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Sierpinski W. Elementary theory of numbers (Warszawa, 1964)(L)(T)(224s).djvu |
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even natural number such that»12" + 2 and да — 11 2n + 1, then the number ro = 2" + 2
satisfies the relations m[2"l + 2 and m— l|2m+l...
Hence, since В is odd, the congruence is solvable only in
the case where D is of the form 4/b+l...
We have thus proved that any root г of congruence E4) must satis-
satisfy one of the following four congruences:
E9)
я = — «0(mod2"),
« =»0(mod2a),
This shows that the number of the roots cannot be greater than four...
Applying the formula, just proved, for m*
= <$, i =1,2, ...,1c, we obtain the formula
But since, by theorem 1, <p{$) = g?-1^—1) holds for * = 1, 2,..., fc,
the following theorem 3 is valid:
Theokem 3...
The equation q>(n+3) =
<p(n), however, has only two solutions, n = 3 and n — 5, for n < 10000...
It follows from F) that if <p(n) = 2-51*, where
h is a natural number, then n must have precisely one odd prime divisor...
We conjecture that for every natural number s > 1 there exist infi-
infinitely many natural numbers m such that the equation cp (я) = то has
precisely s solutions in natural numbers n...
(The argument is that
if jo is a prime anuyl», thenjj— l\(p{n) = m, whence it follows that p— 1 is a natu-
natural power of 2, and so f = F^)...
It is easy to prove that if n — g?1 ??••¦?? w tbe factorization of
the number n into prime factors, then
We also have
d\n
g prime
§ 3...
It follows immediately from theorem 8 that for any natural number
in there exists the least natural number X (m) such that in [ <j*'m' — 1 for
(a,ro) =1A)...
For every natural number s there exists a natural number ms such
that the equation X (n) = ms has more than s solutions in natural num-
numbers ii...
Prove that, if a, b, n are natural numbers a > b, и > 1, then every prime
divisor of number an+bn is of the form 2nh+1, where i is an integer, or is a divisor
of number «ni-{-6ni, where ^ is the quotient obtained by dividing the number и
by an odd number greater than 1...
But by B1) we have
[J
B2)
because, as we know, [i(d) =0 whenever d is divisible by the square of
a natural number >1...
Chapter У, § 7):
For any natural number Ь > 2 there exist infinitely пъану composite
natural numbers n such (hat the relation n | ап~к~1 holds for any integer a
with (a, ii) = 1...
According to the definition
of numbers B6), for & = 1, 3,..., S the relation rk = a!'(vaoAp) holds...
As is easy to notice, numbers m which have property P are precisely those square-
free integers m for which A(m)|ro, where A(m) is the minimum universal exponent
with respect to the modulus m (of...
It can be proved that if и is a prime and m a natural number > 1,
then in order that every integer be an wth power residue foT the modulus
m it is necessary and sufficient that m be a product of different primes,
none of the form «7c+1 (where 7c is a natural пшпЬет) (cf...
In virtue of A) and B), we also have
whence mlogg' < logJV" < (•m4-l)log(? and therefore
logs
:«+i,
which proves
F)
Formulae F) and E) show that if Ж is represented as A) and condi-
conditions B) are satisfied, then the numbers иг- and cn (» = 0,1,..., m)
are uniquely defined by number N...
Prove that for every natural number m there exists a prime -whose represen-
representation as a decimal in the scale of 2 is such that the last digit is 1 and the preced-
preceding m digits are equal to zero...
For example, if n = 3 we have the sequence 3, 9, 81, 65, 61, 37,
58, ..., 16, 37, ...; if n — 5, we have the sequence 5, 25, 29, 85, 89, 145, ...,
58, 89, ...; if я = 7, we have the sequence 7, 49, 97, 130, 10, 1, 1, 1, .....
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