Download OBTAINING SQUARES FROM THE PRODUCTS OF NON

OBTAINING SQUARES FROM THE PRODUCTS OF NON-SQUARE INTEGERS
The difference between two neighboring squares n2 and (n+1)2 is equal to 2n+1 for any integer
n=1,2,3,… . Thus the numbers generated by n2-A can never be perfect squares if A<2n. The
question which now arises is “Is it possible to multiply some of these non-square numbers
together to form a perfect square”?’. The answer is yes and it forms the basis of most sieve
factorization methods for large semi-primes. Let us look at the problem a bit more in detail.
First we construct the following table in which we have calculated n2-1 for the first ten integers
starting with n= 2. We also have written down the exponent vector produced by a prime
number factorization of these numbersn
2
3
4
5
6
7
8
9
10
11
n2-1
3
8
15
24
35
48
63
80
99
120
Exponent Vector (n2-1)
[0 1 0 0 0]
[3 0 0 0 0]
[0 1 1 0 0]
[3 1 0 0 0]
[0 0 1 1 0]
[4 1 0 0 0]
[0 2 0 1 0]
[4 0 1 0 0]
[0 2 0 0 1]
[3 1 1 0 0]
Note the exponent vectors represent the powers of the prime products representing a number.
Thus [0 2 0 1 0 ] represents the number 20∙32∙50∙71∙110=63. What is noticed is that none of the
numbers n2-1 have all even elements in their exponent vectors and hence cannot represent
perfect squares by themselves. However the product of several of these n2-1 numbers can lead
to perfect squares if the sum of the exponent elements are all even numbers. A brief inspection
of the above table shows that the product of 24∙35∙63∙120=6350400 produces the exponent
vector [(3+3+0+0),(1+1+0+2),(1+1+0+0),(1+0+1+0),(0+0+1+1)]=[6 4 2 2] . Halving the elements
in this last vector produces the square root of 6350400 , namely, 2520. There are also other
combinations in the table which produce perfect squares. The product 8∙15∙120 =14400 has
the exponent vector [6 2 2 0] so that [3 1 1 0] produces the square root 8∙3∙5=120.
In matrix form we can represent the above collection of exponent vectors as-
01000
3
2
30000 8 3
01100 15 4
31000
24 5
35 00110 6
M=
for n2-1=
and n= 41000 48 7
02010 63 8
40100 80 9
02001 99 10
31100 120
11
Each row represents the exponent vector of one of the n2-1 numbers starting with n=2 and
going through n=11. Thus the 8th row in the matrix represents the number 80=24∙51. With an
appropriate selection of different numbers n2-1 one can form exponent vectors consisting of all
even elements and hence will produce a perfect square. For the above matrix let us look at the
additional exponent vector produced by adding rows 2 to row 4 to row 6 .This produces the
vectorV = 10200 so that we have the perfect square K=210∙32∙50∙70=9216=(96)2
In finding perfect squares one wants to generally deal only with smooth values of n2-1. That is,
numbers where the highest prime power of each n2-1 is kept as low as possible in order to
minimize matrix size. This can be accomplished by just neglecting those generated numbers
where we have vectors with elements beyond a fixed ithprime. The matrix M above is an
example of where no row exceeds five elements. Hence we have a 5th prime cut-off. Should a
number produced by the generating formula violate a given smoothness condition it is removed
from the list.
Other exponent vectors with even elements generated from the above matrix areV=[6,2,0], V=[8,2,2], and V=[14,2,2]
These vectors produce the perfect squares K=576, 57600, and 3686400, respectively. Their
square roots are 24, 240, and 1920.
The number generator used above is just one of an infinite number of other generators one
may use to generate exponent vectors. Perhaps one of the most interesting is the quadratic
generator − = This is recognized as a Diophantine Equation where x=n, y=sqrt(n2-N) and instead of 1 we
replace it by the larger integer N . Using this formula as our number generator, we get a new
table x=n
xa >sqrt(N)
xb
xc
….
y2=x2-N
xa2-N
xb2-N
xc2-N
….
ExponentVector
[x2a x3a x5a x7a …]
[xb2 xb3 xb5 xb7…]
[xc2 xc3 xc5 xc7…]
….
For N=15 and xa=4, xb=5, etc we find the matrix composed of y2=K=x2-N numbers with no more
than 4 elements per row to be1
4
0000
1010
10 5
M=0101 for x2-15= 21 = 6 49 8
0002
1111
210
15
In this case the first and fourth row indicate the perfect squares K=1 and K=49. Adding together
the second , third and fifths row produces another perfect square defined by V=[2 2 2 2]. It
reads K=10∙21∙210=44100 and sqrt(K)=210. You will notice in these last three result that, if we
define X=5∙6∙15=450, we can write things in the Diophantine form(X-sqrt(K))(X+sqrt(K))=const.N
That is(450-210)(450+210)=240∙660=4080∙15
In terms of modular arithmetic these results can be expressed as(X2-K) mod N =0
Thus (240) mod 15 and (660) mod 15 are both equal to zero.
The vector V can be used to represent any positive integer including large ones such as the
composite Mersenne NumberN=211-1=2047=23∙89
Its exponent vector has all zero elements except for a one at position 9 and a one at position
24. That is-
N=[0000000010000000000000010]
An even more compact form occurs for N=10m where V=[m 0 m ]. Thus the googol which is
defined as G=10100 can be represented by the three element exponent vector V=[100 0 100 ].
Instead of using a generating formula for finding the M matrix, we can also construct matrixes
based directly on all the positive integers 1, 2, 3, etc. but keeping only those values which are
sufficiently smooth. Allowing no elements beyond a finite power for 5 we get the following 22 x
3 matrix for all allowed exponent vectors generated between n=2 and 50100
010
200
001
110
300
020
101
210
011
400
M=
120
201
310
002
030
500
220
301
021
410
102
2
3
4
5
6
8
9
10
12
15
16
for n = 18
20
24
25
27
32
36
40
45
48
50
We can use this matrix M to generate a great number of perfect squares although only a very
few of the rows represent perfect squares to begin with. As an example consider the productK=3∙4∙6∙8∙18∙24∙48=11943936 with the corresponding exponent vector V=[14 6 0]
Its root is sqrt(K)=27∙33=128∙27=3456. Another possibility arises from K=27∙366∙48∙253 whose
exponent vector reads V=[16 16 6 ]. Here
K= 44079842304000000 and sqrt(K) = 28∙38∙53 = 209952000
Finally let us demonstrate how one goes about factoring larger semi-primes. Start with the
number N=24139 whose sqrt(N)=155.3673.. . First we construct the M matrix using the
numbers generator n2-N and restricting ourselves to vectors no longer than five elements. This
produces01201
825
158
1782
161
14001
M=15100 for n2-N=2430 and n=163
11400
3750
167
02102
5790
172
Adding together the elements in rows 3, 4 and 5 produces the exponent vector V=[2 8 6 0 2]
which produces the perfect square K=22∙38∙56∙70∙112=49617562500 with sqrt(K)=222750. The
corresponding X=163∙167∙172=4682012. We thus have(4682012-222750)(4682012+222750)=4459262∙4904762=const(24139)
which produces [4682012^2-49617562500] mod 24139=0. Also we can pick out the
components p and q by performing the following operation involving the greatest common
divisorgcd(24139,4459262)=q=239
and gcd(24139,4904762)=p=101
That is we have factored the semi-prime into 24139=101 x 239. This procedure demonstrates
essentially the route taken by all sieve factorization techniques for breaking large semi-primes
N. The present factoring record lies somewhere around 500 digit length for N and requires
months of computer runs with the fastest existing computers.
It should be pointed out that one can factor the present number N=24139 even faster by using
the variation of the Fermat technique which we discussed in a previous article. One needs to
simply run the one line programfor n from 156 to 180 do {n, n^2-24139,ifactor(n^2-24139)}od;
until a value of n is reached where the exponents of the ifactor terms produce an exponent
vector whose elements are all even. This occurs at n=170 and yields the vectorV=[ 0200000020]
Thus K=32∙232=4761 and sqrt(K)=69. So (170-69)(170+69)=N or p=101 and q=239.
U.H.Kurzweg
April 2012