R=Zsqrt(-5) where Z=intergers.

Define a norm function N on R with values in non negative integers by

N(a+bsqrt(-5))=(a^2)+5b^2

Using N or otherwise show 2 in reals is irreducible.

By finding two distinct factorizations in R of the number 6, show 2 in the reals is not prime.

Im pretty desperate for a correct solution to this question i can revise from, its driving me mad having no notes on it, be very grateful thanks.