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Given that $p$ is a prime number and $1\le r \le p-1$, the binomial coefficient ${p \choose r}$is divisible by $p$. By applying Fermat's Little Theorem and considering the binomial expansion

$$((a^p-a)^p+a)^p = \sum_{r=0}^p {p \choose r}(a^p-a)^{p-r}a^r$$

deduce that $a^{p^2}\equiv a^p\pmod {p^2}$

-Below is my working out.-

By Fermat's Little Theorem,$$((a^p-a)^p+a)^p \equiv a^p\pmod p$$

Also, as the first and the last term of the binomial expantion are not divisible by p $$((a^p-a)^p+a)^p \equiv (a^p-a)^p+a^p\pmod p$$ Therefore, $$(a^p-a)^p+a^p \equiv a^p\pmod p$$ $$(a^p-a)^p\equiv 0\pmod p$$ $$a^{p^2} \equiv a^p \pmod p$$ And then I stuck on $$a^{p^2} \equiv a^p \pmod p \Rightarrow a^{p^2} \equiv a^p \pmod {p^2}$$

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  • $\begingroup$ Do you know the Frobenius endomorphism ? en.wikipedia.org/wiki/Frobenius_endomorphism $\endgroup$ – Jennifer Apr 1 '16 at 15:29
  • $\begingroup$ Did you consider doing the binomial expansion a second time? $\endgroup$ – abiessu Apr 1 '16 at 15:30
  • $\begingroup$ @Jennifer I cant find it in our syllabus $\endgroup$ – Golden Apr 1 '16 at 15:33
  • $\begingroup$ Euler's theorem. $\endgroup$ – zhoraster Apr 1 '16 at 15:59
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Using Fermat's Little theorem, $a^{p-1}=kp+1$ where $k$ is any integer

$\displaystyle a^{p^2}=a^p\cdot(a^{p-1})^p$

Now $\displaystyle(a^{p-1})^p=(1+kp)^p=1+\binom p1 kp++\binom p2(kp)^2+\cdots+(kp)^p$

As $\displaystyle\binom pr$ is integer for $0\le r\le p,$

$\displaystyle\implies(a^{p-1})^p\equiv1\pmod{p^2}$ for $p\ge2$

Multiple both sides by $a^p$

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Euler's phi if $n$ has prime factorization $n = p_1^ip_2^jp_3^k...$

$\phi(n)=(\frac{p_1-1}{p_1}\frac{p_2-1}{p_2}\frac{p_3-1}{p_3}...)$

If $\gcd(a,n)= 1, a^{\phi}\equiv 1(\mod n)$

$\phi(p^2)=p(p-1)$ If $a$ does not equal $p$ or $p^2$, then $a^{p^2} = a^{\phi}a^p \equiv a^p(\mod p^2)$

If $a=p$, or $a={p^2}$ then $a^{p^2}\equiv a^{p}\equiv0(\mod p^2)$

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