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Sierpinski W. Elementary theory of numbers (Warszawa, 1964)(L)(T)(224s).djvu |
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It has been proved that all the numbers with this property
are the numbers n = 2°, a =0,1,2,..., and n = 2^, where a, /?
are natural numbers (cf...
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...
On the other hand, number Qkl as the remainder obtained from
division by m, satisfies the inequalities 0 < gh < m...
If (a,m)=X, then any solution of congruence A9)
is divisible by Ше escponent S to which a belongs with respect to modulus m...
In virtue of the identity
we have p\b1^(ar1 — br1), which, in view of (b1,p) = I, implies that p\ar1—brl for
0 < r < 8, contrary to the assumption that p is a primitive divisor of number af— b^...
The relations 42 ] о (a1806— 1) and 43 \ a (a1806- 1), valid
for any integer a, give by D2, 43) = 1 and 1806 — 42*43 the required relation
1806|а(аш6— 1), which proves that number 1806 has property P...
From the purely theoretical point of view, there exists a method
for establishing -whether a given natural number а Ф 0 is an nth power
residue for a given modulus p...
In fact, according to the definition of indices, we have ginda = a
(modm), g^^ = b(modm) (whenever a and b are relatively prime to да.)...
We say that a natural number N
is expressed as a decimal in ihe scale of g if
A) Ж = ет
where in is an integer > 0 and en (n =0,1,2,...,»») are integers with
the property
B) 0<е„, <з— 1 for n =0,1,..., m and cm #0...
Prove that for every natural number m there exists a prime -whose represen-
representation as a decimal in the scale of 2 is such that the last digit is 1 and the preced-
preceding m digits are equal to zero...
Therefore, by A2), we obtain
the following expansion of number ж into an infinite series:
A3)
where, by A1), numbers о„ are digits in the scale of g...
The number of digits
in the period, as well as the number, iwt less than 0, of digits preceding the
period, is less than the denominator of the rational number in question...
The period (which
starts exactly at the decimal point) is obtained by writing down all the natural num-
numbers from 0 to 99 excluding 98 written as decimals...
The first effective example of an absolutely normal number was given
by me in the year 1916 (Sierpinski [5], see also H...
We have thus proved that anylirrational number x0, 0 < жс < 1, may
be expressed in form B6)...
Consequently, number я must be in an interval whose length is 1/да—
— l/(m+l) = l/m(ro + l)...
Рог example, representing тг as a simple continued fraction we see that ite
second convergent is ^-; therefore the rational у approximates number те better
than any other rational with a denominator < 7...
It is also easy to find all natural numbers D for which the represen-
representation of Y3 as a simple continued fraction has a period consisting of
two terms...
By using theorem 4 it is easy to verify that among all the numbers D
< 1000 there are only 7 numbers, such that Yl> represented as simple
continued fraction has a period consisting of three terms...
Therefore in what follows we may assume that t and n are a solu-
solution of E1) in natural numbers with и > 1...
=_p/2 is impossible since, Ъу assumption, j>
is an odd prime.)
Since D is not divisible by p and in sequence F) the coefficients
at D are natural numbers < (p—1)/2, neither the sum nor the difference
of any two terms of sequence F) is divisible by p...
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