I have the following:
12x+28y=20
I'm trying to find solutions to the equation above defined by: 12x≡20(mod28)
The GCD is d=gcd(28,12)=4 and since 4|20, then there are 4 solutions that exist. (please correct me if I'm wrong).
Using the extending Euclidean Algorithm, we find x0=−2 and y0=1. The general solution is defined by: x0+t(nd)
which in turn gives −2+7t in our case. But how can we have a negative remainder if x=−2(mod7) which can't happen.
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