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Sierpinski W. Elementary theory of numbers (Warszawa, 1964)(L)(T)(224s).djvu |
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« such that n | a" —1 holds if (a, n) = 1 are called Garmichael numbers...
However, we do not know whether there exist infinitely many Mersenne numbers
which are super-Poulet numbers...
In fact, by theorem 5a, for
every integer x we have
ж" = х{тяй.р), да" =ajs(mody), and so on...
Since В
is odd, number z must also be odd, whence, since the square of an odd
Elementary theory of numbers 15
I
22fi
СНДРТЕК V...
Let m be a natural number, I a natural number relatively prime
to in, and r an arbitrary integer...
If ян, m%, ..., щс are natural numbers any two of which
are relatively prime, then
low let та be a natural number >1 and n = !Й1
Й'' its fac-
factorization into prime factors...
As has been verified, all the solutions of the equation in natural
numbers та < 10000 are the numbers n = 1, 3, 15, 104, 164, 194, 256,
496, 584, 975, 2204, 2625, 2834, 3255, 3705, 5186, 5187...
Tbe proof follows from the remark that if n > 1, Ь < n and (», b) = 1, then
l' °a the Other hand> И (п' Ъ) > J' then [~T = °' Therefore the
hjjj [T1
right-hiuid side of the formula is equal to tbe number of natural numbers < » rela-
relatively prime to m, -which, for n > 1, is the value of <p{n)...
Hence <p{(p- Щ) = (p— l)tp(k) (this follows at once from theorem
3 — in fact, if m is a natural number such that any prime divisor of it is a divisor of
a natural number h, then tp(mh) = т/р{Ь))...
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...
Then, in virtue
of A4), we have art — ar^mo&m), and so m | afc—r^) and, since [a, m)
= 1, we have in | rt—rf, which is impossible because rt and Г], as two dif-
different terms of sequence A3), (since * Ф$) are different natural num-
numbers <от...
Thus in order to рготе the theorem we may remove some terms at
the beginning and suppose a> 1...
those for -which d = 6k, where к
n
is a natural number such that 8k \ те, that is ~k \ —...
We call a sequence p,
p+2, p + 6 whose elements ате all primes a triplet of the first category
and a sequence jj,jp + 4, y+6, whose elements are all prime numbers,
a triplet of the second category...
;p — X j (Л—ZI*, which, in virtue of the relations n = dm, p — l = ds,
gives s | Gc — Z)m; so, since (to,s)=1, s | 7c—Z, which is impossible
because 7c and I are two different numbers of the sequence 1,2,...,«...
But, by theorem 11 (with d =
[%,p — 1) =2), there are only \[p—-1) quadratic residues in the se-
sequence 1,2, ...,p — 1...
it can be rewritten
in form A), where the numbers en (n = 0,1, ..., m) are integers which
satisfy inequalities B)...
Prove that for every natural number m there exists a prime -whose represen-
representation as a decimal in the scale of 2 is such that the last digit is 1 and the preced-
preceding m digits are equal to zero...
These are: periods consisting of
one number, which can be any of the numbers 1, 1634, 8208, 9474; a period con-
consisting of the numbers 2178, 6514; a period consisting of seven numbers 13139, 6725,
4338, 4514, 1138, 4179, 9219 (see also Chikawa, Iseki, Kusakabe and Shiba-
mnra [1])...
Thus we obtain an infinite sequence xn (n —
1,2,...) defined by the conditions
(8) ж2 = а—[я], х„+1 = gxn-
These formulae imply
(9) О«да„<
for
«=1,2,.....
In order to рготе the
converse it is sufficient to show that if a sequence of digits сис^,.....
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