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Sierpinski W. Elementary theory of numbers (Warszawa, 1964)(L)(T)(224s).djvu |
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we conclude that a necessary condition for the solvability of congruence
D7) (with В not divisible by p) is that D should be a quadratic residue
for the modulus p...
In fact, the congruence implies that (a, n) =1, and so, if it is valid
for any a < n, then <p(n) = n—1, and consequently n is a prime...
We do not know whether there exist infinitely many natural num-
numbers which are not of the form n—р(та) where и is a natural number...
Moreover, since the num-
ber of numbers in the sequence 1, 2,..., n is equal to та, we obtain the
formula
the
But, clearly, if <J4 runs all over the set of natural divisors of number n,
then — runs all over the same set of natural divisors of n...
In virtue of exercise 3, the number щ is a divisor of a number
m whose digits (in the scale of ten) are equal to 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...
Thus we conclude that q
— 21cp + l, where Tc is a natural number, and so we see that each divisor of the num-
number d is of the form 2&J>+1...
But (я,p) = 1,
and so, by the theorem of Eermat, p \ inP~1—X, whence, by theorem 9,
n\p — 1, i.e...
As we learned in § 6, there exists an integer Ь such
that 0 < A <2>—2 and a = ^(modp) which, in...
We divide the set of sequeuces into clasees by saying that two sequences belong to
the same class if and only if the terms of one are equal to the corresponding terms
of the other...
It is worth noticing that the following expansion into an infinite
product is valid:
Some particular cases of this expansion (for Ъ = 2, b = 3 and some
others) were given by 6...
It can Ъе proved
that if the nimber s of the terms of the period is even, then number
Is is equal to the first index Ъ for which bk+1 = bk; if в is odd, ihen l(s — 1)
is the first index Ъ for which ek+1 = ck p)...
Continued fractions
Now we find all natural numbers D for which tbe representation of Yb
as a simple continued fraction lias a period consisting of one term only...
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...
Accordingly we assume that t and и are a solution of the equation
хг—г1>у* = 1 in natural numbers...
An argument similar to that used in the previous case shows that
number D5) is the (Ъ — l)-th convergent of the simple continued fraction
fqr number \/jD and that It = an, where s is the number of the terms
of the (least) period of the continued fraction for vD and n is a natural
number...
Ifa=l, then,
IB)
by (S), for every integer D not divisible hyj> we have j — I = 1, contrary to the assump-
(D)
tion that j—> is not identically equal to 1 (Z> not being divisible by p)...
+ Д(р_1)B is
exactly the number of the remainders obtained by dividing the numbers
of F) by p, successively...
But (since 2 is odd),
in virtue of II and IV we have
/2o\
which, combined with the formula proved above for I — I, implies the
equality
valid for any odd p and q...
Prom theorem 10 of Chapter V it follows that
any prime of the form 67s+1 has exactly one representation in the form
Зж2+2/2, where x and у are natural numbers...
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