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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: Suppose to the contrary that n is a natural number
greater than 1 such that n\2n— 1 holds...
Then for every integer w we have
If % =m, then f(x) = ап(шо&р), and, since congruence C7) has more
than n roots, there exists an integer x such that f(x) = Ofmodj»), whence
an = 0(modp)...
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)...
As is easy to verify, the
condition is also sufficient and the congruence has four solutions: 1, 3, 6, 7...
This establishes the truth of the following theorem:
Л natural mimber n> X is a prime, if and only if for every natural
number a<n tbe congruence an~l ==l(niodc») holds...
The number of the natural numbers r < I for which (r,l) =1
is oi course <p(l), this being the number of the columns in which all the
I
230
CHAPTER VI...
Since <p{l) — q>B) =1, the equation cp{as) =да, m being odd, is
solvable only in the case where m = 1...
Prove that for any natural number it there exists at least one natural num-
number n such that cp(n-j-k) = g? (n)...
It is also plain that without loss of generality we may suppose that
n > 2 (for in the sequence of odd numbers there exist (as we know) infi-
infinitely many primes)...
We have thus shown that for every natural пшпЬет те > 1 there exists
at least one prime пшпЬет of the form nk + 1, where 7c is a natural num-
number, whence, as we learned above, theorem lla foHows^
As an application of theorem lla we give a proof of the following
theorem of A...
According to the definition
of numbers B6), for & = 1, 3,..., S the relation rk = a!'(vaoAp) holds...
Therefore, for any number x = ga, g1, ...,gp~2, there exists
a number у of the sequence 0,1, 2,...,p—2 such that gy = ai(modjp)...
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...
Therefore in the tables of indices the values of inds; are given only
for natural numbers % less than the modulus (and relatively prime
to it)...
We then have
q}(.p— l)/S, whence <5|(j>— l)/2 and, sinceyls'— 1 (because g belongs to the exponent S
with respect to the modulus p), then a fortiori p\ j№-4/s— 1, i.e...
Therefore in order to prove the theorem it is sufficient to show that
for any natural number Ж and a natural number g > 1 there is at least
one representation A) (conditions B) being satisfied)...
are uni-
uniquely defined by number 3"; so the representation of 3?" in form A) is
unique...
As an immediate consequence of theorem 3* we note that if a number
as has a non-periodic representation as a decimal in a scale of g, then x is
irrational...
JPk and Qk being defined by C), for an arbitrary а„ and positive
%, a2, ¦..,an те have
ft
§ 2...
This, by A9), shows that if A3) is
any representation of x as an infinite simple continued fraction, then
the relations
for и = 1,2,...,
B0)
, = [»„] for я = 0,1,2,.....
A more difficult task is to calculate the probability of the event
that the second quotient is equal to a given natural number m...
We have
h
l7 whence W<
and, consequently, the numbers on both sidee of the last inequality being
[ПвП
Ь-= -ffc —1...
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...



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