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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: We have already mentioned Beeger's theorem that there exist infinitely many
even Poulet numbers...
This proves that the congruence g(xt)
= O(modj)) has at least n different roots, which contradicts the assump-
assumption that theorem 13 holds for polynomials of degree n—1...
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 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...
The equa-
equation y(«) = 2'3, however, has four solutions: n = 7, Й, 14, 18, and the equation
<p(n) = 2-3a also has four solutions: n = 19, 27, 38,54...
So tit2-..tB sa a3 i
1 (modr), whence,
in view of (a, r) = 1, by theorem 8, we infer that aPW ™ l(modr)...
of Carmichael (and so absolutely pseudo-prime) it is necessary and suf-
sufficient that X(m) \m—l (cf...
Thus, as we see, the assumption that for any prime divisor of the
number и"—1 number и belongs to an exponent less than n -with Tespect
to modulus p leads to a contradiction...
Hence, in virtue of ad = l(mody), we have r| s= (as)k = l(mody), which
proves that numbers B6) are roots of congruence B5)...
Therefore, for any number x = ga, g1, ...,gp~2, there exists
a number у of the sequence 0,1, 2,...,p—2 such that gy = ai(modjp)...
Therefore the sets of nth power residues and
dth power residues for the modulus p coincide...
This proves that for a given natu-
natural number Ж (with a fixed natural number g > 1) there is at most one
representation A) such that conditions A) are satisfied...
On the other hand, if n is a natural number with the prop-
property 2S-I < » < 2s, then it has, of course, s digits in the scale of 2...
These are: one term
periods, the term being any of the numbers 1, 153, 370, 371, 407; period consisting
of two numbers, either of 136 and 244 or of 919 and 1459; finally, periods consist-
consisting of three numbers, either of 65, 250, 133 or of 160, 217, 252 (see also Iseki [2])...
In one of
them all cu's except a finite number are equal to zero, in the other from
a certain n onwards all cm's are equal to g—X...
We do not know
whether there exist infinitely many natural numbers и > 2 such that the decimal
of number l/та has the period consisting of n— 1 digits...
Equality E) remains valid
if am is replaced by «„,4 — on each side of the equality (since aro+1 > 0)...
This proves that a rational number cannot be expressed as an infinite
simple continued fraction...
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,.,...
and find, the smallest index s for which, say, bs+1 — br and ee+1 = cx; the
representation of Vl> as a simple continued, fraction is then
/I) = (a0; a1;ai: ...,as)...
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 it follows that
for natural numbers 1c and integers * 5= 0 the simple continued fraction
for the square root of the number D = (Qnt-\-lif+ 2Pnt+l has a period
consisting of n + X terms, each of the first и terms being equal to 2&...
If D фаг-\-1, a being a natural
number, then, if t and и are a solution of equation (SI) in natural numbers,
we also have и > 1 because, if % were equal to 1, we would have t2— D =
= —1, whence D = t'+l, contrary to the assumption concerning num-
number D...



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