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Number theory Difficulty 6.2 National olympiad Prove it

Theorem 1 If (an,m)=1\left(a_{n}, m\right)=1 and
an−1an≡1( mod m),a_{n}^{-1} a_{n} \equiv 1(\bmod m),

then the congruence equation (2) is equivalent to the congruence equation
xn+an−1an−1xn−1+⋯+an−1a1x+an−1a0≡0( mod m)x^{n}+a_{n}^{-1} a_{n-1} x^{n-1}+\cdots+a_{n}^{-1} a_{1} x+a_{n}^{-1} a_{0} \equiv 0(\bmod m)

and they have the same number of solutions.

Solution

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