# congruence property modulo p^2

Assume that such $b$ doesn't exist. Then,
$f(k)=(kp+a)^{p-1}=(p-1)kpa^{p-2}+a^{p-1}=-kpa^{p-2}+a^{p-1}=1(\mbox{mod}\ p^2)$ for all integers $k$,
i.e. $f(k_{1})-f(k_{2})=(k_{2}-k_{1})pa^{p-2}=0(\mbox{mod}\ p^2)$ for all integers $k_{1}, k_{2}$, which is clearly impossible $\rightarrow$ contradiction!