- #1

ramsey2879

- 841

- 0

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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