69th Belarusian Mathematical Olympiad

a) For example, take $$Q(x)=(x_1+x_2-x_3-x_4{)}^2+(x_1-x_2+x_3-x_4{)}^2+(x_1-x_2-x_3+x_4{)}^2+0^2.$$

b) Consider the representation from the problem conditions: $$4(x_1^2+x_2^2+x_3^2+x_4^2)-(x_1+x_2+x_3+x_4{)}^2=P_1^2+P_2^2+P_3^2+P_4^2.(1)$$ The equalities $0=Q(0,0,0,0)=P_1(0,0,0,0{)}^2+\cdots +P_4(0,0,0,0{)}^2$ imply that the constant terms of polynomials $P_1,P_2,P_3$ and $P_4$ are zeroes. First we will show that all these polynomials are linear (i.e. of a degree not exceeding 1).

Consider the monomial order and let $x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}{x}_3^{{\alpha }_3}{x}_4^{{\alpha }_4}$ be the leading monomial over $P_1,P_2,P_3$ and $P_4$. The sum $P_1^2+P_2^2+P_3^2+P_4^2$ contains monomials $x_1^{2{\alpha }_1}{x}_2^{2{\alpha }_2}{x}_3^{2{\alpha }_3}{x}_4^{2{\alpha }_4}$ only with positive coefficients, since they cannot be products of two distinct monomials of $P_i$. Hence such monomial belongs to $Q$, but the leading monomial of $Q$ is $3x_1^2$. Therefore, the leading monomial equals $x_1$, i.e. any term, divisible by $x_1$, equals $a{x}_1$ for some integer $a$. Since we can arrange the variables in the definition of the order arbitrary, similar statement is true for all variables. Thus, polynomials $P_1,P_2,P_3$ and $P_4$ are indeed linear.

Denote $P_i=a_{i1}{x}_1+a_{i2}{x}_2+a_{i3}{x}_3+a_{i4}{x}_4$, $i=1,2,3,4$. Substitute the values $(1,0,0,0)$ of variables to $(1)$, we obtain the equality $a_{11}^2+a_{21}^2+a_{31}^2+a_{41}^2=3$, hence all these coefficients equal 0 or $\pm 1$. Substitutions $(0,1,0,0)$, $(0,0,1,0)$ and $(0,0,0,1)$ lead to similar conditions on the coefficients at $x_2,x_3$ and $x_4$. Wherein, among $a_{ij}$ there are exactly 12 nonzero coefficients. Since $Q(1,1,1,1)=0$, all $P_i(1,1,1,1)=0$, therefore each $P_i$ contains 4, 2 or 0 nonzero coefficients.

The number 12 can be represented as a sum of four integers, which equal to 4, 2 or 0, in two ways: $12=4+4+4+0$ and $4+4+2+2$. Suppose that all $P_i$ are nonconstant polynomials. Then, without loss of generality, let $P_1$ and $P_2$ have 4 nonzero coefficients each, and $P_3$ and $P_4$ have 2 nonzero coefficients each. By rearranging the variables (if necessary) we can make $P_1(1,1,1,1)=x_1+x_2-x_3-x_4$.

Consider the equality $$0=Q(1,1,0,0)-P_1(1,1,0,0{)}^2=P_2(1,1,0,0{)}^2+P_3(1,1,0,0{)}^2+P_4(1,1,0,0{)}^2.$$ Hence $P_i(1,1,0,0)=P_i(1,1,1,1)=0$ for $i\ge 2$. These equalities implies that the coefficients at $x_1$ and $x_2$ has different sign in $P_2$ as well as the coefficients at $x_3$ and $x_4$. By rearranging (if necessary) $x_1$ with $x_2$, and $x_3$ with $x_4$, we can make $P_2=x_1-x_2+x_3-x_4$.

Consider similar equality $$0=Q(1,0,1,0)-P_1(1,0,1,0{)}^2-P_2(1,0,1,0{)}^2=P_3(1,0,1,0{)}^2+P_4(1,0,1,0{)}^2.$$ It implies $P_4(1,0,1,0)=P_3(1,0,1,0)=0$. Recall that $P_3(1,1,0,0)=P_3(1,1,1,1)=0$.

The last three equalities can be written as $$a_{31}+a_{32}=0,a_{31}+a_{33}=0,a_{31}+a_{32}+a_{33}+a_{34}=0.$$ Whence $a_{32}=a_{33}$, $a_{31}=a_{34}$ and $a_{32}=-a_{31}$. Therefore, either all coefficients of $P_3$ are zeroes or none of them are zeroes. But $P_3$ has exactly two nonzero coefficients — a contradiction. So, at least one of $P_1,P_2,P_3$ and $P_4$ is constant.