The original checksum array (PPT) is a triple (A, B, C), where A, B, and C have no public factor and meet the requirements of a 2 + B 2 = C 2.
Obviously, there are infinite arrays (ABC multiplication with N). In the following study, if ABC does not have a public factor, write out some original Arrays:
Case :( 3, 4, 5) (5, 12, 13) (8, 15, 17) (7, 24, 25) (20, 21, 29) (9, 40, 41) (12, 35, 37) (11, 60, 61) (, 53) (, 65) (, 65)
We can see that the parity of A and B is different and C is always odd. (This can be proved to be correct by a bit of skill)
In addition:
3² = 5²-4² = (5-4) (5 + 4) = 1 × 9
15² = 17²-8² = (17-8) (17 + 8) = 9 × 25
35² = 37²-12 ² = (37-12) (37 + 12) = 25 × 49
......
What's amazing is that C-B and C + B always have the number of workers, and C-B and C + B have a public factor. Proof below: Assume that there is a public factor and D is the public factor of C-B and C + B, D is also divisible (C + B) + (C-B) = 2C, (C + B)-(c-B) = 2B, so d is divisible by 2C, 2B, But B and C have a public factor, and assume (A, B, C) it is an array of original elements, so that D is equal to 1 or 2, and because D division (C-B) (C + B) = a². A 2 is an odd number, so d = 1, C-B and C + B have a public factor ., Because (C-B) (C + B) = a², the product of C-B and C + B is the number of bytes, only when both of them are values (the product of prime numbers can be intuitively seen), so that C + B = S 2, c-B = T 2, and
C = (S 2 + T 2)/2, B = (S 2-T 2)/2, a = √ (C-B) (C + B) = ST. this gives the sequence array theorem:
Each primitive concatenation array (A, B, C) (A is an odd number, and B is an even number) can be obtained by the following formula: A = ST, B = (S 2-T 2) /2, c = (S 2 + T 2)/2, where S> T> = 1 is an odd number without a public factor.
When T = 1, we can get many examples above.