THE 2002 VIETNAMESE MATHEMATICAL OLYMPIAD

For $x$, $y$, $u$, $v\in {\mathbb{Z}}^{+}$, we can write the given equation in the form $$(x+y+u+v{)}^2=n^{2}xyuv$$ or in the form $$x^2+2(y+u+v)x+(y+u+v{)}^2=n^{2}xyuv(1)$$ Let $n$ be a positive integer such that (1) has positive integer solutions $(x,y,u,v)$. Let $(x_0,y_0,u_0,v_0)$ be a positive integer solution of (1) such that $x_0+y_0+u_0+v_0$ has the least value. We can suppose w. l. o. g. that $x_0\ge y_0\ge u_0\ge v_0$. It is easily seen that: $$f(x)=x^2-n^2{y}_0^2{u}_0{v}_{0}x+2(y_0+u_0+v_0)x+(y_0+u_0+v_0{)}^2.$$ From (i), (ii) and the theorem of Viet, we deduce that $x_1=(y_0+u_0+v_0{)}^2/x_0$ is also a positive integer root of $f(x)$. Thus $(x_1,y_0,u_0,v_0)$ is a positive integer solution of (1). From the definition of the root $(x_0,y_0,u_0,v_0)$, we have: $$x_1\ge x_0\ge y_0\ge u_0\ge v_0.(2)$$ The theorem on the sign of polynomial of second degree gives: $$\begin{gathered} 0\le f(y_0)=y_0^2-n^2{y}_0^2{u}_0{v}_0+2(y_0+u_0+v_0)y_0+(y_0+u_0+v_0{)}^2\le 16y_0^2-n^2{y}_0^2{u}_0{v}_0 \\ \text{therefore }{n}^2{y}_0^2{u}_0{v}_0\le 16y_0^2,\text{ so }{n}^2\le n^2{u}_0{v}_0\le 16\text{ and }n\in \{1;2;3;4\}. \end{gathered}$$ $(x,y,u,v)=(4,4,4,4),(2,2,2,2),(1,1,2,2),(1,1,1,1)$ are respectively positive integer solutions of (1) for $n=1,2,3,4$. So the demanded values of $n$ are $1,2,3,4$.