We will consider 3 cases: 1)$(\frac{p-1}{2})^{p-1} \not \equiv 1 \mod{p^2}$ and $(\frac{p+1}{2})^{p-1} \not \equiv 1\mod{p^2}$ 2)$(\frac{p-1}{2})^{p-1} \equiv 1 \mod{p^2}$ 3)$(\frac{p+1}{2})^{p-1} \equiv 1 \mod{p^2}$ In the first case, we obviously take $a=\frac{p-1}{2}$. Let us consider the second case. $(\frac{p-1}{2})^{p-1} \equiv 1 \mod{p^2}$ is equivalent to $(p-1)^{p-1} \equiv 2^{p-1} \mod{p^2}$ which is, after reducing $\mod{p^2}$, $p(p-1) + 2^{p-1} - 1 \equiv 0 \mod{p^2}$ From FLT $2^{p-1} - 1$ is divisible by $p$ so, $2^{p-1} - 1 = pk$. So: $p(p-1) + 2^{p-1} - 1 = p(p-1 + k) \equiv 0 \mod{p^2}$ So, $p-1+k \equiv 0 \mod{p}$ which means that $k \equiv 1 \mod{p}$ and $2^{p-1} - 1 \equiv p \mod{p^2}$ Now, we will show that $a=p-2$ satisfies required condition. Indeed: $(p-2)^{p-1} - 1 \equiv 2^{p-2}p(p-1) + 2^{p-1} - 1 \equiv p(2^{p-2}(p-1) + 1) \mod {p^2}$ And since $2^{p-2}(p-1) + 1 \equiv -2^{p-2} + 1 \not \equiv 0 \mod{p}$ above congruence is not $0 \mod{p^2}$. Now we check $p-1$: $(p-1)^{p-1} - 1 \equiv -p(p-1) + 1 - 1 \equiv -p(p-1) \not \equiv 0 \mod{p^2}$ So, in this case we are done. We are left with the case 3) but it's almost the same.

Another solution (I suggest): 1) If $p^2|a^{p-1}-1$ then $(p-a)^{p-1}-1$ is not divisible by $p^2$ We have, if $p^2|(p-a)^{p-1}-1=p^{p-1}-\binom{p-1}{1}.p^{p-2}.a+...-\binom{p-1}{p-2}.p.a^{p-2}+a^{p-1}-1$ then $p^2|\binom{p-1}{p-2}p.a^{p-1}$ which is a contradiction 2) IF $p^2|a^{p-1}-1$ and $p^2|(a+1)^{p-1}-1$ then $(p-a)^{p-1}-1$ and $(p-a-1)^{p-1}-1$ aren't divisible by $p^2$ then we finish the problem because we choose $a'=p-a-1$ then $a'^{p-1}-1$ and $(a'+1)^{p-1}-1$ are not divisible by $p^2$ 3) IF $a^{p-1}-1,(a+1)^{p-1}-1$ there is only one number which is divisible by $p^2$ $(1)$ then we have: i) We have if $p^2|2^{p-1}-1$ then by the statement above, we have $(p-2)^{p-1}-1$ is not divisible by $p^2$ So we choose $a=p-2$ then $(p-2)^{p-1}-1$ is not divisible by $p^2$ and since $p^2|1^{p-1}-1 $ then $(p-1)^{p-1}-1$ is not divisible by $2$ so $a=p-2$ satisfies the problem ii) If $2^{p-1}-1$ is not divisible by $p^2$ then by $(1)$ we have $p^2|3^{p-1}-1$ and $4^{p-1}-1$ is not divisible by $p^2$, .... then $(2k)^{p-1}-1$ is not divisible by $p^2$ and $p^2|(2r+1)^{p-1}-1$ for all $k,r$ $(2)$ We consider the number $(2k)^{p-1}-1$ is not divisible by $p^2$ and $p^2|(p-2k)^{p-1}-1$ with $k$ is an odd number, we have $p^2|(p-2k)^{p-1}-1 \Rightarrow p^2|(2k)^{p-1}-(p-1)p(2k)^{p-2}-1$ $\Rightarrow p^2|(2k)^{p-1}+p(2k)^{p-2}-1$ and since $k$ is odd then $p^2|k^{p-1}-1$ by $(2)$ Thus $p^2|k.2^{p-1}+p2^{p-2}.k^{p-1}-k$ $ \Rightarrow p^2|k.2^{p-1}+p.2^{p-1}-k$ $ \Rightarrow k.(2^{p-1}-1) \equiv 1-p.2^{p-1} \pmod{p^2}$ But $p=5$ the problem is trivial, and $p>5$ we have there is an other odd number $k'$ which $k'\neq k$, similarly, we gain $k'.(2^{p-1}-1) \equiv 1-p.2^{p-1} \pmod{p^2}$ Thus $k.(2^{p-1}-1) \equiv k'.(2^{p-1}-1) \equiv 1-p.2^{p-1} \pmod{p^2}$ then $p^2|(k'-k)(2^{p-1}-1)$ which is clearly impossible because $k,k'<p$ and $p$ is an odd prime and we are considering $2^{p-1}-1$ is not divisible by $p^2$ Then we finished our problem

Another solution : Case $p=5$ is trivial. So let's consider case $p>5$ Let $S=\{k\in\mathbb{N}\mid 1\le k\le p-1, p^2\nmid k^{p-1}-1\}$ thus we need to show that set $S$ contain 2 consecutive number in it. First assume that both $a, p-a\notin S$ $a\in\mathbb{N}$ and $1\le a\le p-1$ so $p^2\mid a^{p-1}-1$ and $p^2\mid (p-a)^{p-1}-1$ But binomial theorem gives us that $\displaystyle{(p-a)^{p-1}\equiv\binom{p-1}{1}pa^{p-2}+1\equiv -pa^{p-2}+1\pmod{p^2}}$ So $p^2\mid pa^{p-2}$ which means $p\mid a$ contradiction! Hence $a\in S$ or $p-a\in S$. It's easy to show that $1\notin S$ and $p-1\in S$ Assume that for every $a\in S$ implies that $a+1\notin S$ By sequences of deduction from what we have proven gives us $S=\{2,4,6,...,p-1\}$. We'll split into 2 cases.