Let $a=$$\frac{1}{p-1}$ $(mod p^2)$.clearly , $p^2\nmid(a+1)^{p-1}-1$ . now , $p^2\mid a^{p-1}-1$ is equivalent to $2p-1$ $=0$$(mod p^2)$ (since $a^{p-1}$ $=$$p^2$$k$ $+$ $(p-1)^2$ $+1$ $=$$-$$2p+2$ $(mod p^2)$ . which is impossible since $p$ $>$$3$

alinazarboland wrote: Let $a=$$\frac{1}{p-1}$ $(mod p^2)$.clearly , $p^2\nmid(a+1)^{p-1}-1$ . now , $p^2\mid a^{p-1}-1$ is equivalent to $2p-1$ $=0$$(mod p^2)$ (since $a^{p-1}$ $=$$p^2$$k$ $+$ $(p-1)^2$ $+1$ $=$$-$$2p+2$ $(mod p^2)$ . which is impossible since $p$ $>$$3$ $a=$$\frac{1}{p-1}$ $(mod p^2)$ implies $a=p^2-p-1$ (mod p) which is wrong according to the fact that $a\in\{1,2,\ldots,p-2\}$

Probably some of the arguments can be simplified. By veryfing $p=5$ and $p=7$ directly (both work with $a=2$), we may assume $p\geq 11$. Observe that $a=1$ is impossible so let us consider only $\{2,3,\ldots,p-2\}$. If $g$ is a primitive root modulo $p^2$ and $x = g^k$, $1\leq k \leq \varphi(p^2) = p(p-1)$, then $x^{p-1} \equiv 1 \pmod {p^2}$ if and only if $g^{k(p-1)} \equiv 1 \pmod {p^2}$, which is if and only if $p(p-1) \mid k(p-1)$, i.e. $p$ divides $k$. So the number of solutions to $x^{p-1} \equiv 1 \pmod {p^2}$ is $\frac{p(p-1)}{p} = p-1$ which implies that the number of solutions in $\left[1,\frac{p^2-1}{2}\right]$ is $\frac{p-1}{2}$ (as if $x$ is a solution, then so is $p^2-x$ since $p-1$ is even). In particular, as $1$ is a solution and $\frac{p^2-1}{2} > p - 1$, the number of solutions in $[2,p-1]$ is at most $\frac{p-3}{2}$. Now if we suppose, for contradiction, that the desired statement does not hold, then in $\{2,\ldots,p-1\}$, $p\geq 11$, among two consecutive elements at least one has to be a solution to $x^{p-1} \equiv 1 \pmod {p^2}$. So the number of solutions in this set is at least $\frac{p-3}{2}$, which equals the abovementioned upper bound. Equality is possible if and only if the solutions in $\left[2,\frac{p^2-1}{2}\right]$ are precisely $3$, $5$, $\ldots$, $p-2$. Moreover, it is clear if $x_1$ and $x_2$ are solutions to $x^{p-1} \equiv 1 \pmod {p^2}$, then so is $x_1x_2$ - hence $(p-2)(p-4) \equiv 8-6p$ and $6p-8$ is also a solution. However, $6p-8 > p - 2$ and $6p - 8 \leq \frac{p^2-1}{2}$ (equivalent to $p(p-12) + 15 \geq 0$ which is true for $p\geq 11$) and this is a contradiction!

ISL 2001 N4