Problem 1648

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:
$$4(x_1^2 + x_2^2 + x_3^2 + x_4^2) - (x_1 + x_2 + x_3 + x_4)^2 = P_1^2 + P_2^2 + P_3^2 + P_4^2. \quad (1)$$
The equalities $0 = Q(0, 0, 0, 0) = P_1(0, 0, 0, 0)^2 + \dots + P_4(0, 0, 0, 0)^2$ 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
$$a_{31} + a_{32} = 0, \quad a_{31} + a_{33} = 0, \quad a_{31} + a_{32} + a_{33} + a_{34} = 0.$$
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.