Since a != 0 and Zprime is a field, a has a multiplicative inverse c. Then substitute x = -bc to see that we've found one root. To prove that there can't be more than one, suppose x and y are both roots. We'll prove that x=y:
ax + b = 0 = ay + bso
ax = ayand thus
c(ax) = c(ay) ( where c is the multiplicative inverse of a )so
x = 1x = (ca)x = c(ax) = c(ay) = ... = y
This argument clearly works in Zn even when n is composite for those values of a that happen to have a multiplicative inverse. We know those are the ones for which gcd(a,n)=1. So for example in Z10[x] the linear polynomials x+b, 3x+b, 7x+b and 9x+b all have just one root for any of the 10 possible values for b. But we saw in class that the polynomial 5x has more than one root, because both 0 and 2 are roots. In fact, in this case that polynomial has five roots: 0, 2, 4, 6 and 8!
We wondered whether in the composite case the average number of roots would turn out to be 1, because even though some linear polynomials had too many roots, others could have none. Lewanda wanted that as a homework problem, so here it is. See how far you can get with it. You might start with the 10 polynomials 5x+b in Z10[x]. Try to discover when those polynomials have any roots at all, and, when they do, how many they have. You might want to look at some examples in Z12[x] too when looking for patterns.
You may skip the starred item 3 on page 125 on Apollonius's Problem.
Courant and Robbins sprinkle many exercises along the way. They are mostly just computations to make concrete some of the more abstract text. I think most of them are worth working on, but they might involve too much tedious algebra and arithmetic. If you find yourself spending time on boring stuff that isn't increasing your understanding of the text, and you do really understand the text, you can say so in your homework and move on to the next exercises.