Finding the Inverse of a Number in a Finite Field

moo5003
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Homework Statement



Suppose that m = 1 mod b. What integer between 1 and m-1 is equal to b^(-1) mod m?

The Attempt at a Solution



m = 1 mod b means that:

m = kb + 1 for some integer k

Let x be the inverse of b mod m, note: x exists since b and m must be coprime due to the previous statement.

xb = 1 mod m

thus: xb = gm + 1 for some integer g.

Now this is were I have little success. I can't seem to manipulate anything to my advantage and I'm unsure how to proceed.

I did find x = (m+1)/b but that is not always an integer. Thanks for any help you can provide.
 
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Well, you don't seem to have made use of the fact that m = 1 mod b...
 
I thought I used that fact when using the statement

m = kb + 1 for some integer k, unless I'm missing something else. Little tired, but I will come back to it tomorrow.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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