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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 fact, we see that it is trivially true for n = 2, and, if да is an
even natural number such that»12" + 2 and да — 11 2n + 1, then the number ro = 2" + 2
satisfies the relations m[2"l + 2 and m— l|2m+l...
Рог
example, the congruence a;2 —1 = 0(mod8) bas four roots: 1, 3, 5, 7;
similarly, the congruence a/2 + 3a!+2 = 0(mod6) has four roots 1, 2, 4, 5,
though the leading coefficient in either of the congruences is relatively
prime to the modulus...
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...
Hence, since В is odd, the congruence is solvable only in
the case where D is of the form 4/b+l...
In order that Ше congruence z% ==D(modm.), where
D is an integer and (D,m) —1, be solvable it is necessary and suffi-
sufficient that 1° В should be a quadratic residue for every modulus that is an
odd prime factor of number in and 2° В should Ъе of the form, 4ft+1 for m
divisible by 4 but not divisible by 8 and of the form 8Jc + l for in divisible
by 8...
Hence, by theorem 3,
In connection with theorem 4 we note that tbere exist infinitely
many natural numbers n such that <p{n) >5»[m-i-l)...
The equation, however, is satisfied also by composite numbers, for exam-
example iv = 12, 14, 20, 44...
Erdos [18], if
for a given natural number s there exists a natural number m such
that the equation <p (n) = m has precisely s solutions (in natural num-
numbers n), then there exist infinitely many natural numbers m with this
property...
So the values for n are 210-3 = 3072, 2»-3 = 2580, 27-17 - 217Й or
23-257 = 2056,
If h = 3, then a— 1 + ft-f/J.^ = 10...
So tit2-..tB sa a3 i
1 (modr), whence,
in view of (a, r) = 1, by theorem 8, we infer that aPW ™ l(modr)...
It is clear that if two numbers are congruent with respect to modu-
modulus m, then they belong to the same exponent with respect to modulus
да; for, if а н= Ь(modm) and for some x formula A9) holds, then bx
= 1 (modm) (since, as we know, the congruence »si(modm) implies
the congruence a? ss ^(modm) for any x = 1,2, ...)•
Tkeokem 9...
Therefore number nn—1 has at least one prime divisor p such that n
belongs to the exponent n with respect to the modulus p...
On the other hand, any of the numbers as that belongs to exponent
6 with respect to modulus p satisfies congruence B5); so it is one of the
numbers B6)...
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...
Accordingly, in the case of и = 3, in order that every integer be
a third power residue foi a prime modulus p it is necessary and suffi-
sufficient that p should not be of the form 3&+1, where й is a natural num-
number, i.e...
We note that an analogous theorem for an nth power residue with
% greater than 2 is not true...
This is equivalent to the rela-
relation 12 1 8y — 4, which, in turn, is equivalent to S]2y—1, i.e...
Therefore, by A2), we obtain
the following expansion of number ж into an infinite series:
A3)
where, by A1), numbers о„ are digits in the scale of g...
The number of digits
in the period, as well as the number, iwt less than 0, of digits preceding the
period, is less than the denominator of the rational number in question...



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