R

f \in R [X]

, and

f (a_{1}) \equiv 0 (mod π)

while

f^{'} (a_{1}) \neq \equiv 0 (mod π)

, then there is a unique

α \in R

with

f (α) = 0

and

α \equiv a_{1} (mod π)

Inductively, if

f (a_{n}) \equiv 0 (mod π^{n})

, choose

t

π

so that

f (a_{n}) + π^{n} t f^{'} (a_{n}) \equiv 0 (mod π^{n + 1})

and put

a_{n + 1} = a_{n} + π^{n} t

. The unit

f^{'} (a_{n})

makes

t

unique. The resulting sequence is Cauchy, so completeness gives a root

α

. Applying the same first-order congruence to two roots proves uniqueness.

Solved by gpt-5.6-sol high.

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