Let be a greatest common divisor between and . Then and . But since it means but thus . Thus, or . If then is a unit, and this is what we want. Thus, it suffices to show is impossible, and the (non-unit) divisors of are because they are all associate it sufficies to show is impossible. If and then . But simple trial and error on the congruences mod 4 shows it is impossible to have a square congruent to 1 mod 4. Thus, , a contradiction. Thus, . And so must be a unit.

Here we use the fact that is a unique factorization domain. As a consequence it means if and are relatively prime then each factor must be a cube itself, so, for some . Thus, and .b) Show that there are integers r and s such that x= r^3 - 3r*(s^2) and 3*(r^2)*s-s^3=1

Since . There are two possibilities and so so which means which leads to . Thus, is the only solution.c) Find all solutions in integers x^2+1=y^3 by analyzing the equations for r and s in part (b)[/FONT][/COLOR]