206 SPECIAL TOPICS IN THEORETICAL ARITHMETIC

1 4+ yu—o = m® (mod m?)

1 4+ yu1 = m! (mod m?)

1 + Yoo = m? (mod m?)
14+ Yorn = mh-1 (mod m*)
14 yix = m* (mod m*+?)

1 4+ 3, = m? (mod e

1 4+ y1 = m™! (mod mn)

The first of this series of statements is tru.

For (n — 1)y, = nz

Hence (n—1)y.>3>n

But n—1]ln §§ 198, 158.
Hence V> 1 § 267.
Therfor Yo = 0 (mod n)

But, sincen = — m, n>>m

Therfor Ya = 0 (mod m)

Hence 1+ v, =1 (modm),

which is equivalent to the first statement.

Now assume that any one of our series of statements except
the last is tru, that is, assume that the kth of these statements
is tru, where 0 = k= n — 2.

Thus we assume

1 4 ypi41 = m*! (mod m*)

Then m(1 4+ Yniy1) = m* (mod m*+1)

But (1 + ynit1)/(1 + yus) = (m + 1)/m
Hence (m 4+ 1)(1 4+ ya) = m(1 + ypis1)

Again mF = m*+ 4 m* (mod m*+1) § 199,
= (m + 1)m*

Therfor (m 4+ 1)(1 4 y.s) = (m 4+ 1)m* (mod m*+)

But m-+1]1m § 198.

Hence m 4+ 1 || m++,

since, if two numbers are prime to each other, any power of
either is prime to any power of the other.
Therfor 1 4+ Yo = m* (mod m*+1)