Feb 042014

Let \(f(x)=x^n+a_{n-1}x^{n-1} +\dotsb+a_1x+a_0\) be a polynomial with integer coefficients, and let \(d_1\),\(\dotsc\), \(d_n\) be pairwise distinct integers. Suppose that for infinitely many prime numbers \(p\) there exists an integer \(k_p\) for which

\begin{equation}f\left(k_p+d_1\right)\equiv f\left(k_p+d_2\right)\equiv\dotsb\equiv f\left(k_p+d_n\right)\equiv0\pmod p.\end{equation}

Prove that there exists an integer \(k_0\) such that

\[f\left(k_0+d_1\right)=f\left(k_0+d_2\right)=\dotsb= f\left(k_0+d_n\right)=0.\]

用 \(P\) 表示 \(\gt n\), 且具有下列性质的质数 \(p\) 所组成的集合: 存在整数 \(k_p\), 使得 \((1)\) 为真.

记 \(u=d_1+d_2 +\dotsb+d_n+a_{n-1}\). 对于 \(p\in P\), 设 \(K_p=nk_p+u\); 对每个 \(\leq n\) 的正整数 \(i\), 设 \(D_i=nd_i-u\). 易见, 所有的 \(\mid D_i-D_j\mid\) 都不会是 \(0\).

\[F(x)=n^nf\Big(\frac xn\Big)=x^n+na_{n-1}x^{n-1}+n^2a_{n-2}x^{n-2}\dotsb+n^{n-1}a_1x+n^na_0.\]

注意到, 对正整数 \(i\)(\(1\leq i\leq n\)), 有

\[F\big(K_p+D_i\big)=n^nf\bigg(\frac{K_p+D_i}n\bigg)=n^nf\big(k_p+d_i\big)\equiv0\pmod p.\]

只要质数 \(p\in P\) 足够大, 任意的 \(\mid D_i-D_j\mid\) 都不被 \(p\) 整除. 既然 \(K_p+D_1\), \(K_p+D_2\), \(\dotsc\), \(K_p+D_n\) \(\bmod p\) 互不同余, 从而它们就是 \(F(x)\equiv0\pmod p\) 的全部解. 然后, Vieta formula 定出

\[\big(K_p+D_1\big)+\big(K_p+D_2\big)+\dotsb+\big(K_p+D_n\big)\equiv-na_{n-1}\pmod p,\]

\[\big(K_p+nd_1-u\big)+\big(K_p+nd_2-u\big)+\dotsb+\big(K_p+nd_n-u\big)\equiv-na_{n-1}\pmod p,\]


\[nK_p\equiv-n\big(a_{n-1}+d_1+d_2+\dotsb+d_n-u\big)=0\pmod p.\]

既然 \(p\gt n\), \(p\mid K_p\).

再次使用 Vieta formula, 当 \(1\leq l\leq n\) 时,

\[(-1)^ln^la_{n-l}\equiv\prod_{1\leq i_1\lt\dotsb\lt i_l\leq n}\big(K_p+D_{i_1}\big)\dotsm\big(K_p+D_{i_l}\big)\pmod p,\]

由 \(p\mid K_p\) 得知

\begin{equation}(-1)^ln^la_{n-l}\equiv\prod_{1\leq i_1\lt\dotsb\lt i_l\leq n}D_{i_1}\dotsm D_{i_l}\pmod p.\end{equation}

只要 \(P\) 中的质数 \(p\) 使得任意的 \(\mid D_i-D_j\mid\) 都不被 \(p\) 整除, 则 \((2)\) 成立. 如此, 就必须有

\[(-1)^ln^la_{n-l}=\prod_{1\leq i_1\lt\dotsb\lt i_l\leq n}D_{i_1}\dotsm D_{i_l}.\]



返回到多项式 \(f\) 以及 \(d_i\),

\[f(x)=\Big(x-d_1-\frac un\Big)\Big(x-d_2-\frac un\Big)\dotsm\Big(x-d_n-\frac un\Big).\]

\(f\) 是首一多项式, 其有理根必是整数. 故 \(\dfrac un\) 是整数.

令 \(k_0=\dfrac un\), 则 \(f(k_0+d_i)=0\) 对所有的 \(1\leq i\leq n\).

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