1
$\begingroup$

Let $u,v,w$ be 3 pairwise coprime integers. Then $$u^3+v^3+3^{5}w^3=2\cdot3^{2}uvw$$ has no non-trivial solutions. How can I prove this?

I have tried to consider many individual cases such as $uvw>0$,$uvw<0$, $max(u,v,w)=u$ etc. a pretty tedious approach. I am certain there must be simpler ones. Any hints?

  • 2
    Here is a simple case, $u>0, v>0, w>0$. Then $$u^3+v^3+243w^3 \ge 3 \sqrt[3]{243}uvw>18uvw$$ By AM-GM inequality. – 2017-01-29
  • 0
    I agree. However, there are quite a few other cases to consider. – 2017-01-29
  • 0
    It may be difficult. – 2017-01-29
  • 0
    i agree. Hence, my question. – 2017-01-29
  • 0
    I could not find this in Dickson's history of the theory of numbers. – 2017-01-29
  • 0
    Why the co-primality restriction? A quick computer search does not even reveal _any_ solution other than $u=-v$. – 2017-01-29
  • 0
    That's correct. the only solutions are $(1,-1,0); (-1,1,0)$. I am trying to figure out why. It seems like it has to do with the size of $3^5w^3$ – 2017-01-29

2 Answers 2

1

$$ u = 1, \; \; v = -1, \; \; w = 0 $$

  • 0
    my apologies. I edited the question. – 2017-01-29
  • 0
    @NumThcurious for a homogeneous function in integers, non-trivial means at least one variable nonzero. It really does. So, where did you get the problem? – 2017-01-29
1

$u = -v, w = 0$.

Generalizing Will Jagy's.

Don't think that either of these helps much.

  • 0
    My apologies, I meant no non-trivial solutions. I edited the question. – 2017-01-29