Olympiad Maths Prep

Track / Stage 6 / 386 of 400 #1386 of 2000

Problem 1386

National olympiad, first round
Algebra Difficulty 6.9 Prove it

5. 133 Let nn be a positive even number. Prove that there exists a positive integer kk, satisfying
k=f(x)(x+1)n+g(x)(xn+1) k=f(x) \cdot(x+1)^{n}+g(x)\left(x^{n}+1\right) \text {, }

where f(x),g(x)f(x), g(x) are some polynomials with integer coefficients. If k0k_{0} denotes the smallest kk satisfying the above equation, express k0k_{0} in terms of nn.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

[Solution] First, we prove the existence of kk.
In fact, when nn is even, ((x+1)n,xn+1)=1\left((x+1)^{n}, x^{n}+1\right)=1. Therefore, there exist rational coefficient polynomials f(x),g(x)f^{*}(x), g^{*}(x), such that
1=f(x)(x+1)n+g(x)(xn+1) 1=f^{*}(x) \cdot(x+1)^{n}+g^{*}(x)\left(x^{n}+1\right) \text {. }

Let kk be a common multiple of the denominators of all the coefficients of f(x)f^{*}(x) and g(x)g^{*}(x), and set f(x)=f(x)= kf(x),g(x)=kg(x)k f^{*}(x), g(x)=k g^{*}(x), then f(x),g(x)f(x), g(x) are integer coefficient polynomials, and
k=f(x)(x+1)n+g(x)(xn+1). k=f(x)(x+1)^{n}+g(x)\left(x^{n}+1\right) .

Thus, the existence of kk is proven.
Next, we find the minimum value of kk, denoted as k0k_{0}.
Let n=2αtn=2^{\alpha} t, where tt is an odd number and α\alpha is a non-negative integer. Let m=2αm=2^{\alpha}. Then, we have
xn+1=(xm)t+1=(xm+1)h(x), x^{n}+1=\left(x^{m}\right)^{t}+1=\left(x^{m}+1\right) \cdot h(x),

where h(x)h(x) is an integer coefficient polynomial.
Let the mm roots of xm+1=0x^{m}+1=0 be
wj=ei2j1mπ,j=1,2,,m,m=2α. w_{j}=e^{i \cdot \frac{2 j-1}{m} \pi}, j=1,2, \cdots, m, m=2^{\alpha} .

If a positive integer kk, and integer coefficient polynomials f(x),g(x)f(x), g(x) satisfy
k=f(x)(x+1)n+g(x)(xn+1), k=f(x) \cdot(x+1)^{n}+g(x)\left(x^{n}+1\right),

then
k=f(wj)(wj+1)n,j=1,2,,m. k=f\left(w_{j}\right)\left(w_{j}+1\right)^{n}, j=1,2, \cdots, m .

Thus, we have
km=j=1mf(wj)j=1m(wj+1)n k^{m}=\prod_{j=1}^{m} f\left(w_{j}\right) \cdot \prod_{j=1}^{m}\left(w_{j}+1\right)^{n} \cdot

Let
σ1=ω1+ω2++ωm,σ2=ω1ω2+ω1ω3++ωm1ωm,σm=ω1ω2ωm. \begin{array}{l} \sigma_{1}=\omega_{1}+\omega_{2}+\cdots+\omega_{m}, \\ \sigma_{2}=\omega_{1} \omega_{2}+\omega_{1} \omega_{3}+\cdots+\omega_{m-1} \omega_{m}, \\ \cdots \cdots \cdots \\ \sigma_{m}=\omega_{1} \omega_{2} \cdots \omega_{m} . \end{array}

By Vieta's formulas, σj\sigma_{j} is an integer, j=1,2,,mj=1,2, \cdots, m. Since j=1mf(wj)\prod_{j=1}^{m} f\left(w_{j}\right) is a symmetric polynomial in ω1,ω2,,ωm\omega_{1}, \omega_{2}, \cdots, \omega_{m} with integer coefficients, it can be expressed as a polynomial in σ1,σ2,,σm\sigma_{1}, \sigma_{2}, \cdots, \sigma_{m} with integer coefficients, and thus it is an integer. Also, because
j=1m(wj+1)n=[j=1m(wj+1)]n=(1+σ1+σ2++σm)n=2n. \begin{aligned} & \prod_{j=1}^{m}\left(w_{j}+1\right)^{n}=\left[\prod_{j=1}^{m}\left(w_{j}+1\right)\right]^{n} \\ = & \left(1+\sigma_{1}+\sigma_{2}+\cdots+\sigma_{m}\right)^{n} \\ = & 2^{n} . \end{aligned}

Therefore, 2nkm2^{n} \mid k^{m}, so 2tk2^{t} \mid k, and k2tk \geqslant 2^{t}.
On the other hand, we set
E(x)=(x+1)(x3+1)(x2m1+1)=(x+1)mF(x) E(x)=(x+1)\left(x^{3}+1\right) \cdots\left(x^{2 m-1}+1\right)=(x+1)^{m} \cdot F(x) \text {. }

For a fixed j{1,2,,m}j \in\{1,2, \cdots, m\}, consider the set
{ωj,ωj3,ωj5,,ωj2m1}, \left\{\omega_{j}, \omega_{j}^{3}, \omega_{j}^{5}, \cdots, \omega_{j}^{2 m-1}\right\},

where the elements are all roots of xm+1=0x^{m}+1=0 and are distinct, hence it is the solution set of xm+1=0x^{m}+1=0. Thus,
E(Wj)=(1+ωj)(1+ωj3)(1+ωj2m1)=(1+ω1)(1+ω2)(1+ωm)=2 \begin{aligned} E\left(W_{j}\right) & =\left(1+\omega_{j}\right)\left(1+\omega_{j}^{3}\right) \cdots\left(1+\omega_{j}^{2 m-1}\right) \\ & =\left(1+\omega_{1}\right)\left(1+\omega_{2}\right) \cdots\left(1+\omega_{m}\right) \\ & =2 \end{aligned}

i.e., ωj\omega_{j} is a root of E(x)2E(x)-2. Therefore, we can set
G(x)(xm+1)+2=E(x)=(x+1)mF(x), G(x)\left(x^{m}+1\right)+2=E(x)=(x+1)^{m} F(x),

raising both sides to the tt-th power, we get
G(x)(xm+1)+2t=(x+1)nFt(x) G^{*}(x) \cdot\left(x^{m}+1\right)+2^{t}=(x+1)^{n} F^{t}(x)

where G(x)G^{*}(x) is some integer coefficient polynomial.
Since
xn+1=(xm+1)h(x), x^{n}+1=\left(x^{m}+1\right) h(x),

where h(x)h(x) satisfies h(1)=1h(-1)=1, we can set
c(x)(x+1)=h(x)1 c(x) \cdot(x+1)=h(x)-1 \text {, }

raising both sides to the nn-th power, we get
Cn(x)(x+1)n=h(x)d(x)+1 C^{n}(x) \cdot(x+1)^{n}=h(x) \cdot d(x)+1

where d(x)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)n1]=(x+1)nU(x)+2t, \begin{aligned} & G^{*}(x) d(x) \cdot\left(x^{n}+1\right) \\ = & G^{*}(x) \cdot\left(x^{m}+1\right) \cdot d(x) \cdot h(x) \\ = & {\left[(x+1)^{n} F^{t}(x)-2^{t}\right] \cdot\left[C^{n}(x)(x+1)^{n}-1\right] } \\ = & (x+1)^{n} \cdot U(x)+2^{t}, \end{aligned}

where U(x)U(x) is some integer coefficient polynomial.
Therefore, there exist integer coefficient polynomials f(x),g(x)f(x), g(x), where
f(x)=U(x),g(x)=G(x)d(x), f(x)=-U(x), g(x)=G^{*}(x) \cdot d(x),

such that
f(x)(x+1)n+g(x)(xn+1)=2t. f(x) \cdot(x+1)^{n}+g(x) \cdot\left(x^{n}+1\right)=2^{t} .

In conclusion, the minimum value of kk is k0=2tk_{0}=2^{t}, where n=2αt,tn=2^{\alpha} \cdot t, t is an odd number, and α\alpha is a non-negative integer.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.