Let $ n$ be an even positive integer. Prove that there exists a positive inter $ k$ such that

\[ k \equal{} f(x) \cdot (x\plus{}1)^n \plus{} g(x) \cdot (x^n \plus{} 1)\]

for some polynomials $ f(x), g(x)$ having integer coefficients. If $ k_0$ denotes the least such $ k,$ determine $ k_0$ as a function of $ n,$ i.e. show that $ k_0 \equal{} 2^q$ where $ q$ is the odd integer determined by $ n \equal{} q \cdot 2^r, r \in \mathbb{N}.$ 

Note: This is variant A6' of the three variants given for this problem.
1. **Expressing the problem in terms of polynomials:**
   We need to show that for an even positive integer \( n \), there exists a positive integer \( k \) such that:
   \[
   k = f(x) \cdot (x+1)^n + g(x) \cdot (x^n + 1)
   \]
   for some polynomials \( f(x) \) and \( g(x) \) with integer coefficients.

2. **Divisibility and polynomial properties:**
   Since \( x+1 \) divides \( x^n - 1 \) for even \( n \), we can write:
   \[
   x^n - 1 = a(x) \cdot (x+1)
   \]
   for some polynomial \( a(x) \). This implies:
   \[
   2 = (x^n + 1) - a(x) \cdot (x+1)
   \]
   Raising both sides to the \( n \)-th power, we get:
   \[
   2^n = (a(x) \cdot (x+1))^n + b(x) \cdot (x^n + 1)
   \]
   for some polynomial \( b(x) \).

3. **Existence of minimal \( k \):**
   This shows that some minimal \( k \) with:
   \[
   k = f(x) \cdot (x+1)^n + g(x) \cdot (x^n + 1)
   \]
   exists. If we subtract this equation \( \lfloor 2^n/k \rfloor \) times from the previous equation, the left-hand side must be 0 (or else the minimality of \( k \) is contradicted). Thus, \( k \) must divide \( 2^n \), implying \( k \) is a power of 2.

4. **Determining \( k_0 \) for \( q = 1 \):**
   When \( q = 1 \), we must have \( k = 2 \). Substituting \( x = 1 \), it is clear that \( k \) must be even. Suppose for contradiction that \( k = 2^c \) for some \( c \geq 2 \). Let \( h(x) = -g(x) \). Our new equation is:
   \[
   2^c = f(x) \cdot (x+1)^{2^r} - h(x) \cdot (x^{2^r} + 1)
   \]
   It is clear that \( \deg f = \deg h \). Let \( f(x) = (x^{2^r} + 1)u(x) + v(x) \) with \( \deg v(x) < 2^r \). We can replace \( f(x) \) and \( h(x) \) with \( v(x) \) and \( h(x) + u(x)(x+1)^{2^r} \), respectively, so \( \deg f = \deg h < 2^r \).

5. **Finite field \( F_2 \) analysis:**
   Consider the equation in the finite field \( F_2 \):
   \[
   f(x) \cdot (x+1)^{2^r} = h(x) \cdot (x^{2^r} + 1)
   \]
   Since \( (x+1)^{2^r} = x^{2^r} + 1 \) in \( F_2 \), we have:
   \[
   (x^{2^r} + 1)(f(x) - h(x)) = 0
   \]
   Since \( x^{2^r} + 1 \) is nonzero in \( F_2 \), we must have \( f(x) - h(x) = 0 \) in \( F_2 \), i.e., \( f(x) = h(x) + 2p(x) \) for some integer polynomial \( p(x) \). Substituting this into the equation, we get:
   \[
   2^c = h(x) \cdot ((x+1)^{2^r} - (x^{2^r} + 1)) + 2p(x) \cdot (x+1)^{2^r}
   \]
   Let \( e(x) = \frac{(x+1)^{2^r} - (x^{2^r} + 1)}{2} \). Note that \( e(x) \) is an integer polynomial, and \( e(1) = 2^{2^r - 1} - 1 \) is odd. Dividing by 2, we get:
   \[
   2^{c-1} = h(x) \cdot e(x) + p(x) \cdot (x+1)^{2^r}
   \]
   Since \( 2^{c-1} \) is even, considering this in \( F_2 \), we get:
   \[
   p(x) \cdot (x+1)^{2^r} = -h(x) \cdot e(x)
   \]
   Since \( e(1) \) is odd, \( (x+1)^{2^r} \) and \( e(x) \) are coprime in \( F_2 \). Thus, \( (x+1)^{2^r} \) divides \( -h(x) \) in \( F_2 \), implying \( h(x) = c(x) \cdot (x+1)^{2^r} + 2d(x) \). This implies \( \deg h \geq 2^r \) or \( h \) is zero in \( F_2 \), both contradictions. Thus, \( k = 2 \).

6. **General case for all \( q \):**
   Suppose we have polynomials \( f(x) \) and \( g(x) \) such that:
   \[
   2 = f(x) \cdot (x+1)^{2^r} + g(x) \cdot (x^{2^r} + 1)
   \]
   Raise both sides to the \( q \)-th power:
   \[
   2^q = f_0(x) \cdot (x+1)^{q \cdot 2^r} + g_0(x) \cdot (x^{2^r} + 1)
   \]
   where \( f_0(x) = f(x)^q \) and \( g_0(x) \) is obtained from binomially expanding and factoring \( x^{2^r} + 1 \) out of all terms but \( f_0(x) \cdot (x+1)^{q \cdot 2^r} \). Replacing \( x \) with \( x^q \):
   \[
   2^q = f_0(x^q) \cdot (x^q + 1)^{q \cdot 2^r} + g_0(x^q) \cdot (x^{q \cdot 2^r} + 1)
   \]
   This completes the construction.

The final answer is \( \boxed{ k_0 = 2^q } \) where \( q \) is the odd integer determined by \( n = q \cdot 2^r \).