How to solve this? If $b in ZZ_8$ is an non-invertible element,demonstrate that $hat2x=b$ have exactly two solutions $x in ZZ_8$ .

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Answer 1 · 2017-04-07T21:28:59

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If $b$ is an even element of $ZZ_8$ then there is at least one $x$ such that:

$hat(2)x = b$

Then:

$hat(2)(x+hat(4)) = hat(2)x+hat(2)*hat(4) = b+hat(8) = b+hat(0) = b$

So there are at least two solutions of:

$hat(2)x = b$

Since there are two solutions for each of the four even elements of $ZZ_8$ , there can be no more than two solutions for any one of them.

We also find:

$hat(4)*b = hat(4)*hat(2)x = hat(8)x = hat(0)x = hat(0)$

So $b$ is a zero divisor and as a result, non-invertible.

Conversely, just checking each of the odd elements of $ZZ_8$ , we find:

$hat(1)*hat(1) = hat(1)$

$hat(3)*hat(3) = hat(9) = hat(1)$

$hat(5)*hat(5) = hat(25) = hat(1)$

$hat(7)*hat(7) = hat(49) = hat(1)$

So these elements are all self inverse.

So every non-invertable element $b$ is even and has exactly two solutions to:

$hat(2)x = b$

How to solve this? If b in ZZ_8b in ZZ_8 is an non-invertible element,demonstrate that hat2x=bhat2x=b have exactly two solutions x in ZZ_8x in ZZ_8.