show this $(x_{1}+x_{2}+\cdots+x_{n})^p\equiv x^p_{1}+x^p_{2}+\cdots+x^p_{n}\pmod p$

Question

Let $x_{1},x_{2},\cdots,x_{n}$ are postive integers,and $p$ is prime number,such $$(x_{1}+x_{2}+\cdots+x_{n})^p\equiv x^p_{1}+x^p_{2}+\cdots+x^p_{n}\pmod p$$

Can this problem be solved using pre-university mathematics?

I am talking about elementary number theoretical solutions. Do they exist?

It follows from the results here and induction on $n$. – Stahl Jun 13 '17 at 03:41 — Stahl, Jun 13 '17 at 03:41
Of course. Keep doing binomial theorem – mathworker21 Jun 13 '17 at 03:41 — mathworker21, Jun 13 '17 at 03:41
this proof have elementary number methods to solve it? – math110 Jun 13 '17 at 03:42 — math110, Jun 13 '17 at 03:42
Of course. Keep doing binomial theorem – mathworker21 Jun 13 '17 at 03:42 — mathworker21, Jun 13 '17 at 03:42

score 4 · Answer 1 · answered Jun 13 '17 at 03:44

4

If you know how to prove $$(a_1+a_2)^p \equiv a_1^p + a_2^p\pmod{p} $$ (through Fermat's little theorem) you just have to apply induction since, for instance:

$$ (a_1+a_2+a_3)^p = (a_1+(a_2+a_3))^p \equiv a_1^p+(a_2+a_3)^p \equiv a_1^p+a_2^p+a_3^p\pmod{p}.$$

answered Jun 13 '17 at 03:44

Jack D'Aurizio

353,855

if $x_{i}$ not integer? – math110 Jun 13 '17 at 04:21
1

To the O.P. : Expand $(a_1+a_2)^p$ using the Binomial Theorem, Observe that $p$ divides $\binom {p}{j}$ when p is prime and $1\leq j\leq p-1.$ – DanielWainfleet Jun 13 '17 at 04:22

score 2 · Answer 2 · answered Jun 13 '17 at 04:17

$$(****)(x_{1}+x_{2}+\cdots+x_{n})^p\equiv \\x^p_{1}+x^p_{2}+\cdots+x^p_{n}+\sum_{\large{k_i=1,2,3,...,p-1\\k_1+k_2+...+k_n=p}}^{}\dfrac{p!}{k_1!k_2!...k_n!}x_1^{k_1}x_2^{k_2}...x_n^{k_n}\pmod p$$ now :note that $p$ is prime so $$\dfrac{p!}{k_1!k_2!...k_n!}=\dfrac{p(p-1)!}{k_1!k_2!...k_n!}\\k_1,k_2,...k_n<p \to \\k_1!k_2!...k_n!|(p-1)! \\\ \to \\ \dfrac{p(p-1)!}{k_1!k_2!...k_n!}=p.\dfrac{(p-1)!}{k_1!k_2!...k_n!}=pQ$$ npw look again to (****)$$(x_{1}+x_{2}+\cdots+x_{n})^p\equiv \\\equiv x^p_{1}+x^p_{2}+\cdots+x^p_{n}+\sum_{\large{k_i=1,2,3,...,p-1\\k_1+k_2+...+k_n=p}}^{}pQ.x_1^{k_1}x_2^{k_2}...x_n^{k_n}\pmod p\\\equiv x^p_{1}+x^p_{2}+\cdots+x^p_{n}+p\times\sum_{\large{k_i=1,2,3,...,p-1\\k_1+k_2+...+k_n=p}}^{}Q.x_1^{k_1}x_2^{k_2}...x_n^{k_n}\pmod p\\\equiv x^p_{1}+x^p_{2}+\cdots+x^p_{n}+0\times\sum_{\large{k_i=1,2,3,...,p-1\\k_1+k_2+...+k_n=p}}^{}Q.x_1^{k_1}x_2^{k_2}...x_n^{k_n}\pmod p\\x^p_{1}+x^p_{2}+\cdots+x^p_{n}\pmod p$$

show this $(x_{1}+x_{2}+\cdots+x_{n})^p\equiv x^p_{1}+x^p_{2}+\cdots+x^p_{n}\pmod p$

2 Answers2