[Solution] First, we prove the existence of k.
In fact, when n is even, ((x+1)n,xn+1)=1. Therefore, there exist rational coefficient polynomials f∗(x),g∗(x), such that
1=f∗(x)⋅(x+1)n+g∗(x)(xn+1).
Let k be a common multiple of the denominators of all the coefficients of f∗(x) and g∗(x), and set f(x)= kf∗(x),g(x)=kg∗(x), then f(x),g(x) are integer coefficient polynomials, and
k=f(x)(x+1)n+g(x)(xn+1).
Thus, the existence of k is proven.
Next, we find the minimum value of k, denoted as k0.
Let n=2αt, where t is an odd number and α is a non-negative integer. Let m=2α. Then, we have
xn+1=(xm)t+1=(xm+1)⋅h(x),
where h(x) is an integer coefficient polynomial.
Let the m roots of xm+1=0 be
wj=ei⋅m2j−1π,j=1,2,⋯,m,m=2α.
If a positive integer k, and integer coefficient polynomials f(x),g(x) satisfy
k=f(x)⋅(x+1)n+g(x)(xn+1),
then
k=f(wj)(wj+1)n,j=1,2,⋯,m.
Thus, we have
km=j=1∏mf(wj)⋅j=1∏m(wj+1)n⋅
Let
σ1=ω1+ω2+⋯+ωm,σ2=ω1ω2+ω1ω3+⋯+ωm−1ωm,⋯⋯⋯σm=ω1ω2⋯ωm.
By Vieta's formulas, σj is an integer, j=1,2,⋯,m. Since ∏j=1mf(wj) is a symmetric polynomial in ω1,ω2,⋯,ωm with integer coefficients, it can be expressed as a polynomial in σ1,σ2,⋯,σm with integer coefficients, and thus it is an integer. Also, because
==j=1∏m(wj+1)n=[j=1∏m(wj+1)]n(1+σ1+σ2+⋯+σm)n2n.
Therefore, 2n∣km, so 2t∣k, and k⩾2t.
On the other hand, we set
E(x)=(x+1)(x3+1)⋯(x2m−1+1)=(x+1)m⋅F(x).
For a fixed j∈{1,2,⋯,m}, consider the set
{ωj,ωj3,ωj5,⋯,ωj2m−1},
where the elements are all roots of xm+1=0 and are distinct, hence it is the solution set of xm+1=0. Thus,
E(Wj)=(1+ωj)(1+ωj3)⋯(1+ωj2m−1)=(1+ω1)(1+ω2)⋯(1+ωm)=2
i.e., ωj is a root of E(x)−2. Therefore, we can set
G(x)(xm+1)+2=E(x)=(x+1)mF(x),
raising both sides to the t-th power, we get
G∗(x)⋅(xm+1)+2t=(x+1)nFt(x)
where G∗(x) is some integer coefficient polynomial.
Since
xn+1=(xm+1)h(x),
where h(x) satisfies h(−1)=1, we can set
c(x)⋅(x+1)=h(x)−1,
raising both sides to the n-th power, we get
Cn(x)⋅(x+1)n=h(x)⋅d(x)+1
where d(x) is some integer coefficient polynomial.
From (1) and (2), we get
===G∗(x)d(x)⋅(xn+1)G∗(x)⋅(xm+1)⋅d(x)⋅h(x)[(x+1)nFt(x)−2t]⋅[Cn(x)(x+1)n−1](x+1)n⋅U(x)+2t,
where U(x) is some integer coefficient polynomial.
Therefore, there exist integer coefficient polynomials f(x),g(x), where
f(x)=−U(x),g(x)=G∗(x)⋅d(x),
such that
f(x)⋅(x+1)n+g(x)⋅(xn+1)=2t.
In conclusion, the minimum value of k is k0=2t, where n=2α⋅t,t is an odd number, and α is a non-negative integer.