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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: This shows that the number a?"'3— 1
is divisible by p and by 3; consequently, since (p,3) =1, it is also
divisible by 3p...
In virtue of the theorem
of Fermat, p}2P~l— 1 (this is because да and, consequently, p are odd)...
This proves the existence of a one-to-one correspon-
correspondence between all the roots (non-congruent with respect to the modulus
m) of congruence D0) and all the pairs и, ¦» consisting of the roots of con-
congruences D1) and D2), respectively...
Moreover, z and z1 are not congruent with
respect to the modulus p", since, if they were, we would have p" | 2г, which,
since p is odd, would give pa | % and hence p \ B, contrary to the assump-
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...
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...
The equation q>(n+3) =
<p(n), however, has only two solutions, n = 3 and n — 5, for n < 10000...
The argument is that if gj and g2 were two different odd prime divisors
of the number n, then, by E), (&—l)ft,—l) 1 <p(n) = 2-5'k and so i | <p(n),
which is impossible...
It can be calculated that
m9 = l,ma=2,m4 — i,m5 = 8,m6 = 12,m7 = 32,ma = 36,щ = 40, mu
= 24, ma = 48, mlt = 160, m13 = 396, mu = 2268, mls = 704...
If a natu-
natural number x satisfies the congruence Xе = S(modlO), then any of the terms of the
arithmetical progression х + Ш [h = 0, 1, 2, ...) just obtained also satisfies it...
< q3 is the factorization of number m into prime factors, then
and that for any natural number jb there exist natural numbers that
belong to the exponent X[m) with respect to modulus m (cf...
Since p \ nn—1, -we
have 5 | n, whence we infer that nj5 is a natural number >1 (because
6 < n)...
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)...
But, by theorem 11 (with d =
[%,p — 1) =2), there are only \[p—-1) quadratic residues in the se-
sequence 1,2, ...,p — 1...
For example number
is normal in the scale of 10; number ^ is normal in the scale of 2 but
it is not normal in the scale of 3...
Representation of numbers by decimals
In particular, for a — 3, we obtain ax -— 3, as = 7, a3=47?
щ — 2207, % = 4870847 and so on...
but, since i4"+2) > an+2, we have
whence, in virtue of lim -
¦Ч-1-Г ^н-j)
= #»» we infer
Consequently sjn>an for any n = 0,1,2,.....
йот this we infer that the probability of
the event that the first quotient of the representation of a real number
m as a continued fraction is equal to m, is l/»i(»i+l)...
Continued fractious of quadratic irrationals 293
The representation of YD as a continued fraction is usually written
in the form Yd = («oj ai> an ¦¦¦> fflsh *be bar above the terms indicat-
indicating that they form a period...
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)...

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