I have found an unusual method of finding the factors of numbers. I can't be sure it always works, since I don't understand it fully. It involves using two numbers, which are products of at least two factors each, in a comparative way, creating a lookup system having columns of pairs that are remainders of division by sequential numbers. Can anyone explain why it works, or help find all the ways in which it gives answers? Following is a description, and worksheet pics will be attached.
Method of factorization:
If you take a number N, the product of two primes, then find another product close to it (N2), which is the sum of a series of numbers starting at 1, you can find the factors of N using a type of Chinese remainder system.
(It may not be necessary to use a series of numbers for the comparison product, but it is useful at least for finding the factors of a nearby number. In the examples used, the product N2 is a little more than N; being less than N should also work, but I'm not sure. It does not seem to matter if N2 is the product of 2 prime numbers, but also, I am not sure. It also helps to have a difference between N and N2 that is greater than at least one of the factors in N and N2, if you want to get directly to the results. This means a number higher than the square root, as a general minimum; but the high factor of N2 seems sufficient. The difference between N and N2 also is good, if it is larger than the high factor of N2. I have chosen most numbers with N-N1 greater than the high factor of N2)---
Starting with N=7x47=329 at the top of one column, make a list of sequential numbers (starting at 2 or 3), and going up to a target number, to the side of that column, serving as reference and calculation numbers.
To get N2, I multiplied sqrt of 329 times sqrt of 2. This results in 25.65, so round up to the next odd number. The result is 27x14 (the factors of the sum of a series of numbers from 1 through 27). In this case, since I know that 14 is a multiple of 7, I increased the factors to 29x15, so as to not stack the deck. N2 =435. The difference between N and N2 is 106.
For each sequential number (Ns), divide it into N, then take the remainder and list it each to each sequential number, in the column with N at the top of it.
Repeat the process for N2. To be specific, N2 (or N) divided by Ns, minus the whole number portion of the quotient, multiplied by Ns. The result will be 3 columns; the list of sequential numbers, the list of remainders for N, and the list of remainders for N2.
(note: the reverse remainder also is useful, that is, the remainder of Ns left after the remainder process. In this case, subtract the quotient from the whole part of the quotient +1. I won't be using the reverse quotient much in this explanation, if at all.
Now, take the pairs of results from the Ns lines for Ns of factors of N2. For lines 29 and 15, the pairs are 10,0 and 14,0. Since N2 factors are not prime, we can also look at lines for Ns 3, 5, 87, and 145.
There are multiple ways to get results here. Ns lines of a factor (or other referenced number) minus 1 give results for factors of N, as well as the factor lines.
First way: Lines 28 and 14 (N2 factors-1) have pairs of 21,5 and 7,1. 7 is one of the factors of N, and 21 is a multiple.
Second way: take the pairs for lines 14 and 29 and look at them as whole numbers (100 and 140), then go to the Ns lines for 100 and 140 (or calculate them as needed for remainder pairs). For Ns 100 and 140 the pairs are 29,35 and 49,15. One number of each pair is a factor of N2; the other numbers are multiples of 7. since we don't get the luxury of looking at large numbers and knowing their factors, go to the Ns lines for 49 and 35. Being multiples of 7 the remainders for the N column will refer you to 35 and 14; the N column of those rows are 14 and 7. They both devolve (divide) down to row 7, which has a zero in the N column, meaning it is a factor of N. (the numbers in the N2 columns for these Ns rows are also instructive. For Ns 49 and 35, the N2 column numbers are 15 and 43. 15 is a factor of N2, so we ignore it (or use it as verification that we are on the right path). Ns 43 pair is 28,5. Row 28 is pair 21,15. Rows 21 and 15 have pairs 14,15, and 14,0; as before, going to row 14 has an N value of 7.
This column system has the property that every row's pairs, each number in the pair can be added to the Ns of the row and get a useful number. Adding the row number once often give a multiple of a factor of that row. In some cases, the Ns number has to be added multiple times. This in itself is not all that impressive; any number can be added multiple times to another number, and eventually get the multiple of another number.
Also, the validity of using the pairs of numbers as whole numbers can be shown by the many ways in which the following answers or references are either recursive or give a correct factor answer. In this example using 329 and 435 for N and N2, the difference is 106. The pair for Ns 106 is 11,11. Row Ns 11 has pair 10,6 (106). Row Ns 53 also has pair 11, 11. That refers back to 106, which is 2x53. Starting almost anywhere (maybe anywhere) eventually gets back to the factors of N and N2. Using Ns of the factors of N2 as a starting point seems to be the fastest way.
It has also been found that taking the pair numbers, and adding and subtracting from the Ns row number, will provide factors or other references.
Example 2.
Example 3.
N=305713, N2=783x392=306936, difference =1223
row 783 and 392 are supplanted for brevity to 782 and 391 (Ns-1)
row 391 has pair of -49,1 (using a negative quotient in this case)
491 has pair of 311, 61
solving for 311 has no remainder in N column (=0), so is a factor of 305713
Example 3a.
N=305713, N2=797x399, difference =12290
row 399 is supplanted by 398, row 398 has pair of 49,1
491 has pair of 311, 61. 311 is factor of N.
rows 399 and 398 have N column values of 79 and 49
using rows 79,49 and 78,48 gives vertically configured values in N column of 62,2 and 31,1. May be a coincidence, may be a valid answer.
note: I did find one answer where the Ns number was 457, and the pair of number pairs generated added up to a factor of N. I didn't write that one down due to distraction, and would have a hard time finding where it was.
I am including photos of worksheets. As an explanation of some of the row configurations, where you see a number followed by a comma in a row, and one of the numbers is negative, I listed both the remainder and the negative remainder. In some cases, a third number is present, which is the remainder+the Ns number, and will be marked as a +number. For pages with multiple columns (more than 2), the first column is for N, and the other columns are for multiple N2 variations. It was confusing to work with, and not a very useful shortcut.
The idea is that for each N, an N2 can be identified, then the factors of N2 (and maybe the difference of N-N2) can be used to identify the Ns rows getting most directly to the factors and factor multiples of N. It seems most often that the Ns-1 numbers are usually the most direct route.
