MHB Prove that the sum of 6 positive integers is a composite number

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SUMMARY

The discussion centers on proving that the sum of six positive integers, denoted as \( S = a + b + c + d + e + f \), is a composite number under specific conditions. It is established that if \( S \) divides both \( abc + def \) and \( ab + bc + ca - de - ef - df \), then \( S \) must be composite. The proof leverages properties of divisibility and the nature of composite numbers, concluding that the conditions provided ensure \( S \) cannot be prime.

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Let $a,\,b,\,c,\,d,\,e,\,f$ be positive integers and $S=a+b+c+d+e+f$. Suppose that the number $S$ divides $abc+def$ and $ab+bc+ca-de-ef-df$, prove that $S$ is composite.
 
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All the coefficients of

$\begin{align*}f(x)&=(x+a)(x+b)(x+c)-(x-d)(x-e)(x-f)\\&=Sx^2+(ab+bc+ca-de-ef-fd)x+(abc+def)\end{align*}$

are multiples of $S$. Evaluating $f$ at $d$, we get that $f(d)=(a+d)(b+d)(c+d)$ is a multiple of $S$.

So this implies that $S$ is composite, since $a+d,\,b+d,\,c+d$ are all strictly less than $S$.
 

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