Remainder factorization of uneven integers. A new insight into this problem


#1

Larger examples such as factoring numbers greater than say Q[8] = 437 need an algorithm, since this is too much to do by hand.
Best shown by an example. Factorization of Q[5] = 35.
Taking the square root of 35 gives 5.91
A reasonable assumption is that the factors are each in the same binary period, N = 2, in this example, which is the smallest case of two prime factors, each having one remainder. in larger example where the prime factors are near to one another, the prime factors would most often be in the same period, or in adjacent periods.
Writing the factors: Q[2, 1] = 2^{2} + 2^{1} + s[1].2^{0} and
Q[2, 2] = 2^{2} + 2^{1} + t[1].2^{0}
multiplying gives;
Q[5] = 2^{5} + 2^{2}.[s[1] + t[1] + 1] + 2^{1}[s[1] + t[1]] + s[1].t[1]
In remainder Factorization, it is essential to write every term in exponents of prime two as above.
Another matter to notice is that every uneven integer that can be factored [primes cannot] is either of the form; [4.k + 1] or [4.k + 3], written [1, 4] or [3, 4].
Q[5] is of the form: [3, 4], since: [35 - 3]/4 = 8, but [35 - 1]/4 does not go.
Since Q[5] is [3, 4], then as written in terms of the unknown remainders, we can subtract three from this equation and see what happens next.
[35 - 3]/4 = [2^[5} + 2^{2}.[s[1] + t[1] + 1] + 2^{1}.[s[1] + t[1] - 1]
+ 2^{0}.[s[1].t[1] - 1]
noting that three was not subtracted from the first term on the right, thirty-two.
The terms on the right in 2^{5} and 2^{2} are divisible by four, no problem. The terms: D[5] = [2^{1}.[s[1] + t[1] - 1] + 2^{0}.[s[1].t[1] - 1]] are divisible by four only for certain selections of values of s[1] and t[1], are are not divisible by four for other selections of values of s[1] and t[1]. Calling the above function D[5] - " D " for divisibility. The subscript five is for the association with Q[5].
Since there are only two unknown remainders in this example, s[1] and t[1], then there are four possible selections;
Decimal s[1] t[1] D[5] Divisible by four ?
0 - 1 - 1 - 6 no
1 - 1 + 1 - 4 yes
2 + 1 - 1 - 4 yes
3 + 1 + 1 2 no
This means that the prime factors of thirty-five are given by
s[1] = - 1 and t[1] = + 1 OR s[1] = + 1 and t[1] = - 1.

The prime factors are therefore: Q[2, 1] = 2^{2} + 2^{1} + [ - 1].2^{0} and
Q[2, 2] = 2^{2} + 2^{1} + [ + 1].2^{0}
OR just commute the minus one with the plus one in, the coefficients of 2^{0}. The factors are [4 + 2 - 1] and [4 + 2 + 1].
The important matter to notice is the difficulty in dividing by four of the terms in 2^{1} and 2^{0}, after having incorporated a subtraction of one or as may be the case three into these terms.

The above is the smallest instance of remainder factorization involving two unknown prime factors.
The smallest example, one of three such examples is the factorization of Q[8] = 437, a [1, 4] integer.
In this case, the iterated equations; Q[8] = 2.Q[7] + r[7]
Q[7] = 2.Q[6] + r[6]

Down to: Q[1] = 2.Q[0] + r[0],
The remainders r[0] … r[6] when counted give N = 8, the number of the binary period.
The successive quotients whether expressed as numbers or in terms of the unknown remainders are given by:
Q[j - 1] = [Q[j] - r[j- 1]]/2

For example Q[8] = 2.Q[7] + r[7]
that is 437 = 2.219 + [ - 1], where r[7] = - 1 and Q[7] = 219.
The remainder in the iterated equations above is given by:

 r[j - 1] = i^{1 + Q[j]}

Rearranging the iterated equation: 219 = [437 - [ - 1]]/2
The quotients expressed numerically are equated to the quotients expressed in terms of the unknown remainders and the divisibility function found, having discovered that the numerical quotient is either [1, 4] or [3, 4] and subtracted one or three from the terms in 2^{1} and 2^{0}.
This example is far to lengthy to do here, but noting that the square root of 437 give the period in which the two prime factors reside.
An easier example is Q[7] = 143. Taking the square root to two decimal places is sufficient to establish the period of the prime factors, assumed to be in the same period in common. This becomes more likely to be true as the period N is increased, whilst the prime factors are near to one another.
[I do not know what a tag is.]