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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: Thus we see
that conjecture H implies the existence of infinitely many absolutely
pseudoprime numbers...
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...
However, we do not know whether there exist infinitely many Mersenne numbers
which are super-Poulet numbers...
But since, as we have assumed, В = 8fc + l, the congruence
г* = i)(mod23) is solvable; hence, by induction we see that (for D = SJs +1)
congruence E4) is solvable for the natural numbers a > 3...
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...
Therefore /3=1, whence
p =2-53i:+l which is impossible since the number 5~k = (a*J is con-
congruent to 1 with respect to the modulus 3, whence 3 | p, so p =3 and
this is clearly false...
(It can be proved that the numbers 10, 26, 34 and 50 are not of this
form.) We do not know whether every odd number is of this form...
Thus we see that the equation if (n) = 2-36t+I, where 7; is
a natural number, has precisely two solutions, n = 36J:+I and n = 2-36i+2...
Using the formula
i
7Г»'
proved in Chapter IV, § 10, one can prove that the ratio of number ]?<p{n)
to number 3<г2/тт2 tends to 1, as x increases...
This proves that the elements of sequence A6) and those of sequence
A3) are identical apart from the order...
On the other hand, it is clear that congruence A9) has natu-
natural solutions only in the case where (a,m) =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...
If q = 3, then 2-1 =¦ 0(mod9), whence 22p = I(mod9) and, by theorem 9,
I
248
СЯАРТЕВ VI...
Fw any natural number n th?re exist infinitely many
prime numbers of the form nfc + 1, where к is a nai/nral number...
We have thus shown that for every natural пшпЬет те > 1 there exists
at least one prime пшпЬет of the form nk + 1, where 7c is a natural num-
number, whence, as we learned above, theorem lla foHows^
As an application of theorem lla we give a proof of the following
theorem of A...
Hence, in virtue of ad = l(mody), we have r| s= (as)k = l(mody), which
proves that numbers B6) are roots of congruence B5)...
But, as we know, since (b,m) =1, the congruence f = i(modm) has
precisely one root among the numbers 0,1, ...,<p[m) — 1, and this is
indb; we conclude that inda = indi...
Thus the congruence turns into the congruence
ox = 11 (mod 12), which is satisfied only for я = 7, provided the as's
are taken out of the sequence 0,1, ...,11...
Proof, This follows immediately from the fact that for every natural number
n there exists a natural number m > n such that f(m) = я...
It is easy to prove that this condition
is equivalent to saying that ж is a rational number equal to an irreducible
fraction whose denominator is a product of primes each of which is a divi-
divisor of g...
Here are the first 30 of the quotients:
3, 7, 16, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, 1, 84, 2, 1,
1, 15, 3, 13, 1, i...
For natural numbers n > 1 we write
B7) bn+1 — ancn~~bn, cn+1 = en_1 — alen
We are going to prove that for n > 1 the equality
B8) ...



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