Wednesday, 22 February 2017

elementary number theory - Solve axequivbmodm without Linear Congruence Theorem or Euclid's Algorithm?


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Origin page 5. The overhead doesn't look like Linear Congruence Theorem or anything from Euclid's Algorithm. page 4 tries to delineate
Casting Out the Modulus? Is this it?



So if d|m, then I rewrite \begin{align} ax & \equiv b \mod m \\dx + x & \equiv \\ \implies x & \equiv \end{align}



Is this errorless? Did I blight anything?

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