will have degree at most $ n$, say $ f$. The polynomial $ P(X)$ takes only strictly positive values (being a perfect square $ +1$), so without loss of generality we may assume that both $ f(X)$ and $ g(X)$ take only strictly positive values for all real values of $ X$.

If we substitute now the integers $ a_i$ for $ X$ we get that

$\displaystyle 1=P(a_i)=f(a_i)g(a_i)$

for all $ i\in\{1,\ldots,n\}$, and since both $ f$ and $ g$ have integer coefficients it follows that

$\displaystyle f(a_i)=g(a_i)=1$

for all $ i$. Thus both polynomials $ f$ and $ g$ take the same value for $ n$ distinct values of $ X$, and since $ \deg(f)\le n$ it follows that the only possibility is that $ \deg(f)= \deg(g)=n$.

Taking the derivative of $ P(X)=f(X)g(X)$ and substituting then $ a_i$ for $ X$ leads to

$\displaystyle P'(a_i) = 0 = f'(a_i)g(a_i) + f(a_i)g'(a_i) = f'(a_i) + g'(a_i) $

In other words the degree $ n-1$ polynomial $ (f(X)+g(X))'$ vanishes at $ n$ distinct values of $ X$ and thus must be identically zero. Hence $ f(X)+g(X)$ is a constant polynomial, and by the above we must have

$\displaystyle f(X)+g(X)\equiv 2 $

Since $ f$ and $ g$ take only (positive) integral values for all integral value of $ X$, it follows that

$\displaystyle f(X)\equiv 1\qquad {\textrm{and}}\qquad g(X)\equiv 1, $

which is a contradiction since both $ f$ and $ g$ are polynomials of degree $ n\ge 1$.