- #1
tejolson
- 8
- 0
Is this of interest to the mathematics community?
((sqrt(2*x^2-1))^4+(sqrt(2*x^2-1))^2*2+1)/(x^4*4)=1
((sqrt(2*x^2-1))^6+(sqrt(2*x^2-1))^4*3+(sqrt(2*x^2-1))^2*3+1)/(x^6*8)=1
This is just a warped Pell Number. But it turns out it is true for all numbers, not just Pell Numbers. I don't think it's supposed to prove anything worthwhile.
I also have a Pell Number of 2,373,210 digits. Which means I have the 2,373,210th perfect square that is also a triangular number. It also means I have the 2,373,210th leg/leg primitive Pythagorean triangle. And it means I have 2 triangular numbers that square to make another Primitive Triangle. I don't know how many decimals this square root of two will go to but I'm sure the other methods are better than this one. I guess I can keep going and stop it at a number that is relatively prime, but I doubt that would be of any use. I'm sure there is more that I don't know about.
((sqrt(2*x^2-1))^4+(sqrt(2*x^2-1))^2*2+1)/(x^4*4)=1
((sqrt(2*x^2-1))^6+(sqrt(2*x^2-1))^4*3+(sqrt(2*x^2-1))^2*3+1)/(x^6*8)=1
This is just a warped Pell Number. But it turns out it is true for all numbers, not just Pell Numbers. I don't think it's supposed to prove anything worthwhile.
I also have a Pell Number of 2,373,210 digits. Which means I have the 2,373,210th perfect square that is also a triangular number. It also means I have the 2,373,210th leg/leg primitive Pythagorean triangle. And it means I have 2 triangular numbers that square to make another Primitive Triangle. I don't know how many decimals this square root of two will go to but I'm sure the other methods are better than this one. I guess I can keep going and stop it at a number that is relatively prime, but I doubt that would be of any use. I'm sure there is more that I don't know about.