I have attached the problem, and my work so far. Thanks in advance!

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- Apr 20th 2009, 04:40 PMminivan15congruence property modulo p^2
I have attached the problem, and my work so far. Thanks in advance!

- May 13th 2009, 08:50 AMfardeen_gen
Assume that such $\displaystyle b$ doesn't exist. Then,

$\displaystyle f(k)=(kp+a)^{p-1}=(p-1)kpa^{p-2}+a^{p-1}=-kpa^{p-2}+a^{p-1}=1(\mbox{mod}\ p^2)$ for all integers $\displaystyle k$,

i.e. $\displaystyle f(k_{1})-f(k_{2})=(k_{2}-k_{1})pa^{p-2}=0(\mbox{mod}\ p^2)$ for all integers $\displaystyle k_{1}, k_{2}$, which is clearly impossible $\displaystyle \rightarrow$ contradiction!