Có: $a^{2}+ab+b^{2}=c^{2}+cd+d^{2}$
$\leftrightarrow a^{2}+2ab+b^{2}-ab=c^{2}+2cd+d^{2}-cd$
$\leftrightarrow (a+b)^{2}-ab=(c+d)^{2}-cd$
$\leftrightarrow (a+b)^{2}-(c+d)^{2}=ab-cd$
$\leftrightarrow (a+b+c+d)(a+b-c-d)=ab-cd$
$\rightarrow ab-cd \ \vdots \ a+b+c+d$
$\rightarrow ab-cd+c.(a+b+c+d) \ \vdots \ a+b+c+d$
$\rightarrow (a+c)(b+c) \ \vdots \ a+b+c+d \ \ \ (1)$
Giả sử $a+b+c+d$ là số nguyên tố
Khi đó, $(1)\rightarrow$ \(\left[ \begin{array}{l}a+c \ \vdots \ a+b+c+d\\b+c\ \vdots\ a+b+c+d\end{array} \right.\)
$\rightarrow$ \(\left[ \begin{array}{l}a+c \geq a+b+c+d\\b+c \geq a+b+c+d\end{array} \right.\)
(Vô lí vì $a+b+c+d > a+c,b+c \geq 1$)
Vậy $a+b+c+d$ là hợp số
