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Sierpinski W. Elementary theory of numbers (Warszawa, 1964)(L)(T)(224s).djvu |
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definition of д implies that 3 < p— 1 which gives 1 < S <p, contrary to the defi-
definition of the prime p...
In fact, by theorem 5a, for
every integer x we have
ж" = х{тяй.р), да" =ajs(mody), and so on...
If n is a prime, then, of course, every natural number less than n
is relatively prime to n; accordingly for prime n,
A)
<Р(П) = )! — 1...
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...
In fact, we do not know any natural
number in such that the equation <p(n) = m has precisely one solution
in natural numbers n...
И /ij = 2, then n + j32 = 9, $, = 4 or 8, whence a = 5 or 1 and so »
= 2s-5-17 or 2-О-257...
Therefore one of the numbers Fmv J»2, •••> Fj,k
must be equal to Jf5, which is impossible, since -F5 is a composite number...
Using the formula
i
7Г»'
proved in Chapter IV, § 10, one can prove that the ratio of number ]?<p{n)
to number 3<г2/тт2 tends to 1, as x increases...
The number P
is relatively prime to m because anyone of its factors is relatively prime
to m...
It is easy to verify that among the integers x such that 0 < x < 20 only numbers V
anu 13 satisfy the congruence...
of Carmichael (and so absolutely pseudo-prime) it is necessary and suf-
sufficient that X(m) \m—l (cf...
By the corollary to theorem 9,
the terms of the sequence
B3)
1, 2, 3, ..., p-1
belong (modp) to the exponents which are divisors of number <p(p)
= p—l...
Therefore tbe numbeT of №th power residues for a given
modulus p is understood as the number of mutually non-congruent (modp)
nth power residues for the modulus p...
As an immediate corollary to theorem 14 we have the following
proposition: in order that for a given natural number n every integer be an
n-ih power residue for a given prime modulus p it is necessary and suffi-
sufficient that n be relatively prime to p — 1...
Consider an exponential congruence
a1 = Ь(топ.р),
where a, b are integers not divisible by the prime p...
In virtue of A) and B), we also have
whence mlogg' < logJV" < (•m4-l)log(? and therefore
logs
:«+i,
which proves
F)
Formulae F) and E) show that if Ж is represented as A) and condi-
conditions B) are satisfied, then the numbers иг- and cn (» = 0,1,..., m)
are uniquely defined by number N...
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...
If m is the
least natural number such that en — g—1 for n > m, then in the case
of m = 1, by A3), we would have x = M+l, which is impossible...
Consequently ттг is a natural number and, since, by (8),
we have mxnJrl = gmxn—m [<№,] for any n =1,2, ..., then, by induction,
we infer that all the numbers тх„ are integers and, moreover, by (9),
that they satisfy the inequalities 0 < гож„ < m for n = 1,2,.....
However, as regards most of the com-
commonly used numbers, we either know them not to be normal or we are
unable to decide -whether they are normal or not...
For natural numbers n > 1 we write
B7) bn+1 — ancn~~bn, cn+1 = en_1 — alen
We are going to prove that for n > 1 the equality
B8) ...
In fact, by C6), we have YT> = (a; b, 2a), where b Ф 2a,
Hence Yd = a-\ ¦-\-, -= and, consequently, D = a2-) ...
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