Show that $$ \sum\limits_{i=1}^{p} i^{p} $$ is divisible by $p$ for all primes $p > 2$.
I think this has something to do with Fermat's Theorem and I have tried using congruences modulus to do it.
Asked
Active
Viewed 274 times
5
-
elcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. – José Carlos Santos Aug 26 '18 at 09:46
-
See https://math.stackexchange.com/questions/529790/1k-2k-cdots-p-1k-equiv-begincases-1-mod-p-text-if-p-1 – lab bhattacharjee Aug 26 '18 at 10:26
2 Answers
5
Guide:
Notice that $p$ is odd.
$$1^p + (p-1)^p \equiv 1^p+(-1)^p \equiv 1-1 \equiv0 \pmod{p}$$
$$2^p + (p-2)^p \equiv 2^p+(-2)^p \equiv 2^p-2^p \equiv0 \pmod{p}$$
Something to think about:
- How important do you think $p$ being prime is here, or is $p$ being odd positive number sufficient.
Siong Thye Goh
- 149,520
- 20
- 88
- 149
3
And yes, it can be solved using LFT as well, since $$i^p\equiv i \pmod{p}$$ thus $$\sum\limits_{i=1}^{p}i^p \equiv \sum\limits_{i=1}^{p}i \pmod{p} \tag{1}$$ but $$\sum\limits_{i=1}^{p}i = \frac{p(p+1)}{2}$$ and $\forall p>2$ primes are odd, thus $2 \mid p+1$ which means $$\sum\limits_{i=1}^{p}i = \frac{p(p+1)}{2} \equiv 0 \pmod{p}$$ and from $(1)$ $$\sum\limits_{i=1}^{p}i^p \equiv 0 \pmod{p}$$
rtybase
- 16,907