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[tex] f(x)=0mod(p) [/tex]

then if we have as a first approximation [tex] f(x_{n+1})=0mod(p) [/tex]

then using a linear interpolation: [tex] f(x_{n})+f'(x_{n})(x_{n+1}-x_{n})=0mod(p) [/tex] or [tex] x_{n+1}=(x_{n}+\frac{f(x_{n}}{f'(x_{n})})0mod(p/f'(x_{n}) [/tex]

So we have 'modified' Newton method for solving Polynomial congruences. :Bigrin: