Proving $(n+1)^p\equiv n^p+1\pmod{p^3}$
问题内容
Let $n$ be a positive integer and let $p>3$ be a prime number such that $p\mid n^2+n+1$.
Prove that $(n+1)^p\equiv n^p+1\pmod{p^3}$.
Set $P(x)=\dfrac{(x+1)^p-x^p-1}{p}$.
First use $p\mid n^2+n+1$ to show that $n^3\equiv1\pmod p$ and hence $p\equiv1\pmod3$. Now let $\omega$ be a primitive third root of unity, so $\omega^2+\omega+1=0$. Using $p\equiv1\pmod3$, show that $P(\omega)=P(\omega^2)=0$. Hence $x^2+x+1\mid P(x)$, so write $P(x)=(x^2+x+1)Q(x)$ with $Q(x)\in\mathbb Z[x]$. I don't know how to continue. Could anyone help me? Thanks in advanced
回答 (1)
We have $$ P'(x)=\frac{p(x+1)^{p-1}-p x^{p-1}}{p}=(x+1)^{p-1}-x^{p-1}. $$
Now evaluate $P'$ at $\omega$ gives $$ P'(\omega)=(\omega+1)^{p-1}-\omega^{p-1}. $$ Since $\omega+1=-\omega^2$, we get $$ P'(\omega)=(-\omega^2)^{p-1}-\omega^{p-1}. $$ Because $p>3$ is prime, $p-1$ is even, so $(-1)^{p-1}=1$. Hence $$ P'(\omega)=\omega^{2(p-1)}-\omega^{p-1}. $$ Since $p\equiv 1\pmod 3$, we have $p-1\equiv 0\pmod 3$, so $\omega^{p-1}=1$. Therefore, $$ \omega^{2(p-1)}=(\omega^{p-1})^2=1. $$ Thus $$ P'(\omega)=1-1=0. $$ The same calculation works for $\omega^2$ (since $\omega^2$ also satisfies $\omega^2+1=-\omega$), so $P'(\omega^2)=0.$
Since $P(\omega)=0$ and $P'(\omega)=0$, $\omega$ is a multiple root of $P(x)$. Similarly, $\omega^2$ is a multiple root.
Thus in $\mathbb C[x]$, $$ (x-\omega)^2(x-\omega^2)^2 \mid P(x). $$ But $$ (x-\omega)(x-\omega^2)=x^2+x+1, $$ so $$ (x^2+x+1)^2 \mid P(x) \quad\text{in } \mathbb C[x]. $$
Because $x^2+x+1$ is monic and $P(x)\in\mathbb Z[x]$, the quotient must have integer coefficients. Hence $$ P(x)=(x^2+x+1)^2 R(x) $$ for some polynomial $R(x)\in\mathbb Z[x]$.
Evaluate at $x=n$ gives $$ P(n)=(n^2+n+1)^2 R(n). $$ By hypothesis, $p\mid n^2+n+1$. Therefore, $$ p^2 \mid (n^2+n+1)^2, $$ so $$ p^2 \mid P(n). $$ Since $pP(n)=(n+1)^p-n^p-1$, we get $$ p^3 \mid (n+1)^p-n^p-1. $$ Hence $$ (n+1)^p\equiv n^p+1 \pmod{p^3}. $$