208 SPECIAL TOPICS IN THEORETICAL ARITHMETIC

Assume that any one of the statements except the last is
tru, that is, assume that the kth statement is tru, where
1=Zk=n-—1

Thus Ve =m"*(m + 1) (gn +1) — 1
Then 14y, =m"*(m + 1)*(gn 4+ 1)
Hence (m 4+ 1)1 + y1) = m"*(m + 1)*(gn + 1)

But (1 4+ ye)/(L + 1) = (m 4 1)/m

Hence m(1 + ye1) = (m + 1)(1 + )

Therfor m(1 4+ yi41) = m**(m 4 1)*(gn + 1)

And 1 4+ Y1 = m"* 2 (m + 1)*(gn + 1)
Therfor Wiy = m"—’“‘l(m + Df(gn+1) — 1

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

The last statement in the series is equivalent to the statement
yn=(m + 1)*gn + 1) — 1
Then z2=m— 1)y, :n
= ot 1) Yen A 1) —~ 112m

We have thus completed our direct proof.

For the converse suppose now that ¢ is an integer, that

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Fegut —mt—nll,
Vi = m"*(m + 1)*(gn + 1) — 1, where 1 =k = n,

andthatz = (n — 1)[(m + D) (gn +1) — 1] : n

Then evidently x and vy, are integers.
To show that z also is an integer, since m = # = 0 (mod #),

(m 4+ 1)"Ygn+1) —1 =111 — 1 (mod »)
=)
Hence m + )" Y(gn + 1) — 1>
Then evidently z is an integer.
It may now be shown by actual substitution that these
values of x, y;, and z satisfy the equations of § 488.

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