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Given the polynomial $P(x)=(x^2-7x+6)^{2n}+13$ where $n$ is a positive integer. Prove that $P(x)$ can’t be written as a product of $n+1$ non-constant polynomials with integer coefficients.
Source of the problem
Vietnam MO 2014 Problem 2
Topic
Polynomial, Algebra
7/10
Suggested Book
Excursion in Mathematics by Bhaskaracharya Prathistan

Do you really need a hint? Try it first!

The idea is that if a polynomial can be factorized, then we must check if a polynomial has real roots or not. The first observation is that the given polynomial is always > 0 as it is of the form $(f(x))^2 + 13$. Hence, it has no real roots. So, if the polynomial can be factorized, then obviously the factors will not be linear.  An odd degree polynomial has atleast one real root. Hence, none of the factors will be odd degree.
So, we have seen the factors will not be linear. Hence, let’s investigate the factors and their properties if the polynomial can be factorized. Let us approach the method of contradiction. Assume that the polynomial can be factorized into n+1 factors with integer coefficients. Now, we know that the factors must have atleast degree 2, in fact only even degrees i.e. 2, 4, 6, … Now, let’s see if all the factors have a degree more than 2, then the degree of the whole polynomial must be $\geq 4n + 4$, but the degree of the polynomial is 4n. So, two of the factors must be of degree 2.
Hence there exist $a,b, c, d$ such that $(x^2+ax+b)(x^2+cx+d) \mid P(x)$. Note that $f(x)=x^2+ax+b > 0$ and $g(x)=x^2+cx+d > 0$ for all $x \in \mathbb R$, because they cannot have a real root. Now, we have to make use of the idea that the coefficients are integers. So, some idea of divisibility must come in. The given polynomial is $P(x)=(x^2-7x+6)^{2n}+13$ . Try to find out some values of x, for which P(x) is easily determined without n. See, x = 1 and 6 works as $x^2 – 7x + 6 = (x-1)(x-6)$.
So, we saw that x = 1 and x = 6 are good values and we get that P(1) = P(6) = 13. Also, we know that f(1).g(1) | P(1) = 13. So, one of them must be 1. Say, f(1) = 1 = 1 + a + b.  Also, f(6) = 36 + 6a – a = 36 + 5a.  Now, f(6) | P(6) = 13. So, f(6) = 36 + 5a = 1 or 13.  This gives rise to a single case of a = -7; where $f(x) = x^2 -7x + 7$. But f(x) has real roots implying that P(x) has real roots. Hence contradiction. QED.

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