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Sierpinski W. Elementary theory of numbers (Warszawa, 1964)(L)(T)(224s).djvu |
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since atj, ж2,..., жй+1 are different roots of congruence C7), implies that
p\g{xi) for i = 2, 3,..., w-j-l...
If p iB an odd prime, then the congruence a? = l(mod^j) has preci-
precisely two roots, 1 and p— 1 (cf...
If n is a prime, then, of course, every natural number less than n
is relatively prime to n; accordingly for prime n,
A)
<Р(П) = )! — 1...
For every natural number s there exists a natural number
m such that the equation <p(n) = m lias more Hum s different solutions in
natural numbers я...
in any case, p = 2 • 36&+x -{-1, which is impossible since, by к > 1, we have p > 7
and, in virtue of the theorem of jFermat, 3'sl(mod?), whence p = 2-36'г+1 -f 1 =
2-3+1 = 0(mod7), so 7\p...
Thus in order to рготе the theorem we may remove some terms at
the beginning and suppose a> 1...
It is clear that if two numbers are congruent with respect to modu-
modulus m, then they belong to the same exponent with respect to modulus
да; for, if а н= Ь(modm) and for some x formula A9) holds, then bx
= 1 (modm) (since, as we know, the congruence »si(modm) implies
the congruence a? ss ^(modm) for any x = 1,2, ...)•
Tkeokem 9...
of Carmichael (and so absolutely pseudo-prime) it is necessary and suf-
sufficient that X(m) \m—l (cf...
Now we are going to find the least exponents of b that appear
in the numerator and in the denominator of this quotient...
if either q = p 01 q — p-\-2
or q =j)-(-6 and the numbers p,p-\-2,p + 6 weie prime, then, in the
first case, the number p-\-2 = q+2 would be composite, in the second
case, the number p + G = q + ± would be composite, and...
A glance at the proof of the theorem shows that if g is a primitive
root of a prime p, then all the primitive roots of p that belong to sequence
B3) are to Ъе found among the remainders yielded by the division by p
of those terms of tbe sequence
whose exponents are relatively prime to p — 1...
Consequently, number m must be a prod-
product of different prime factors; being greater than 2, the product must contain a prime
odd factor p...
Therefore the sets of nth power residues and
dth power residues for the modulus p coincide...
Thus the problem of solving binomial congruences reduces to that of sol-
solving linear congruences...
But since
p\g2 —1 is impossible (because g is a primitive root of p), p\g2 -r 1 is
valid, i.e...
Therefore in order to prove the theorem it is sufficient to show that
for any natural number Ж and a natural number g > 1 there is at least
one representation A) (conditions B) being satisfied)...
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...
It can be proved that each positive real number has precisely one
representation in this form and that a sufficient and necessary condition
for as to be an irrational number is that lim ~kn = +oo (Sierpinski [3])...
We have thus proved that anylirrational number x0, 0 < жс < 1, may
be expressed in form B6)...
Consequently,
for да = 1 the probability is equal to \, for m = 2 it is \; for да = 3
it is only j?, and so on...
But Bau a?+l) =1 (since ax is even); therefore number a0 — ax/2 is
divisible by aj + 1 and this results in the equality ao — a1j2 = (o'+ljft,
where ft is an integer...
Taking into account the fact that the period (к; 2h) = l/jfe2-j- X has one
term only, we see that the proof of theorem 6 is completed...
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