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Old 2015-07-13, 12:24   #4
R.D. Silverman
 
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Nov 2003

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Quote:
Originally Posted by irina View Post
For prime number A, there is only one value B, such that what А + В2 = С2
В = (А-1)/2
С = (А+1)/2
А = С2 – В2 = (С-В)*(С+В)
С – В = 1
If the number of semiprime A = k1 * k2, then there are at least two values, such that А + В2 = С2
1. 1) А = С2 – В2 = (С-В)*(С+В);
k1 = C-B; k2= C + B
B2 = (n + trunc (sqrt (A))2 – A;
n – natural number [1; +∞);
C = n + trunc (sqrt (A))
2. 2) А = С2 – В2 = (С-В)*(С+В); С – В = 1
В = (А-1)/2 (B- maximum)
С = (А+1)/2

Example, А = 21
B1 = (А-1)/2 = (21-1)/2 = 10;
С = (А+1)/2 = (21+1)/2 = 11
21 + 102 = 112
If semiprime A, then there is at least one value В2< B1:
Sqrt (21) = 4,58257..
Trunc (4,58257) = 4
B2 = (n + 4)2 – 21
for n =1 B2 = (1+4)2 – 21 = 4; B = 2;
C = n+ 4 = 1 + 4 =5
A = С2 – В2 = (С-В)*(С+В) = (5 – 2)*(5+2) = 3*7
А=k1 * k2= 3*7
This is nothing but trivial algebra. Where is the ALGORITHM?
You have failed to specify any kind of procedure. All you have done is
assert the existence of some values satisfying some relations.
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