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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: But, since {a, 3p) = 1, the number a
is not divisible by 3 and the number p — 1 is even (since the number p is an
odd prime), therefore 3 | a3'"-1' —1...
Let p be the least prime divisor of the num-
number n and 5 the least natural number for which p\^~ 1...
+ яп_1<г+я„ satisfies the conditions of theorem 13,
and so it has at most та—т < та different roots, contrary to the assumption...
Since the relation in \ ax2+bx-\-c is equivalent to the relation 4дт |
[4a(aai2+6iB+o), congruence D3) is equivalent to the congruence
D4)
0 (mod 4да»)...
Thus we
see that each root of congruence D7) is congruent with respect to the
modulus p" either to я or to —г...
Hence ж4 = pl1+1pl2+I...¦
a-.-Pe and consequently
Hence, looking at the formula for pt—l and recalling the definition of
m, we see that <p{щ) — mfor t = l, 2,...,«...
Erdos one may additio-
additionally assume that any two numbers of the sequence %, % as are relatively prime
(cf...
И 4 = 3, then ft-f j32 = 10, whence
we easily infer that Д, = 2, $, = 8, and this gives n = б'267...
Formulae (8) and C7)
of Chapter IT give together the formula
(9)
cp[n) =
d\n
valid for all natural numbers n...
In virtue of exercise 3, the number щ is a divisor of a number
m whose digits (in the scale of ten) are equal to 1...
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...
Prove the following theorem of Permat:
If p is a prime > 3, t\en any natural divisor > 1 of number Bp-f-1)/3 is of the
form 2Ьр -\- \, where Ъ is a natural number...
But, since Р„ = 1,
this leads to a contradiction because, since b>2, numbers 6—1 and
6Z —6 + 1 are different from numbers 1 and 62—1...
For each natural divisor д of number p — 1 denote by f(8)
the number of those elements of sequence B3) which belong to exponent
6 with respect to modulus p...
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...
In order that a number a, which is not divisible by p,
be a quadratic residue for an odd prime p, it is necessary and sufficient
that тйа be even...
Clearly, it is not necessary to calculate number 2" for every natural expo-
exponent n; for, if r-ic is the remainder obtained by dividing 2k by 13, then the
remainder yielded by 2k+l divided by 13 is equal to the remainder of 2rk...
If g < 10, the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are taken as the
symbols to denote the numbers of C)...
Thus we have proved theorem 1 and, at the same time, we have
found an algorithm for finding the representation of Ж as a decimal in
the scale of g...
Hence 3% = a—ea
and so 'ffajjl < |a>| + e0 < W + lffl —!; whence \x^ «? (И + |?| —l)/lffl- H
{\x\+\g[-l)l\g\ > Щ, then |al + |ff|-l > \g\ 'A, i.e...
all its digits
are zero from a certain » onwards) if and only if a? is the quotient of an
integer by a power of number g...
This means that of any two consecutive convex-gents to x, the second
gives a better approximation than the first, Formula A0) shows that
>0
<0
for even n,
for odd n,
which means that the even convergent are less than я, whereas the odd ones
are greater than x...
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)...



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