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Sierpinski W. Элементарная теория чисел (Warszawa, 1964)

Sierpinski W. Elementary theory of numbers (Warszawa, 1964)(L)(T)(224s).djvu

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Date Jan 16, 2005

Cites: If ft < a, the congruence D7) is equivalent to the equation
D8) i =p"(B1+tp°-")!
where t is a suitably chosen integer and the number JD1-{-tp''->' is not
divisible by p (because Dt is not divisible by p)...
Hence, since В is odd, the congruence is solvable only in
the case where D is of the form 4/b+l...
On the other hand, it is easy to verify that each number given by any of
the congruences E9) satisfies congruence E4) (whenever it is true for s0)
and, since for a > 3 any two of these numbers are not congruent with
respect to the modulus 3", we see that they are different roots of con-
congruence E4)...
It follows that
the least natural number n which satisfies the equation <p(ri) = у(та+1)
= <р(я-|-2) is the number 5186...
We show this
by proving that this is the case for the numbers m — 2-б2*, where Tc
= 1,2,..., for instance...
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...
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...
Numhers which belong to a given exponent
Number 7 is also a primitive root of number 10 since 71 = 7, 72 = 9,
73 = 3, 74 = 1 (modlO)...
For j = 0,1,2, ..., s, s+1 we have X(&{2"h+1)) = 2sk, so
putting ms = 2sJc we obtain the desired result...
We are going
to prove that if p is any of those numbers, then number те = Тер is a com-
composite number, whose existence thet heorem asserts...
Our aim is to find numbers rk which belong to the expo-
exponent д with respect to the modulus p...
Number д, as a primitive
root of m, belongs to the exponent <p(m) with respect to the modulus in...
As we know, 8 must be a divisor of number p — 1, whence number
(p— 1)/S is a natural number > 1, and so it has a prime divisor q...
Moreover, the
representation is infinite and has infinitely many digits different from
As an immediate consequence of theorem 3* we note that if a number
as has a non-periodic representation as a decimal in a scale of g, then x is
In fact, we have ax — [jKjJ = I 1; therefore in order that
Lx— a0J
% = m it is necessary and sufficient that m <l/(«— «„) < m+1, i.e...
Therefore formulae C5) imply tbe relations
Since m1 > 1 and 1/osi > 1, these relations give
C6) e,=2ao=2|Y5], ol=o,_1, as=a._2! ..., a,.,...
Hence, by B7) and B8), we obtain the following algorithm for repre-
representing number ]/D ав a simple continued fraction:
We set я» = [Vz>], &i = a0, el=D — al and we find the numbers
an-u bn and en successively using the formulae
On = ¦
Now we boh at the sequence
(ua, o2), (bs,c3), (Ь4,с4), .....
It is also easy to find all natural numbers D for which the represen-
representation of Y3 as a simple continued fraction has a period consisting of
two terms...
Hence, in particular, for t = X, we obtain
^12 = C; 2~7б), V^il = F; 2,2,12),
/180 = A3; 2,2,2,26), ^926= f30; 2,2,2,2,60)...
The representation of a real quadratic irrational as a simple continued
fraction is periodic...
be an infinite sequence of natural numbers among which
there are infinitely many numbers different from 1...
If p = 5Й—1 {To being a natural number) is a prime, then Ъ must
be even (since otherwise p would be an even number > 2, and thus com-
For example, for so = 1, у — 2, we have
2-7s = 34 + l4+24, 2-7 =32 + l2 + 22;
for x = 2, у = 1 we have
2-132 = 34+14+44, 2-13 = 3a+l2 + 42...

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