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Sierpinski W. Elementary theory of numbers (Warszawa, 1964)(L)(T)(224s).djvu |
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case m = i the following theorem holds: if f(x) is a polynomial of degree
n with integral coefficients such that the leading coefficient is relatively prime
to in, then the congruence f(x) = O(mod)») has at most n different roots
(cf...
They cannot both be divisible
by 4, since if they were, number 2г would be divisible by 4, and so 2 | s,
which is impossible...
This shows that in each of tbe rp{l) columns, the
terms of which are relatively prime to Z, there are <p(m) numbers rela-
relatively prime to m...
It is worth observing that conjecture H implies the existence of infinitely
many primes x, у such that y(re): cp(y) = a:b Jot a given pair of natural numbers
a, b (cf...
Formulae (8) and C7)
of Chapter IT give together the formula
(9)
cp[n) =
d\n
valid for all natural numbers n...
For this purpose it is
sufficient to prove that
1° any term of sequence A6) is a natural number relatively prime
to m and less than m,
2° the elements of sequence A6) are different...
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...
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...
We prove that if from the set of primes we remove those primes
which belong to triplets of the first or of the second category, then
infinitely many primes still remain in the set...
Since p— ljm and (p— 1,p) = 1,
we have p— 112; so, in view of p > 3 (since p is an odd prime), we conclude that
p — 3 and consequently m = 2-3 = 6...
Therefore the sets of nth power residues and
dth power residues for the modulus p coincide...
Consequently, all the solu-
solutions of the congruence are numbers of the form 7-J-12A, where й
= 0,1,2,.....
Hence 3% = a—ea
and so 'ffajjl < |a>| + e0 < W + lffl —!; whence \x^ «? (И + |?| —l)/lffl- H
{\x\+\g[-l)l\g\ > Щ, then |al + |ff|-l > \g\ 'A, i.e...
Then ж = rn-\-ljgn, and so, by
(li), ж is the quotient of an integer by a power of пшпЬет д...
Infinite fractions
273
about cn's (n = 1,2,...) which remains true is that they satisfy the
inequalities 0 < cn < g and that they are integers...
Then the en's satisfy condition A1), whence
it follows that infinite series A3) is convergent and its sum x is a real
number...
A more difficult task is to calculate the probability of the event
that the second quotient is equal to a given natural number m...
is periodic Moreover, the period of the sequence in
pure and, if it consists of s terms ax, а„ ...,а„ then s <2D, a, = 2[/D]
and the sequence аг, аг,..., ад_г is symmetries
§ 4...
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...
For example, for Ь = 1 and n = 1, 2, 3, 4, 5, 6 we find for
< = 0,1,2, ..., respectively (cf...
Since, by the property of the numbers
of sequence G) proved above, terms at different places are different, we
infer that the numbers of G) are (in a certain order) equal to the num-
numbers 1,2, ...,{p —1)/2...
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