Finding a non-zero divisor in mod 5.?

Question

I posted my question in detail at Cramster.com at
http://answerboard.cramster.com/math-topic-5-146463-0.aspx

Any help/suggestions/hints are greatly appreciated.
Thank you for your time!

Answer 1

Remember that i^2 = -1, and that

(a+bi)(a-bi) = a^2 + b^2.

So, taking a=2, b=1, we have

(2+i)(2-i) = 2^2 + 1^2 = 5 = 0 mod 5.

Finally, -i is congruent to 4i. Then

(2+i)(2+4i) = 0 mod 5.

As for the set you wrote down, it started out okay, then it went in a strange direction. It should look like this:

{0, 1, 2, 3, 4, i, 1+i, 2+i, 3+i, 4+i, 2i, 1+2i, 2+2i, 3+2i, .... }

Hope this helps.

Answer 2

non-zero divisor of what?
maybe you mean non-zero divisor of 0 !

Also, is not a divisor of 0 in mod 5. It is a divisor of 0 in the ring Z_5[i].