How To Solve A Modular Arithmetic Equation

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I would love to see more videos on modular arithmetic, I always had trouble understanding it and your videos are being very helpful!

barberickarc

A question: how do we know that there are only two solutions?
I know, it's a second degree polynomial, but integers (mod 49) is NOT a field, but has zero dividers, so strange things might happen. Consider, for example, that x^2=1 (mod 15) has 4 different solutions, namely 1, 14 (=-1), AND 4 and 11.

I think this should be discussed!

karl

You could also just lift the mod 7 solutions using Hensel's lemma (basically Newton's method):
2 - (2^2 + 2 + 1)/(2*2 + 1) = 2 - 7 * 10 = -68 = 30 mod 49
4 - (4^2 + 4 + 1)/(2*4 + 1) = 4 - 17 * 11 = -227 = 18 mod 49

sbares

I thought of an odd thing:
Since GCD(a, b) = K*a + J*b
and c # d (mod n) means c = d + l*n
All variables are integers of course.
So l*N turns into K*a+J*b

Could you combine them to make an equation a # b (mod n, m)?
Does that makes sense?

pauljackson

X-3=4(mod7) ????
I did not understand how you get this

BenSamuel-tf

you multiply by 4, you have (2x+1)^2=144(mod

hazalouldi

X-3=4(mod7) ????
I did not understand how you get this

BenSamuel-tf

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