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Sierpinski W. Elementary theory of numbers (Warszawa, 1964)(L)(T)(224s).djvu |
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since atj, ж2,..., жй+1 are different roots of congruence C7), implies that
p\g{xi) for i = 2, 3,..., w-j-l...
Then the difference between these numbers, equal to (ft—fc)Z, is divis-
divisible by m, whence, in view of (I, m) = 1, m | h—Ъ, which is impossible
since 0 < ft— Ti<m...
But, since §1 < p;i (for gx < q2), we would
obtain ft, > i so j3a > 8 and p1J-p.3 > 12, which is impossible...
Formulae (8) and C7)
of Chapter IT give together the formula
(9)
cp[n) =
d\n
valid for all natural numbers n...
For this purpose it is
sufficient to prove that
1° any term of sequence A6) is a natural number relatively prime
to m and less than m,
2° the elements of sequence A6) are different...
The theorem we have just proved implies that in any infinite arithmetical
progression there are infinitely many terms which have the same prime factors (cf...
Another fact
worth reporting is that there exists an increasing infinite sequence of
natural numbers м* (fc =1,2,...) such that bmX{nk)k(nk) =0...
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 call a sequence p,
p+2, p + 6 whose elements ате all primes a triplet of the first category
and a sequence jj,jp + 4, y+6, whose elements are all prime numbers,
a triplet of the second category...
According to the definition
of numbers B6), for & = 1, 3,..., S the relation rk = a!'(vaoAp) holds...
;p — X j (Л—ZI*, which, in virtue of the relations n = dm, p — l = ds,
gives s | Gc — Z)m; so, since (to,s)=1, s | 7c—Z, which is impossible
because 7c and I are two different numbers of the sequence 1,2,...,«...
n— 1.) There are known primes whose decimals in the scale of 2 consist of digits all
equal to zero with the exception of the first and the last digits...
Hence, since by A4) number о"У„ is an inte-
integer, we see that gnrn = [gV], this being also true for n = 0 provided
» is defined as [»]...
A number > 4 whose digits (in
the scale of 10) are all equal to 4 is divisible by 4 but not divisible by 8...
= <?, this coincides with the ordinary representation
of ж as a decimal in tbe scale of </...
№>w we are going to
consider slightly moie general continued fractions of the form
A)
where и is a given natural number, a0 a real number and ах,аг, ..., an
positive numbers...
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...
On the other hand, if for some natural numbers a0 and at Ф 2a0
number D of D1) is natural, then, since 2a0a1+l is odd, number a\+l
(as a divisor of it) must also be odd; so number ax is even and, since num-
number D of D1) is an integer and, consequently, —\-^ 1 =
——^— is an integer, number «i + l divides number (aa—a1j2Ja1...
Hence, in particular, for t = X, we obtain
^12 = C; 2~7б), V^il = F; 2,2,12),
/180 = A3; 2,2,2,26), ^926= f30; 2,2,2,2,60)...
If the period of the simple continued fraction for num-
number VD consists of an even number s of terms, then the numerator and the
denominator of the (ns — l)-th convergent, n — 1, 2, ..., form a solution
of the equation m*—Dy2 = 1 in natural numbers...
We note that if the coatimied fraction F2) has a well-defined value,
then it may happen that some of its convergents do not have this property...
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