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Sierpinski W. Elementary theory of numbers (Warszawa, 1964)(L)(T)(224s).djvu |
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remainder r with 0 < r < S, then » = hB + r, whence 2"— 1 = 2M2r- 1...
In
view of the remark in the corollary to theorem 14, the solution of congruence
D6) reduces to the solution of the congruences
D7) »2 = D(modj?a),
where p is a prime and a is a natural number...
И i-j-г is not divisible by p, then p" | t—z, that is, t = г(шойр"), if t — z
is not divisible by p, then p -\-«, whence * s= — «(modj)°)...
The number of the natural numbers r < I for which (r,l) =1
is oi course <p(l), this being the number of the columns in which all the
I
230
CHAPTER VI...
This suggests
a conjecture that there are no odd numbers n, except for the pairs of twin
primes »,»+2 for which the equality <p(n+2) = p(») + 2 holds...
It is easy to verify that among the integers x such that 0 < x < 20 only numbers V
anu 13 satisfy the congruence...
И s 12" — 1, then 29 = l(modg) and so, by theorem 9,
S I p, where д denotes the exponent to which number 2 belongs with
respect to modulus q...
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...
Since the number of them is cp{m), this being equal to the number of the
numbers relatively prime to m which appear in the sequence 1,-2, ...,m,
then for any integer a relatively prime to m there exists precisely one
number у in the sequence 0,1,2,..., q>(m) — 1 such that g" = a;(modm)...
Moreover, the
representation is infinite and has infinitely many digits different from
g—1...
We do not know
whether there exist infinitely many natural numbers и > 2 such that the decimal
of number l/та has the period consisting of n— 1 digits...
He for-
formulated the conjecture that the number whose decimal is obtained by
writing 0 for the integers and the consecutive prime numbers (instead of
consecutive natural numbers) to the right of the decimal point, i.e...
№>w we are going to
consider slightly moie general continued fractions of the form
A)
where и is a given natural number, a0 a real number and ах,аг, ..., an
positive numbers...
On the other hand, if for some natural numbers a0 and at Ф 2a0
number D of D1) is natural, then, since 2a0a1+l is odd, number a\+l
(as a divisor of it) must also be odd; so number ax is even and, since num-
number D of D1) is an integer and, consequently, —\-^ 1 =
——^— is an integer, number «i + l divides number (aa—a1j2Ja1...
Continued fractions of quadratic irrationals
303
sisting of n + 1 terms, each of the first и terms being equal to 1 (cf...
In fact, if the only thing we know is that 0 < x < 1/10100, then we may conclude
that l/ж > 10100, i.e...
Now, subtracting the last equality from the
equality P—JDu2 = 1, we obtain
D6) t(u' — t)= u(t'—Du)...
We note that if the coatimied fraction F2) has a well-defined value,
then it may happen that some of its convergents do not have this property...
is defined by induction and its terms are non-negative inte-
integers such, that ап<Ъп, as well as the sequence n>x, <ot, .....
p—1 ff—1
ЖишЬег • is odd if and only if each, of the numbers p
2 2
and 2 is of the form 44 + 3; hence equality V may be expressed by saying
If two different odd primes p and q are of the form iJt + 3, then
: least one of them is of the form 4& + 1, then I —I =1 — 1...
Speziali [1] have tabulated the representations in the form Ьх%-\-у*
of primes of the form 67c+l which are less than 10000...
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