Excercise | |||||
---|---|---|---|---|---|
verification of above statements (first few primes) (note: for ranges of p see above) |
|||||
c | p^{2} | n = sqrt(c - p^{2}) | k = (p-n-1)/2 |
PT([n,k]) = (a,b,c) or error reasoning |
ptc(c) |
1,2,3 | trunc(sqrt(c)) = 1 < 2 ==> error | 0 | |||
5 | 2^{2} = 4 | 1 | 0 | PT([1,0]) = (4,3,5) | 1 |
7 |
trunc(sqrt(7)) = 2 and 2^{2} + 1^{2} = 5 < 7 ==> error |
0 | |||
11 | 3^{2} = 9 | sqrt(2) | n not integer ==> error | 0 | |
13 | 3^{2} = 9 | 2 | 0 | PT([2,0]) = (12,5,13) | 1 |
17 | 4^{2} = 16 | 1 | 1 | PT([1,1]) = (8,15,17) | 1 |
19,23 | 4^{2} = 16 | sqrt(3),sqrt(7) | n not integer ==> error | 0 | |
29 | 5^{2} = 25 | 2 | 1 | PT([2,1]) = (20,21,29) | 1 |
31 | 5^{2} = 25 | sqrt(6) | n not integer ==> error | 0 | |
37 | 5^{2} = 25 | sqrt(12) | n not integer ==> error | 1 | |
6^{2} = 36 | 1 | 2 | PT([1,2]) = (12,35,37) | ||
41 | 5^{2} = 25 | 4 | 0 | PT([4,0]) = (40,9,41) | 1 |
6^{2} = 36 | sqrt(5) | n not integer ==> error | |||
verification of above statements (some composite numbers) | |||||
25 |
25 <= 4^{2} + 3^{2} = 25 lowest possible p is 4 trunc(sqrt(25)) = 5 highest possible p is 5 ptc(25) = 2 ptc(5) = 2 [ptc(5)] = 2 counts: 5 * PT([1,0]) = 5 (4,3,5) |
1 | |||
4^{2} = 16 | 3 | 0 | PT([3,0]) = (24,7,25) | ||
35 |
35 <= 5^{2} + 4^{2} = 41 lowest possible p is 5 trunc(sqrt(35)) = 5 highest possible p is 5 ptc(35) = ptc(5*7) = [ptc(5)] + (2* ptc(5)]+1)ptc(7) = 1 + 0 = 1 counts: 7 * PT([1,0]) = 7 (4,3,5) |
0 | |||
5^{2} = 25 | sqrt(10) | n not integer ==> error | |||
45 |
45 <= 6^{2} + 5^{2} = 61 lowest possible p is 6 trunc(sqrt(45)) = 6 highest possible p is 6 ptc(45) = ptc(3^{2}5) = [2 ptc(3)] + (2*[2 ptc(3)]+1)ptc(5) = 0 + 1 = 1 |
1 | |||
6^{2} = 36 | 3 | 1 | PT([3,1]) = 9 (4,3,5) | ||
65 |
65 <= 7^{2} + 6^{2} = 85 lowest possible p is 7 trunc(sqrt(65)) = 8 highest possible p is 8 ptc(65) = ptc(5*13) = ptc(5) + (2*ptc(5)+1)ptc(13) = 1 + 3 = 4 includes 5 * PT([2,0]) = 5 (12,5,13) and 13 * PT([1,0]) = 13 (4,3,5) |
2 | |||
7^{2} = 49 | 4 | 1 | PT([4,1]) = (56,33,65) | ||
8^{2} = 64 | 1 | 3 | PT([1,3]) = (16,63,65) |