Math Newsletter number 13;Wednesday, October 20, 2010.
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Special factoring of positive integers

Suppose z is the product of two integers, x and y.
x y = z
Suppose x = A + C and y = A + B
(A + C)(A + B) = z
A**2 + A C + A B + B C = z
A (A + B + C) + B C = z

Illustration:

7*11 = 77
A = 1; B = 10; C = 6; A+B+C = 17: 1*17 + 10*6 = 77
A = 2; B = 9; C = 5; A+B+C = 16: 2*16 + 9*5 = 77
A = 3; B = 8; C = 4; A+B+C = 15: 3*15 + 8*4 = 77
A = 4; B = 7; C = 3; A+B+C = 14: 4*14 + 7*3 = 77
A = 5; B = 6; C = 2; A+B+C = 13: 5*13 + 6*2 = 77
A = 6; B = 5; C = 1: A+B+C = 12: 6*12 + 5*1 = 77
A = 7; B = 4; C = 0: A+B+C = 11: 7*11 + 4*0 = 77
A = 8; B = 3; C = -1: A+B+C = 10: 8*10 - 3*1 = 77
A = 9; B = 2; C = -2: A+B+C = 9: 9*9 - 2*2 = 77
A = 10; B = 1; C = -3: A+B+B = 8: 10*8 - 1*3 = 77
A = 11; B = 0; C = -4: A+B+C = 7: 11*7 - 0*4 = 77

If we notice that z is of form A D + B C, where
D = A + B + C,
then
z = (A + C) * (A + B).

A highly composite number slightly less than 77, is 72.
77 - 5 = A * (A + 1 + 5)
72 = A * (A + 6)
72 = 6 * 12
A = 6
B = 5
C = 1
x = A + C = 6 + 1 = 7
y = A + B = 6 + 5 = 11

For each number b, and odd positive integer m > 1,
((b**2)**m - 1)/(b**2 - 1)
= (b**2)**(m-1) + (b**2)**(m-2) + .... + b**2 + 1

Also

((b**2)**m - 1)/(b**2 - 1)
= ((b**m)**2 - 1)/(b**2 - 1)
= [(b**m -1)(b**m + 1)]/[(b-1)(b+1)]
= [(b**m-1)/(b-1)][(b**m+1)/(b+1)]
=(b**(m-1) + b**(m-2) + ...+b + 1)
(b**(m-1) - b**(m-2) + ...+b - 1)

Illustration:

(4**3 - 1)/(2**2 - 1) = [(2**3-1)/(2-1)][(2**3+1)/(2+1)]
4**2 + 4 + 1 = 7 * 3 = (2**2 + 2 + 1) * (2**2 - 2 + 1)

21 = 7 * 3
21 + 4**3 + 4**4 = 341;
7 + 2**3 + 2**4 = 31;
2**4 - 2**3 + 3 = 11
341 = 31 * 11

341 + 4**5 + 4**6 = 5120+341 = 5461;
31 + 2**5 + 2**6 = 127
2**6 - 2**5 + 11 = 43
127 * 43 = (130 - 3)*(40 + 3)
= 130 * 40 - 3 * 40 + 3 * 130 - 9
= 5200 - 120 + 390 - 9
= 5200 + 390 - 120 - 9
= 5590 - 120 - 9
= 5470 - 9
= 5461

Numbers of the form 1 + b**2 + b**4 + b**6 + ... b**(2k),
where k > 1, is even, are easily factored.

Let z be odd, and be the product of two integers, x and y.
x y = z and x and y are odd. Let y be >= x.
Define t = (y + x)/2. Define s = (y - x)/2
(t - s) = [(y+x)/2] - [(y-x)/2] = 2x/2 = x
(t + s) = [(y+x)/2] + [(y-x)/2] = 2y/2 = y

This is just another viewing of the basic identity,
(t**2 - s**2) = (t-s)(t+s).

If we notice that z is the difference of two squares,
then we can immediately factor z.

91 = 100 - 9 = 10**2 - 3**2 = (10-3)(10+3) = 7 * 13
221 = 15**2 - 2**2 = (15 -2)(15+2)=13*17
341 = 21**2 - 10**2 =(21-10)(21+10)=11*31
899 = 30**2 - 1**2 = (30-1)(30+1) = 29*31
1591 = 40**2 - 3**2 = (40-3)(40+3) = 37*43

B = 2 m n
A = 2 m**2
C = n**2

(A p**2 + C)**2 - (B p)**2
= (A**2 p**4 + [2 A C - B**2] p**2 + C**2
= 4 m**4 p**4 + [4 m**2 n**2 - 4 m**2 n**2] p**2 + n**4
= 4 (m p)**4 + n**4

Reversing this,

If we see that z is of the form,

4 (m p)**4 + n**4,

then

z factors as

(2 m**2 p**2 - 2 m n p + n**2)
(2 m**2 p**2 + 2 m n p + n**2)

m=1; p=1; n=3.
4 * 1**4 + 3**4
= (2*1*1 - 2*1*3*1 + 3**2)(2*1*1 + 2*1*3*1 + 3**2)

4 + 81 = (2-6+9)(2+6+9)

85 = 5 * 17

629 = 5**4 + 4
m = 1; p = 1; n = 5
x = 2*1*1 - 2*1*5*1 + 5**2 = 2 - 10 + 25 = 17
y = 2*1*1 + 2*1*5*1 + 5**2 = 2 + 10 + 25 = 37

37 * 17 = (40 - 3)(20 - 3) = 40*20 - 3*40 -3*20 + 9
= 800 - 120 - 60 + 9
= 800 - 180 + 9
= 629