Let x=mc+n and y=pc+q, where 0<n,p<c and m, n, p and q are integers, then xy=mpc^2+c(mq+np)+nq. Dividing this product by c: mpc+mq+np+nq/c. By definition nq is not divisible by c. We can find values of m, n, p and q for all values of x and y, so, since nq is not divisible by c, neither are x and y.
If ab and a+b are even then we can write: ab=2m and a+b=2n, where m and n are integers. a=2n-b can be substituted into the other equation giving b(2n-b)=2m. So 2nb-b^2=2m and b^2=2(nb-m). Therefore b^2 must be even, implying b must also be even and, since a=2n-2p=2(n-p) where 2p=b, a must be even. <\p>