-1
$\begingroup$

Do there exist positive integers $a, b, c, d, e, f$ such that $a^{2}b^3 + c^{2}d^3 = e^{2}f^3$ where $b, d, f$ are pairwise coprime ?

Addendum: From the comments and Matt. F's answer, there clearly are infinitely many solutions. But what are their parametrizations ?

$\endgroup$
8
  • 2
    $\begingroup$ $1^{3}+2^{3} = 3^{2}$ ($a=b = c = f =1$ and $d = 2, e = 3$). $\endgroup$ Aug 24, 2020 at 9:19
  • 5
    $\begingroup$ And even infinitely many: take $c=f=x+1,d=e=x, b=1, a=x(x+1) $, for $x=2$ this is Matt F.'s example. $\endgroup$ Aug 24, 2020 at 9:44
  • $\begingroup$ @FedorPetrov, okay, can the solutions be parametrized ? $\endgroup$
    – Q_p
    Aug 24, 2020 at 10:26
  • 2
    $\begingroup$ $a^{2}b^3 + c^{2}d^3 = e^{2}f^3\implies \Big(\dfrac{ab^3}{c}\Big)^2 - (bf)^3\Big(\dfrac{e}{c}\Big)^2 = -(bd)^3$ $$$$ This have Pell form. Some solutions $(a,b,c,d,e,f)$=(137819, 7, 7, 5, 491218, 3), (1522899144, 7, 27, 4, 1044610624, 9), (5925421773487638370, 7, 70, 3, 3470294476762229557, 10). $\endgroup$ Aug 24, 2020 at 10:30
  • $\begingroup$ $r^3s^2(r+s)^2+r^2s^3(r+s)^2=r^2s^2(r+s)^3$ provided $\gcd(r,s)=1$. $\endgroup$ Aug 24, 2020 at 12:27

2 Answers 2

4
$\begingroup$

Yes: $6^2 1^3+3^2 2^3=2^2 3^3$

$\endgroup$
1
$\begingroup$

We can get the parametric solutions using known solution $(a,b,c,d,e,f)$ for fixed $(b,d,f).$
For instance, we get a parametric solution using $(a,b,c,d,e,f)=(6,1,3,2,2,3).$
$1^3(48m^2-6n^2-48mn)^2 + 2^3(24m^2-3n^2+12mn)^2 = 3^3(16m^2+2n^2)^2$
$m,n$ are arbitrary.

             m  n     a     c     e
             
             1  1     2    11     6
             1  2     6     3     2
             1  3   150    33    34
             1  4    10     1     2
             1  5   114     3    22
             2  1    30    39    22
             2  3   150   141    82
             2  5   146    47    38
             3  1   282   249   146
             3  2    30    69    38
             3  5   438   321   194
             4  1   190   143    86
             4  3   138   501   274
             4  5    38    61    34
             5  1   318   219   134
             5  2    58    59    34
             5  3   426   753   418
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.