Prove that there exists an integer $a$ with $1 \leq a \leq p-2$ such that neither $a^{p-1}-1$ nor $(a+1)^{p-1}-1$ is divisible by $p^2$. 
Let $p \geq 5$ be a prime number. Prove that there exists an integer $a$ with $1 \leq a \leq p-2$ such that neither $a^{p-1}-1$ nor $(a+1)^{p-1}-1$ is divisible by $p^2$.

All my progress:

*

*We have by FLT $\{1^p,2^p,\dots, {p-2}^p\}\equiv \{1,2,\dots , p-2\}
   \mod p.$


*We also have by FLT, $p|a^{p-1}-1$


*We want to show that $v_p(a^{p-1}-1)=1=v_p({a+1}^{p-1}-1).$


*I also got $(p-a)^{p-1}\equiv a^{p-1}+a^{p-2}\cdot (p-1)\cdot
   a^{p-1}+p\cdot a^{p-2}$
Also, if we can show $ \{a^{p-1}-p \mod p^2:  1 \leq a \leq p-2\}=$ $\{0,p,\cdots, p(p-1)\} $ in some form, and then we will be done.
Would be helpful if one can send hints in place of a solution.

Identify the integers $\{1\le n< p^2:p\nmid n\}$ with the cyclic group $(\mathbb{Z}/p^2\mathbb{Z})^\times$. Precisely $p-1$ of these have order dividing $p-1$. Call these integers $a_1<\ldots<a_{p-1}$.
Let $m$ be the largest integer with $a_m\le p-1$. Note that $a_1=1$. If your claim is false, every pair $\{1,2\},\{3,4\},\ldots,\{p-2,p-1\}$ contains at least one $a_i$, whence $m\ge \frac12(p-1)$.
However, note that for $p\ge 3$ we can construct $3m-1$ elements of order dividing $p-1$ as follows:
$$1=a_1^2<a_1a_2<a_2^2<\dots<a_{m-1}a_m<a_m^2<p^2-a_m<\dots<p^2-a_1<p^2.$$
Hence, $\frac13p\ge m\ge \frac12(p-1)$, which gives a contradiction for $p\ge 5$.
