- #1
ramsey2879
- 841
- 3
I discovered the following general solution to primitive forms of y^2 + 2x^2 = a^2
a = 3 + 4n(n+1)
y = 4n(n+1) - 1
x = 4n + 2
Moreover let P = {11, 17, 19, 41, 43 ...} = the set of prime divisors of numbers of the form 3+4n(n+1) without the 3, and let Q = { -1, 7, 17, 23 ...} = the set of prime divisors of numbers of the form 4n(n+1) -1 including the -1. My conjecture is that for any member "p" of set P there is a corresponding member "q" of set Q such that q^2 + 72x^2 = p^2 where x is an interger, e.g.
11^2 = 7^2 + 72*1
17^2 = -1^2 + 72*2^2
19^2 = 17^2 + 72*1
41^2 = 23^2 + 72*4^2
43^2 = 7^2 + 72*5^2
...
a = 3 + 4n(n+1)
y = 4n(n+1) - 1
x = 4n + 2
Moreover let P = {11, 17, 19, 41, 43 ...} = the set of prime divisors of numbers of the form 3+4n(n+1) without the 3, and let Q = { -1, 7, 17, 23 ...} = the set of prime divisors of numbers of the form 4n(n+1) -1 including the -1. My conjecture is that for any member "p" of set P there is a corresponding member "q" of set Q such that q^2 + 72x^2 = p^2 where x is an interger, e.g.
11^2 = 7^2 + 72*1
17^2 = -1^2 + 72*2^2
19^2 = 17^2 + 72*1
41^2 = 23^2 + 72*4^2
43^2 = 7^2 + 72*5^2
...
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