SOLUTION: Let $m$ and $n$ be non-negative integers. If $m = 6n + 2$, then what integer between $0$ and $m$ is the inverse of $2$ modulo $m$? Answer in terms of $n$.

Question 1207719: Let $m$ and $n$ be non-negative integers. If $m = 6n + 2$, then what integer between $0$ and $m$ is the inverse of $2$ modulo $m$? Answer in terms of $n$.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52775)   (Show Source): You can put this solution on YOUR website!
.
Let m and n be non-negative integers. If m = 6n + 2, then what integer between 0 and m
is the inverse of 2 modulo m? Answer in terms of n.
~~~~~~~~~~~~~~~~~~~~~~

Notice that m = 6n+2 is an even integer number, for any integer number n. Let's assume that an integer x between 0 and m is the inverse of 2 modulo m. It means that 2*x = 1 mod m, which is the same as to say that 2x - 1 is a multiple of m : 2x - 1 = k*m. But 2x is an even number, and k*m is an even number, since "m" is even. Therefore, this equality 2x - 1 = km is not possible with integer x and "k". Hence, there is NO any integer between 0 and m which is inverse of 2 modulo m. ANSWER. There is NO any integer between 0 and m which is inverse of 2 modulo m.

Solved.

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What I proved in my post, in terms of abstract algebra is THIS general statement:

        In the ring of integers modulo m,   Z/m,   where " m " is an even number,
        the class {2 mod m} is NOT an invertible element.

In opposite, in such a ring, the class {2 mod m} is a divisor of zero;
and it is well known fact of abstract algebra that in a ring a divisor of zero can not be invertible.

Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

I'll follow a similar line of thinking as the other tutor, but show a slightly different pathway.

Let x be some integer in the set {0,1,2,3,...,m-2,m-1} such that 2x = 1 (mod m), assuming such a value exists.
x is the multiplicative inverse of 2 in mod m

This converts to 2x-1 = km for some integer k.
The general rule is that if a = b (mod n) then a-b is a multiple of n, i.e. a-b = kn for some integer k.
Rephrased: a & b generate the same remainder when dividing by n.

That slight tangent aside, let's revisit 2x-1 = km to apply a substitution.

2x-1 = k*m
2x-1 = k(6n+2) ......... plug in m = 6n + 2
2x-1 = 2k(3n+1)
2x - 2k(3n+1) = 1
2(x - k(3n+1) ) = 1
2(some integer) = 1

The left hand side on the last line is always even, but the right hand side is odd. This contradiction proves that there aren't any integer solutions to 2x-1 = km where m = 6n+2.
By extension, there aren't any solutions to 2x = 1 (mod m) either.
2 does not have a multiplicative inverse in mod m.

Answer: No solutions.

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