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Sierpinski W. Элементарная теория чисел (Warszawa, 1964)

Sierpinski W. Elementary theory of numbers (Warszawa, 1964)(L)(T)(224s).djvu

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Date Jan 16, 2005

Cites: Carmichael [2], [3],
Sispanov [1], Duparc [1], Knodel [1], Erdos [17], Sierpinski [12],
pp...
Now, looking
at the factorization of number я into prime factors, we see that n [ 2"— 2...
But,
since /(«)== O(moda) and f(v) = 0(modb), we have /(#) = O(moda)
and f(x) = 0(modJ); consequently, since (a, b) =1 and ab = in, f(x)
— 0 (mod да)...
Prom this
we conclude that a necessary condition for the solvability of congruence
D7) (with В not divisible by p) is that D should be a quadratic residue
for the modulus p...
Thus we
see that each root of congruence D7) is congruent with respect to the
modulus p" either to я or to —г...
On the other hand, it is easy to verify that each number given by any of
the congruences E9) satisfies congruence E4) (whenever it is true for s0)
and, since for a > 3 any two of these numbers are not congruent with
respect to the modulus 3", we see that they are different roots of con-
congruence E4)...
has a divisor d
such that 1 < d < «, then in the set 1, 2, ..,,« there are at least the two
numbers, n and Д, that are not relatively prime to n; therefore tp[n)
<n—2...
Applying the formula, just proved, for m*
= <$, i =1,2, ...,1c, we obtain the formula
But since, by theorem 1, <p{$) = g?-1^—1) holds for * = 1, 2,..., fc,
the following theorem 3 is valid:
Theokem 3...
It follows from F) that if <p(n) = 2-51*, where
h is a natural number, then n must have precisely one odd prime divisor...
The question arises wbether for every natural number g there exists
a natural number m such that the equation <p (it) —m has precisely s
solutions in natural numbers...
If p =? 3, that is if p > 3, then the number n cannot be divisible by y2 because,
if it were, p\q>(n) = г-З6"^1, which for p > 3 is impossible...
Thus we see that in the sequenee 1,2, ..., nfor every natural divisor
d of a natural number n there are precisely <p{njd) natural numbers m such
that, for any m, (m, «.)=$...
We are going to prove that the numbers
A6)
6l l Si! • • • J в<р(т)
and numbers A3) are identical in certain order...
If x = д is the least natural solution of congruence A9), then we
say that number a belongs to exponent 8 with respect to modulus in...
We have
thus proved that the condition (Jc, 6) = 1 is both necessary and suffi-
sufficient in order that number rk should belong to the exponent 8 with respect
to the modulus p...
are uni-
uniquely defined by number 3"; so the representation of 3?" in form A) is
unique...
The algorithm for representing a real number яг as a decimal may
also be applied in the case where g is a real number > 1...
We do not know
whether there exist infinitely many natural numbers и > 2 such that the decimal
of number l/та has the period consisting of n— 1 digits...
Representation of numbers by decimals
In particular, for a — 3, we obtain ax -— 3, as = 7, a3=47?
щ — 2207, % = 4870847 and so on...
Continued fractions
But, since an+i >1, by replacing n by n+1 in A1) we obtain
1
A4)
\x-S,
¦n+ll
Tbe relation §„+1 = Qaan+Qn_1 > 6» applied to A3) and A4) gives the
evaluation
(IS)
|я-Д,+1| < \x-Rn\, valid for any n = 1, 2, 3,.....
Similarly, since the third conver-
convergent is 355/113, this numher approximates -к better than any rational with a
denominator < 113...
If Bis a natural number which is not the square of a natu-
natural number, then in the representation of YJ) as a simpb continued fraction,
fte sequeiwe %, a2,.....



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