3

$$\frac{1}{u^2}+\frac{1}{v^2}=\frac{1}{w^2}$$

I want to generate all primitive solutions up to $u \le N$. Is there a parametric solution?

By brute force, I got these solutions:

$(15, 20, 12),(20, 15, 12),(65, 156, 60),(136, 255, 120),(156, 65, 60),(175, 600, 168),(255, 136, 120),(580, 609, 420),(600, 175, 168),(609, 580, 420)$.

It seems that $c$ is a multiple of $12$. Another observation is $u^2+v^2=t^4$ where $t$ is the hypotenuse of a triangle.

piepie
  • 547

2 Answers2

3

With $A$ being even, $r>s>0,$ $\gcd(r,s) = 1$ and $r+s$ odd,

$$ A = 2rs\left( r^2 + s^2 \right) \; , \; \; B = r^4 - s^4 \; , \; \; C = 2rs\left( r^2 - s^2 \right) \; . \; \; $$

The proof starts with $\gcd(A,B,C) = 1,$ goes on with $\gcd(A,B) = g, \; \;$ $\gcd(g,C) = 1,$ $A = ga, \; \; B = gb, \; \; $ $\gcd(a,b)=1,$ and goes on from there...

Mon Mar 12 20:54:12 PDT 2018
  A: 20    B: 15     C:  12  r: 2  s: 1
  A: 156    B: 65     C:  60  r: 3  s: 2
  A: 136    B: 255     C:  120  r: 4  s: 1
  A: 600    B: 175     C:  168  r: 4  s: 3
  A: 580    B: 609     C:  420  r: 5  s: 2
  A: 1640    B: 369     C:  360  r: 5  s: 4
  A: 444    B: 1295     C:  420  r: 6  s: 1
  A: 3660    B: 671     C:  660  r: 6  s: 5
  A: 1484    B: 2385     C:  1260  r: 7  s: 2
  A: 3640    B: 2145     C:  1848  r: 7  s: 4
  A: 7140    B: 1105     C:  1092  r: 7  s: 6
  A: 1040    B: 4095     C:  1008  r: 8  s: 1
  A: 3504    B: 4015     C:  2640  r: 8  s: 3
  A: 7120    B: 3471     C:  3120  r: 8  s: 5
  A: 12656    B: 1695     C:  1680  r: 8  s: 7
  A: 3060    B: 6545     C:  2772  r: 9  s: 2
  A: 6984    B: 6305     C:  4680  r: 9  s: 4
  A: 20880    B: 2465     C:  2448  r: 9  s: 8
  A: 2020    B: 9999     C:  1980  r: 10  s: 1
  A: 6540    B: 9919     C:  5460  r: 10  s: 3
  A: 20860    B: 7599     C:  7140  r: 10  s: 7
  A: 32580    B: 3439     C:  3420  r: 10  s: 9
  A: 5500    B: 14625     C:  5148  r: 11  s: 2
  A: 12056    B: 14385     C:  9240  r: 11  s: 4
  A: 20724    B: 13345     C:  11220  r: 11  s: 6
  A: 32560    B: 10545     C:  10032  r: 11  s: 8
  A: 48620    B: 4641     C:  4620  r: 11  s: 10
Mon Mar 12 20:54:12 PDT 2018
Will Jagy
  • 139,541
2

Wikipedia calls the projective curve $$\{[a:b:c]| a^2b^2=c^2(a^2+b^2)\}$$ the cruciform curve.

Consider the birational map on the projective plane induced by the standard quadratic transformation $$\begin{align}[a:b:c]&=[yz:xz:xy] & [x:y:z]&=[bc:ac:ab]\text{;}\end{align}$$ via this map, the cruciform curve is birational to $$\{[x:y:z] | x^2+y^2=z^2\}\text{,}$$ i.e., the circle. But the circle is birational to the projective line via stereographic projection: $$\begin{align}[x:y:z]&=[t^2-u^2:2tu:t^2+u^2]& [t:u]&=[x+z:y]=[y:z-x]\text{.} \end{align}$$ Consequently, the cruciform curve is also birational to the projective line: $$\begin{align}[a:b:c]&=[2tu(t^2+u^2):(t^2-u^2)(t^2+u^2):2tu(t^2-u^2)]\\ [t:u]&=[b(a+c):ac]=[ac:b(a-c)] \text{.}\end{align}$$

K B Dave
  • 7,912