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

$$f\left(k_p+d_1\right)\equiv f\left(k_p+d_2\right)\equiv\dotsb\equiv f\left(k_p+d_n\right)\equiv0\pmod p.$$

Prove that there exists an integer $$k_0$$ such that

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

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

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

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

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

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

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

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

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

$F(x)=\big(x-D_1\big)\big(x-D_2\big)\dotsm\big(x-D_n\big).$

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

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

This site uses Akismet to reduce spam. Learn how your comment data is processed.