# On the equation $a^{2}b^3 + c^{2}d^3 = e^{2}f^3$

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 ?

• $1^{3}+2^{3} = 3^{2}$ ($a=b = c = f =1$ and $d = 2, e = 3$). – Geoff Robinson Aug 24 '20 at 9:19
• 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. – Fedor Petrov Aug 24 '20 at 9:44
• @FedorPetrov, okay, can the solutions be parametrized ? – Pres10 Aug 24 '20 at 10:26
• $a^{2}b^3 + c^{2}d^3 = e^{2}f^3\implies \Big(\dfrac{ab^3}{c}\Big)^2 - (bf)^3\Big(\dfrac{e}{c}\Big)^2 = -(bd)^3$  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). – Dmitry Ezhov Aug 24 '20 at 10:30
• $r^3s^2(r+s)^2+r^2s^3(r+s)^2=r^2s^2(r+s)^3$ provided $\gcd(r,s)=1$. – Gerry Myerson Aug 24 '20 at 12:27

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

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