SOLUTION: x, y, z ∈ Z³ 13/x² + 1996/y² = z/1997 x, y, z = ?

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Question 1210348: x, y, z ∈ Z³
13/x² + 1996/y² = z/1997
x, y, z = ?
Found 4 solutions by Edwin McCravy, mccravyedwin, ikleyn, AnlytcPhil:
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!



Since 1996 is divisible by perfect squares 1 and 4, let 
y2 = ± either of these.

, and where y2 = 1 or 4 







For ,







So that gives these solutions

(x,y,z) = (1,1,4011973), (1,-1,4011973), = (-1,1,4011973), = (-1,-1,4011973) 

For ,







So that gives these solutions

(x,y,z) = (1,2,2018967), (1,-2,2018967), = (-1,2,2018967), = (-1,-2,2018967) 

in addition to these we already found:

(x,y,z) = (1,1,4011973), (1,-1,4011973), (-1,1,4011973), (-1,-1,4011973)


I doubt there are any other solutions besides these 8, but I don't know that for
sure.

Maybe another tutor can find others or show that there are no others.

Edwin
Answer by mccravyedwin(406)   (Show Source): You can put this solution on YOUR website!

Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
 .
x, y, z ∈ Z³
13/x² + 1996/y² = z/1997
x, y, z = ?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


I will list below two families of    solutions in    numbers  (x,y,z).

    (a)   (x,y) = (+/-1, +/-1) ---> z = (13+1996)*1997 = 4011973.        4  solutions.

    (b)   (x,y) = (+/-1, +/-2) ---> z = (13+499)*1997 = 1022464.         4  solutions.


Why they are the solutions - it is obvious:  it is enough to look at denominators.

I don't know if where are other solutions.

Edwin correctly recognized and pointed 4 solutions of family  (a).

Edwin made an error pointing other  4  his solutions.
In my notations,  they are  4  solutions  (b),  with (or after) my correction.


/////////////////////////////////////////////


Here is an addition to the set of solutions found by Edwin

    (x,y,z) = (+/-7, +/-7, 81877).


Indeed,  left side of the original equation is    + = = = 41,

and right side is    = 41.



Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!
1210348
I kept on using the same technique, trying powers of 2 for x and y and
found all these solutions, and corrected the one I mis-punched my
calculator on.



(x,y,z) = (1,1,1422464), (1,-1,1422464), (-1,1,1422464), (-1,-1,1422464).

These others give 4 solutions in the same way: 











So, except for the 1st solution above, we could generalize on the others this
way.  Maybe Ikleyn can prove this generalization for them:

 
for n = 0,1,2,3,4 

But I'm still not sure there are any other solutions. I said I was sure I had
all of them before, and then found these and had to eat my words. LOL

Edwin
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