which way should i follow to solve the problem regearding Fermat's little theorem?

Question

The question is if $p$ is a prime number and $a$, $b$ are positive integer, then we have to prove that $$(a+b)^p \equiv {a+b} \mod p $$

I did it in two ways-

$(i)$ By Fermat's little theorem I have, $$a^{p-1}\equiv1\mod p$$ and $$b^{p-1}\equiv1\mod p$$

$$\therefore a\equiv1^{\frac{1}{p-1}}\mod p$$, and $$b\equiv1^{\frac{1}{p-1}}\mod p$$ $$\therefore a+b\equiv2^{\frac{1}{p-1}}\mod p$$ $$(a+b)^{p-1}\equiv2\mod p$$, hence $$(a+b)^p\equiv2(a+b)\mod p$$.
My question is how to remove that constant $2$ ?

$(ii)$ Again by Fermat's little theorem,

$$a^p\equiv a\mod p$$, $$b^p\equiv b\mod p$$ $$\therefore a^p+b^p\equiv {a+b}\mod p$$ $$(a+b)^p-p.N\equiv {a+b}\mod p$$, where $N=a^{p-1}.b+\frac{(p-1)}{2}.a^{p-2}.b^2+.....+\frac{(p-1)}{2}.b^{p-2}.a^2+b^{p-1}.a$ $$(a+b)^p\equiv {a+b}\mod p$$

My question is - would be it correct if I use the formula $a^p+b^p=(a+b)^p-p.N$, where $$N=a^{p-1}.b+\frac{(p-1)}{2}.a^{p-2}.b^2+.....+\frac{(p-1)}{2}.b^{p-2}.a^2+b^{p-1}.a$$. In particular for $p=2$ $$a^2+b^2 = (a+b)^2-2.a.b$$, where $2ab$ is divisible by $p=2$ and the remainder remains as $a^2+b^2$.

Any help is appreciated.

Answer 1

Hint

By Binomial theorem $$(a+b)^p=\sum_{k=0}^p\binom{p}{k}a^kb^{p-k}.$$ If $k\notin\{0,p\}$, one can easily prove that $p\mid \binom{p}{k}$.

The proof is complete.

Answer 2

If you know already Fermat's little theorem, there is nothing to prove. Then you know that $c^p\equiv c\bmod p$ for all $c$, and you can set $c:=a+b$ to derive your claim $(a+b)^p\equiv a+b\bmod p$.

Answer 3

You have an incorrect step where you essentially write

$$1^{1/(p-1)}+1^{1/(p-1)} = 2^{1/(p-1)}$$

Exponents don't work like that. Also, you are assuming that $a$ and $b$ are relatively prime to $p$ when you use Fermat's little theorem.