$\begin{array}{l}P = (a+b)(b+c)(c+a) - abc\\ \Leftrightarrow P = (a+b+c - c)(b+c)(c+a) - abc\\ \Leftrightarrow P = (a+b+c)(b+c)(c+a) - c(b+c)(c+a) - abc\\ \Leftrightarrow P = (a+b+c)(b+c)(c+a) - c(ab + bc + ca + c^2 + ab)\\ \Leftrightarrow P = (a+b+c)(b+c)(c+a) - c[2ab + c(a+b+c)]\\ \Leftrightarrow P = (a+b+c)(b+c)(c+a) - c^2(a+b+c) - 2abc\\ \text{Ta có:}\\ \quad a+ b+ c \quad \vdots \quad 4\\ \Rightarrow (a+b+c)(b+c)(c+a) - c^2(a+b+c) \quad \vdots \quad 4\qquad (1)\\ \text{Mặt khác:}\\ Do\,\,a+ b+ c \quad \vdots \quad 4\\ \Rightarrow \text{a,b,c có ít nhất 1 số chẵn}\\ \Rightarrow abc \quad \vdots \quad 2\\ \Rightarrow 2abc \quad \vdots \quad 4\qquad (2)\\ (1)(2) \Rightarrow(a+b+c)(b+c)(c+a) - c^2(a+b+c) - 2abc \quad \vdots \quad 4\\ Hay \,\,P \quad \vdots \quad 4\end{array}$
