Home / lib / M_Mathematics / MT_Number theory / | ||
Sierpinski W. Elementary theory of numbers (Warszawa, 1964)(L)(T)(224s).djvu |
Size 7.6Mb Date Jan 16, 2005 |
integer a, then, of course, for (a, n) = 1 we have n \ a"-3—1...
Now, looking
at the factorization of number я into prime factors, we see that n [ 2"— 2...
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 s=plsl and so, by D8), 4 = 1>i+tpa~1'- This yields the con-
congruence
Thus we see that the solution of congruence D7) reduces to the solution
of a congruence of the same type, the right-hand side of which is not
divisible by p...
(This, as proved above, is true for a — ?.) Then there exists
an integer у such that y* = D(mod2°) and, since D is odd, y, of course,
must also be odd...
The number of the roots of the congruence is equal to 2д+;1, where I is
the number of odd prime factors of the number m and p = 0 for m not divis-
divisible by 4, ft = 1 for m divisible by 4 but not divisible by 8, and, finally, /л = 2
for m divisible by 8...
The equation cp(n) = <jj(»+1) in natural numbers n has been a sub-
subject of interest for several authors (cf...
г„(т) (mod m),
that is, the congruence P = a^^P (mourn) which is, clearly, equivalent
to m\P{av{^—l), whence, since (P,m)=l, we obtain m\a^m)— 1...
, Prove that if m, a, r are natural numbers with (a, r) = I and Z is any infinite
set of terms of the arithmetical progression a-{- hr (k = 1, 2,...), then the progression
contains terms which are products of more than in, different numbers of the set Z...
On the other hand, it is clear that congruence A9) has natu-
natural solutions only in the case where (a,m) =1...
< 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...
Therefore, for every
natural number к | re/5, X is the greatest exponent such that <px \ nsk—X...
But jJ— 1 = 3 is
impossible because p2 is a prime, so p2 — ^ ~ ® ^ valid, whence p2 ~ 7 and conse-
consequently m = 2- 3 ¦ 7 = 42...
Consequently, tbey
are the only moduli m for which the congruences a s= fr(modm) and e s= d{niodm)
imply ac = ^(modm) for any positive integers а,Ъ,с,й...
Thus we conclude that the numbers gn, /",..., p™ divided by у yield
different remainders...
In this way we find 2 = 2, 22 = 4, 23 = 8, 2" = 3, 2s = 6, 2s =12,
2' = 11, 2s = 9, 2" = 5, 210 = 10, 2" = 7, 213 = I(modl3)...
Consequently, all the solu-
solutions of the congruence are numbers of the form 7-J-12A, where й
= 0,1,2,.....
In fact, we have ax — [jKjJ = I 1; therefore in order that
Lx— a0J
% = m it is necessary and sufficient that m <l/(«— «„) < m+1, i.e...
Рог example, representing тг as a simple continued fraction we see that ite
second convergent is ^-; therefore the rational у approximates number те better
than any other rational with a denominator < 7...
Consequently, there
exist numbers Jc and s < 2D such that
C1)
since
for n =1,2,,..,
C1) gives xk+1 = xk+s+1 and, more generally, a>n = x^n for я > Тс...
But Bau a?+l) =1 (since ax is even); therefore number a0 — ax/2 is
divisible by aj + 1 and this results in the equality ao — a1j2 = (o'+ljft,
where ft is an integer...
In fact, if the only thing we know is that 0 < x < 1/10100, then we may conclude
that l/ж > 10100, i.e...
© 2007 eKnigu | ||