The polynomial 
$$
Q(x_1,x_2,\ldots,x_4)=4(x_1^2+x_2^2+x_3^2+x_4^2)-(x_1+x_2+x_3+x_4)^2
$$
is represented as the sum of squares of four polynomials of four variables with integer coefficients.
[b]a)[/b] Find at least one such representation
[b]b)[/b] Prove that for any such representation at least one of the four polynomials isidentically zero.

[i](A. Yuran)[/i]
Let's solve the problem step-by-step.

### Part (a)
We need to represent the polynomial 
\[ Q(x_1, x_2, x_3, x_4) = 4(x_1^2 + x_2^2 + x_3^2 + x_4^2) - (x_1 + x_2 + x_3 + x_4)^2 \]
as the sum of squares of four polynomials of four variables with integer coefficients.

1. **Expand and simplify the given polynomial:**
   \[
   Q(x_1, x_2, x_3, x_4) = 4(x_1^2 + x_2^2 + x_3^2 + x_4^2) - (x_1 + x_2 + x_3 + x_4)^2
   \]
   \[
   = 4(x_1^2 + x_2^2 + x_3^2 + x_4^2) - (x_1^2 + x_2^2 + x_3^2 + x_4^2 + 2(x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4))
   \]
   \[
   = 4(x_1^2 + x_2^2 + x_3^2 + x_4^2) - (x_1^2 + x_2^2 + x_3^2 + x_4^2) - 2(x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4)
   \]
   \[
   = 3(x_1^2 + x_2^2 + x_3^2 + x_4^2) - 2(x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4)
   \]

2. **Find a representation as a sum of squares:**
   We need to find polynomials \( P_1, P_2, P_3, P_4 \) such that:
   \[
   Q(x_1, x_2, x_3, x_4) = P_1^2 + P_2^2 + P_3^2 + P_4^2
   \]
   One possible representation is:
   \[
   Q(x_1, x_2, x_3, x_4) = (x_1 + x_2 - x_3 - x_4)^2 + (x_1 - x_2 + x_3 - x_4)^2 + (x_1 - x_2 - x_3 + x_4)^2
   \]
   To verify, expand each square and sum them up:
   \[
   (x_1 + x_2 - x_3 - x_4)^2 = x_1^2 + x_2^2 + x_3^2 + x_4^2 + 2(x_1x_2 - x_1x_3 - x_1x_4 - x_2x_3 - x_2x_4 + x_3x_4)
   \]
   \[
   (x_1 - x_2 + x_3 - x_4)^2 = x_1^2 + x_2^2 + x_3^2 + x_4^2 - 2(x_1x_2 - x_1x_3 + x_1x_4 + x_2x_3 - x_2x_4 - x_3x_4)
   \]
   \[
   (x_1 - x_2 - x_3 + x_4)^2 = x_1^2 + x_2^2 + x_3^2 + x_4^2 - 2(x_1x_2 + x_1x_3 - x_1x_4 - x_2x_3 + x_2x_4 - x_3x_4)
   \]
   Summing these:
   \[
   Q(x_1, x_2, x_3, x_4) = 3(x_1^2 + x_2^2 + x_3^2 + x_4^2) - 2(x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4)
   \]
   This matches our simplified form of \( Q(x_1, x_2, x_3, x_4) \).

### Part (b)
We need to prove that for any such representation, at least one of the four polynomials is identically zero.

1. **Assume a general representation:**
   Suppose \( Q(x_1, x_2, x_3, x_4) = P_1^2 + P_2^2 + P_3^2 + P_4^2 \) where \( P_i \) are polynomials with integer coefficients.

2. **Evaluate at specific points:**
   Consider \( x_1 = x_2 = x_3 = x_4 = a \):
   \[
   Q(a, a, a, a) = 4(a^2 + a^2 + a^2 + a^2) - (a + a + a + a)^2 = 16a^2 - 16a^2 = 0
   \]
   Thus, \( P_1(a, a, a, a)^2 + P_2(a, a, a, a)^2 + P_3(a, a, a, a)^2 + P_4(a, a, a, a)^2 = 0 \).

3. **Conclude that at least one polynomial is zero:**
   Since the sum of squares of non-negative terms is zero, each term must be zero:
   \[
   P_1(a, a, a, a) = 0, \quad P_2(a, a, a, a) = 0, \quad P_3(a, a, a, a) = 0, \quad P_4(a, a, a, a) = 0
   \]
   Since this holds for any \( a \), each \( P_i \) must be identically zero when evaluated at \( (a, a, a, a) \). Therefore, at least one of the polynomials \( P_i \) must be identically zero.

\(\blacksquare\)

The final answer is \( \boxed{ Q(x_1, x_2, x_3, x_4) = (x_1 + x_2 - x_3 - x_4)^2 + (x_1 - x_2 + x_3 - x_4)^2 + (x_1 - x_2 - x_3 + x_4)^2 } \)