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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...
Thus we have shown that to each pair («, »), where « is a root of
congruence D1) and v is a root of congruence D2), there corresponds a root
of congruence D0)...
In fact, Euler was tbe first to investigate this function and its properties
in the year 1760...
We do not know
whether there exist composite natural numbers n for which у (n) | n— 1 (x)...
Hence ж4 = pl1+1pl2+I...¦
a-.-Pe and consequently
Hence, looking at the formula for pt—l and recalling the definition of
m, we see that <p{щ) — mfor t = l, 2,...,«...
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...
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...
Congruence B5),
however, is of <5th degree and satisfies the conditions of Lagrange's the-
theorem (theorem 13, § 8, Chapter T), so it cannot have any other solu-
solution than that given by the 5 numbers B6)...
Thus we conclude that the numbers gn, /",..., p™ divided by у yield
different remainders...
Not all prime
divisors of Ж are of the form бй + l since, if they were, their product
would be of this form, which is trivially untrue since Ж is not of this
§ 7, An nth power residue
259
form...
Therefore in the tables of indices the values of inds; are given only
for natural numbers % less than the modulus (and relatively prime
to it)...
Thus the problem of solving binomial congruences reduces to that of sol-
solving linear congruences...
IE each number of the sequence
0, 1, 2, ..., </-l
C)
is denoted by a special symbol, the symbols are called the digits
formula A) can be rewritten in the form
and
There yn is the digit which denotes the number cn...
Let n denote a natural number and let n= oo + fti^ + .-.+afc—iJ/* he the repre-
representation of n as a decimal in the scale of g...
In order to prove that the representation of Ж in form A), condi-
conditions G) being satisfied, is unique, it is sufficient to note that Ж divided
by \g\ leaves the remainder e0, (Ж-соIд divided by \g\ leaves the remain-
remainder Й! and so on...
Moreover, the
representation is infinite and has infinitely many digits different from
g—1...
Thus we come to the conclusion that the denominators of the
rational numbers »0,a!j,a;2,.....
Since, in view of theorem 3, the sequence %, aa must be symmetric, we
have ax = a2 and, moreover, аг Ф 2a0 since otherwise the period of the
simple continued fraction for YT> would consist of one term at...
Therefore in what follows we may assume that t and n are a solu-
solution of E1) in natural numbers with и > 1...
Here are the solutions in the least natural numbers of the equation
x^—Dy* =1 for D < 40...
+ Д(р_1)B is
exactly the number of the remainders obtained by dividing the numbers
of F) by p, successively...
Hence
¦PI
By A3), these two formulae show that the formula
is valid for any two different odd primes p and q...
To calculate I—I we look at equalities B1) and
find the number m of the pairs Pi_x and, eJPi in which both P^_i and BiPt
are of the form 44+3...
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