This is a proof from Abstract Algebra by Beachy and Blair. I had thought I was following it until I got to the last sentence. How does taking any solution and adding multiples of n/d to it produce d many solutions?
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$a\left(x+k\dfrac{n}d\right)\equiv ax+k\dfrac ad n\equiv ax\equiv b\pmod n$, and these are distinct $\pmod n$ for $k=0 ,1,...d-1$ – J. W. Tanner Jan 22 '23 at 16:04
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See the more conceptual presentation here in the linked dupe. – Bill Dubuque Jan 22 '23 at 18:12