OBTAINING SQUARES FROM THE PRODUCTS OF NON-SQUARE INTEGERS The difference between two neighboring squares n 2 and (n+1) 2 is equal to 2n+1 for any integer n=1,2,3,. Thus the numbers generated by n 2 -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 n 2-1 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 numbers- n n 2-1 Exponent Vector (n 2-1) 2 3 [01000] 3 8 [30000] 4 15 [01100] 5 24 [31000] 6 35 [00110] 7 48 [41000] 8 63 [02010] 9 80 [40100] 10 99 [02001] 11 120 [31100] Note the exponent vectors represent the powers of the prime products representing a number. Thus [0 2 0 1 0 ] represents the number 2 0 3 2 5 0 7 1 11 0 =63. What is noticed is that none of the numbers n 2-1 have all even elements in their exponent vectors and hence cannot represent perfect squares by themselves. However the product of several of these n 2-1 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 30000 01100 31000 M= 00110for n 2-1= 41000 02010 40100 02001 31100 3 8 15 24 35 48 63 80 99 120 2 3 4 5 6 7 8 9 and n= 10 11 Each row represents the exponent vector of one of the n 2-1 numbers starting with n=2 and going through n=11. Thus the 8th row in the matrix represents the number 80=2 4 5 1. With an appropriate selection of different numbers n 2-1 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 vector- V = 10200 so that we have the perfect square K=2 10 3 2 5 0 7 0 =9216=(96) 2 In finding perfect squares one wants to generally deal only with smooth values of n 2-1. That is, numbers where the highest prime power of each n 2-1 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 5 th 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 are- V=[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(n 2 -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 y 2 =x 2 -N ExponentVector x a >sqrt(n) x 2 a -N [x 2a x 3a x 5a x 7a ] x b x 2 b -N [x b2 x b3 x b5 x b7 ] x c x 2 c -N [x c2 x c3 x c5 x c7 ]... For N=15 and x a =4, x b =5, etc we find the matrix composed of y 2 =K=x 2 -N numbers with no more than 4 elements per row to be- 0000 1010 1 10 4 5 M= 0101 0002 1111 for x 2-15= 21 49 210 = 6 8 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- That is- (X-sqrt(K))(X+sqrt(K))=const.N (450-210)(450+210)=240 660=4080 15 In terms of modular arithmetic these results can be expressed as- (X 2 -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 Number- N=2 11-1=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=10 m where V=[m 0 m ]. Thus the googol which is defined as G=10 100 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 50-100 010 200 001 110 300 020 101 210 011 400 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 product- K=3 4 6 8 18 24 48=11943936 with the corresponding exponent vector V=[14 6 0] Its root is sqrt(k)=2 7 3 3 =128 27=3456. Another possibility arises from K=27 36 6 48 25 3 whose exponent vector reads V=[16 16 6 ]. Here K= 44079842304000000 and sqrt(k) = 2 8 3 8 5 3 = 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 n 2 -N and restricting ourselves to vectors no longer than five elements. This produces- 01201 14001 825 1782 158 161 M= 15100 11400 02102 for n 2 -N= 2430 3750 5790 and n= 163 167 172 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=2 2 3 8 5 6 7 0 11 2 =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 divisor- gcd(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 program- for 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 vector- V=[ 0200000020] Thus K=3 2 23 2 =4761 and sqrt(k)=69. So (170-69)(170+69)=N or p=101 and q=239.
U.H.Kurzweg April 2012