MATHEMATICAL RECREATIONS 207

But this statement is the £ 4+ 1th in the series of statements.
Therfor by mathematical induction the statements are all
GEIT:

Next we will show that x — 1 = — m" — n (mod m"+!)

Since 1 4+ y, = m™ ! (mod m"),

y1 = m™* — 1 (mod m®)
Hence ny; = n-m* ! — n (mod n-mn)
But, since n = — m,

n-m"! = —m*and n-m* = — mrtt > mrtt
Therfor ny, = — m* — n (mod m"+)
Butx — 1 = ny,

Therfor x — 1= — m" — n (mod m**+)

which was to be proved.
Since this last statement is tru, there exists an integer ¢
such that

x=qgmtt—mr —n-+1

We will now show that, if 1 = 2 = #,

yie = m"*(m + 1)*(gn + 1) — 1
that is, that y = moi(m + 1)%gn + 1) — 1
yo = m"*(m+ 1) (gn + 1) — 1
y3=m"3(m~+ 1)%(gn + 1) — 1
Ve = m" *(m + 1) (qgn + 1) — 1
Yerr = m**(m + 1)¥(gn + 1) — 1
Yna = mi(m + 1)"2(gn + 1) — 1
Yo = m'(m + 1)»(gn + 1) — 1

The first of this series of statements is tru.
For x —1=qgm"*' —m"—n=m"(gm — 1) —n
=—m(gn+1) —n=nm"Y gn+1) —n
and m=x—1) :n=m"(gn+ 1) — 1,
which agrees with the first statement above, since (m + 1) = 1