Euler Theorem. (2)

     If numbers a and b are co-prime, j(ab) = j(a)j(b).

     Proof. Let's consider the table:

     In every column of this table there are situated numbers which give the same remainder when divided into b. If this remainder isn't equal to 1 and is divisible by b, then all numbers of this column have different from 1 common divisor with b and, consequently aren't co-prime with ab. Let's cross these columns out. Then the quantity of columns left, will be equal to j(b). In each of these columns there are found numbers of the following kind: k, k+b, k+2b, ... , k+(a-1)b, a numbers in all. Anologousely to how it was done in the small Fermat theorerm, it's proved, that all of them give different remainders when divided into a. Consequentely, there are exactly j(a) numbers amoung them, co-prime with a and at the same time co-prime with ab. So we have j(b) columns and j(a) numbers in each column, all of them co-prime with ab, consequentely j(ab) = j(a)j(b), what was to be proved.