Competition problem — Number theory

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Number theory Difficulty 6.1 National olympiad Prove it

Example 5 Let $a, b, c, d, e, f$ all be positive integers, and the number $S(=a+b+c+d+e+f)$ is a common divisor of $a b c+d e f$ and $a b+b c+c a-d e-e f-f d$ . Prove: $S$ is a composite number.

Solution

Prove that for the polynomial
$\begin{aligned} f(x) & =(x+a)(x+b)(x+c)-(x-d)(x-e)(x-f) \\ & =S x^{2}+(a b+b c+c a-d e-e f-f d) x+(a b c+d e f) . \end{aligned}$

From the given conditions, it is known that for any $x \in \mathbf{Z}$ , $S \mid f(x)$ .
In particular, $S \mid f(d)$ , i.e.,
$S \mid (d+a)(d+b)(d+c).$

Combining the fact that the numbers $d+a, d+b, d+c$ are all less than $S$ , it follows that $S$ is a composite number. The proposition is proved.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.