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The equation in integers $$x^2y^2+y^2z^2+z^2x^2=N^2$$ has solutions such as $$(x,y,z,n)\in \{(1,1,2,3),(1,3,4,13),(5,6,11,91) \}$$ Can one find a family of infinitely many? The problem is related with the tetrahedron of vertices $$(x,0,0),(0,y,0),(0,0,z),\ \text{and}\ \ (0,0,0)$$ having faces of integer/rational areas (Heronian faces). This is almost a Heronian tetrahedron. https://en.wikipedia.org/wiki/Heronian_tetrahedron

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    $\begingroup$ If $(x,y,z,n)$ is a solution, so is $(cx,cy,cz,c^2n)$ for any positive integer $c$. Does that count as an infinite family? Solutions do not seem to be particularly rare. A Python search with $1\le x \le y \le z \le 100$ helped me find over $3000$ solutions. The largest (by $n$) of these was $(92,96,99,15852)$ $\endgroup$
    – paw88789
    Sep 23 at 1:47
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    $\begingroup$ Given a solution $(x,y,z,N)$, then $(xy/N,yz/N,zx/N)$ is a rational point on $a^2+b^2+c^2=1$. Conversely, if $a=p_1/q_1$, $b=p_2/q_2$, and $c=p_3/q_3$ satisfy $a^2+b^2+c^2=1$, then $$(x,y,z,N)=(p_1 q_2 p_3,p_1 p_2 q_3,q_1 p_2 p_3,p_1 p_2 p_3 q_1 q_2 q_3)$$ is an integral solution to the original equation. This shows you how to find all solutions (there are certainly infinitely many), because one can parametrize rational points on the sphere, for example $a = 2u/(1+u^2+v^2)$, $b= 2v/(1+u^2+v^2)$, $c=(u^2+v^2-1)/(u^2+v^2+1)$ with $u,v \in \mathbf{Q}$. $\endgroup$ Sep 23 at 2:15
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    $\begingroup$ So, for example, you could take $$(x,y,z,N) = (2ABD^2(B^2C^2 + A^2D^2 - B^2D^2) (B^2C^2 + A^2D^2 + B^2D^2), 4AB^3CD^3(B^2C^2 + A^2D^2 + B^2D^2), 2B^2CD(B^2C^2 + A^2D^2 - B^2D^2) (B^2C^2 + A^2D^2 + B^2D^2), 4AB^3CD^3(B^2C^2 + A^2D^2 - B^2D^2) (B^2C^2 + A^2D^2 + B^2D^2)^3).$$ $\endgroup$ Sep 23 at 2:20
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    $\begingroup$ math.stackexchange.com/questions/2507125/… $\endgroup$
    – individ
    Sep 23 at 5:35
  • $\begingroup$ That is a beautiful parametrization, thanks! So the problem was asked before. $\endgroup$ Sep 23 at 13:12

2 Answers 2

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For example, if $$ u^2 - 13 v^2 = 36 $$ you may take your $(x,y,z) = (2,3,v)$ while your $N$ becomes $u$

One sequence of $v$ that are coprime with $6$ are

$$ 1, \; \; 1909, \; \; 2477881, \; \; 3216287629, \; \; ... $$ extended by $$ v_{j+2} = 1298 v_{j+1} - v_j$$

So, we may fix two elements, call them $a,b.$ To have quadruple $(a,b,v,u)$ we are asking for all solutions to Pell type $$ u^2 - (a^2 + b^2)v^2 = a^2 b^2 $$ If we have $a^2 + b^2 = w^2$ this is a finite set of points because the left hand side factors. Otherwise:

Take the fundamental solution to $R^2 - (a^2 + b^2)S^2 = 1,$ meaning the smallest positive $R$ with $S \neq 0.$ Then the $v$ values come in a finite number of interlaced sequences of type $ v_{j+2} = 2R v_{j+1} - v_j$

Let me put in all the solutions for my original 13: There are 10 sequences of solutions under the Fibonacci type recursion I mentioned for $v.$ Given line numbered $k,$ the next two are line $k+10$ and line $k+20,$ and the precise recursion becomes $ v_{k+20} = 1298 v_{k+10} - v_k$

jagy@gost:~/Desktop/Cplusplus$ ./Pell_Target_Fundamental
Usage: ./ Pell_Target_Fundamental  d t s target    time_in_seconds 
jagy@gost:~/Desktop/Cplusplus$ ./Pell_Target_Fundamental   13  649  180  36 1
  Automorphism matrix:  
    649   2340
    180   649
  Automorphism backwards:  
    649   -2340
    -180   649

  649^2 - 13 180^2 = 1

 w^2 - 13 v^2 = 36 =  2^2 3^2

Fri 23 Sep 2022 08:28:18 AM PDT

1. x:  6      y: 0  SEED   KEEP +- 
2. x:  7      y: 1  SEED   KEEP +- 
3. x:  19      y: 5  SEED   KEEP +- 
4. x:  33      y: 9  SEED   KEEP +- 
5. x:  58      y: 16  SEED   KEEP +- 
6. x:  202      y: 56  SEED   BACK ONE STEP  58 ,  -16
7. x:  357      y: 99  SEED   BACK ONE STEP  33 ,  -9
8. x:  631      y: 175  SEED   BACK ONE STEP  19 ,  -5
9. x:  2203      y: 611  SEED   BACK ONE STEP  7 ,  -1
10. x:  3894      y: 1080  SEED   BACK ONE STEP  6 ,  0
11. x:  6883      y: 1909
12. x:  24031      y: 6665
13. x:  42477      y: 11781
14. x:  75082      y: 20824
15. x:  262138      y: 72704
16. x:  463353      y: 128511
17. x:  819019      y: 227155
18. x:  2859487      y: 793079
19. x:  5054406      y: 1401840
20. x:  8934127      y: 2477881

Fri 23 Sep 2022 08:28:19 AM PDT

 w^2 - 13 v^2 = 36 =  2^2 3^2

6,  7,  19,  33,  58,  202,  357,  631,  2203,  3894,  
6883,  24031,  42477,  75082,  262138,  463353,  819019,  2859487,  5054406,  8934127,  


0,  1,  5,  9,  16,  56,  99,  175,  611,  1080,  
1909,  6665,  11781,  20824,  72704,  128511,  227155,  793079,  1401840,  2477881,  


jagy@gost:~/Desktop/Cplusplus$ 
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  • $\begingroup$ Good point! That shows the existence of other infinite families than the one below. $\endgroup$ Sep 23 at 13:07
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Here's an infinite family: $(a,b,a+b,a^2+ab+b^2)$.

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  • $\begingroup$ That is a nice identity! Good eye! My examples are all like this! $\endgroup$ Sep 23 at 12:56
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    $\begingroup$ Thanks. As I mentioned in a comment, I generated a bunch of data with Python. I noticed this pattern from the data (and then verified it algebraically). Not all solutions fall into this family (e.g., the one noted in my comment: $(92,96,99,15852)$ ). $\endgroup$
    – paw88789
    Sep 23 at 13:06
  • $\begingroup$ It seems like the general parametrization was already published here math.stackexchange.com/questions/2507125/… $\endgroup$ Sep 23 at 13:18
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    $\begingroup$ @Eugen Ionascu: Interesting, but there is no verification or explanation of that parameterization being the full solution. Without looking at it too hard, it seems curious that the solution you linked seems to have $4$ free variables ($k,t,s,p$). I would have guessed, since the original problem has $4$ variables, that the parametrized solution would have (at most) $3$ free variables. $\endgroup$
    – paw88789
    Sep 23 at 13:47
  • $\begingroup$ Yes, I think we need to check into it and maybe have a proof. $\endgroup$ Sep 23 at 17:31

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