$u^3+v^3+3^{5}w^3=2\cdot3^{2}uvw$ has no non- trivial solutions if $u,v,w$ are pairwise co-prime.

Question

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?

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. — S.C.B., Jan 29 '17 at 01:51
I agree. However, there are quite a few other cases to consider. — , Jan 29 '17 at 01:54
I could not find this in Dickson's history of the theory of numbers. — marty cohen, Jan 29 '17 at 05:43
Why the co-primality restriction? A quick computer search does not even reveal any solution other than $u=-v$. — Tito Piezas III, Jan 29 '17 at 10:23
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 $3^5w^3$ — , Jan 29 '17 at 12:35

score 1 · Answer 1 · answered Jan 29 '17 at 03:45

1

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

answered Jan 29 '17 at 03:45

Will Jagy

139,541

my apologies. I edited the question. – Jan 29 '17 at 12:33
@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? – Will Jagy Jan 29 '17 at 15:58

score 1 · Answer 2 · answered Jan 29 '17 at 04:35

1

$u = -v, w = 0$.

Generalizing Will Jagy's.

Don't think that either of these helps much.

answered Jan 29 '17 at 04:35

marty cohen

107,799

My apologies, I meant no non-trivial solutions. I edited the question. – Jan 29 '17 at 12:33

$u^3+v^3+3^{5}w^3=2\cdot3^{2}uvw$ has no non- trivial solutions if $u,v,w$ are pairwise co-prime.

2 Answers2