2003 Vietnamese Mathematical Olympiad

• First, it is easy to show the Lemma: For arbitrary integers $a,b$, we have: $$a^2+b^2\equiv \{\begin{array}{llc}2;& 1\equiv 5(\mathrm{mod}8)& \text{if }a\equiv \pm 1(\mathrm{mod}4)\\ 1;& 0\equiv 4(\mathrm{mod}8)& \text{if }a\equiv 0(\mathrm{mod}4)\\ 5;& 4\equiv 0(\mathrm{mod}8)& \text{if }a\equiv 2(\mathrm{mod}4)\end{array}$$ • Returning to the problem: - For $n=3$: it is easily seen that the given system has integral solution $(x=-2;y_1=0;y_2=1;y_3=0)$. - For $n=4$: Suppose that the given system has an integral solution $(x,y_1,y_2,y_3,y_4)$. Then, as $x+1,x+2,x+3,x+4$ form a complete system of residues mod 4, from the above mentioning lemma, there exists an integer $m$ such that $$m\in \{2;1;5\}\cap \{1;0;4\}\cap \{5;4;0\}=\varnothing .$$ This contradiction proves that the given system has no integral solution when $n=4$. And then it is evident that the given system of equations has no integral solution for every $n\ge 4$. - So $n=3$ is the greatest integer sought for.