Divisibility , prime numbers and Fermat’s little theorem [duplicate]

I found somthing about the divisibility with Fermat’s Little Theorem and i prove it . I want to know if it is interrseting or not, because it is trivial ? I want also to know if it is already known as a theory or something like that, when it is the case what are the researches should i do to know if it already exists , thank you and tell me what do you think ?

if $$(a+b)=p$$ and $$p$$ is a prime number then $$(a^p+b)$$ , $$(a+b^p)$$ and $$(a^p+b^p)$$ are divisibile by $$p$$. sorry my English is not that good.

• Keep in mind that $a^p\equiv a \pmod p$, by little Fermat. All of your claims follow from that immidiately. – lulu Sep 17 at 22:10
• These follow immediately by little Fermat $(x^p\equiv x)$ and/or congruence arithmetic rules using $\,\ a^p\equiv a,\ b\equiv -a\Rightarrow\, b^p\equiv (-a)^p\equiv -a^p\pmod{p}\ \$ – Bill Dubuque Sep 17 at 22:18
• If it got you to work and figure something out it's interesting and not trivial. It is true because $a^p\equiv a\pmod p$ and $b^p\equiv b\pmod p$. So $a^p+b^p \equiv a + b^p\equiv a^p + b \equiv a+p \equiv p \equiv 0 \pmod p$ so $p$ divides $a^p+b, a+b^p, a^p + b^p$. Just because something is straightforward and "obvious" doesn't mean its not interesting or trivia... but, no, this is not "significant". Makes a nice exercise for a student though. – fleablood Sep 17 at 22:21
• so it exists already? – adam Sallaoui Sep 17 at 22:37

theory of Fermat is : if P is prime then $$a^p=a+np$$ ; a and n are integers.
$$a+b=p$$;
$$(a^p+b)/p=(a^p-a+p)/p=K*p$$; with $$k$$ an integerfor the other forms also about the same approach .......