Fermat-like equation $c^n=a^{2n}+a^n b^n + b^{2n}$ Hello,
Is there a solution to the following equation:
$c^n=a^{2n}+a^n b^n + b^{2n}$
where $a,b,c \in \mathbb{N}^*$, and $n$ is an integer $\geq 2$.
The problem is due to Antoine Balan.
Thanks in advance.
 A: I'd like to add another line of proving this (and similar problems); the proof is not yet formally complete but the method as such seems helpful for many such problems, so I show my reasoning here:     
[update] After rereading my own answer I think now, that that line of attack might be more incomplete for this problem than initially thought; that the holes cannot be simply filled with one or two little more thoughts; so possibly I'll retract the whole approach later. Thanks for the upvotings anyway and apologies for distractions ... [/update]
The original answer went as below:     

Let us define the following notations      
a) {n,p} the exponent to which a prime p occurs in the primefactorization of some number n
b) the iverson-bracket [n:p] which evaluates to 1 if p divides n and to 0 if not      
c) $ \small f(a,b,n)=a^n - b^n  $ , the short form for expressions where a,b are thought to be constant and n is seen as varying
d) $ \small \lambda_p $ the smallest k>0 such that p divides $ \small f(a,b,k) $ ,
e) $ \small w_p $ the exponent, to which p occurs in $ \small f(a,b,\lambda_p) $ , formally  $ \small w_p=\{ f(a,b,\lambda_p),p \}   $      
We use that definitions to rewrite the analysis of the problem in terms of the Euler-totient-theorem and the concept of the order of cyclic subgroups modulo some primes p.
Here the idea is to compare the odd primefactors in the canonical primefactorizations of the lhs and rhs in the conveniently rewritten problem
$$ \small c^n-(ab)^n =^? a^{2n}+b^{2n} =  {a^{4n}- b^{4n} \over a^{2n}- b^{2n} } .     $$
It will be sufficient to compare the odd primefactors $ \small p, q \in \text{odd primes} $ only; so we refer to possible exponents of 2 with some anonymous exponents s and t only. Then the lhs is with $ \small q \in \text{odd primes} $
$$ \small f(c,ab,n)=2^s \prod q^{ [n:\lambda_q] (w_q+\{n,q\})} $$
and the rhs is with some exponent t at the primefactor 2 :
$$ \small a^{2n}+b^{2n}={ a^{4n}-b^{4n} \over a^{2n}-b^{2n} }={f(a,b,4n)\over f(a,b,2n)}  $$
and
$$ \small 
{f(a,b,4n)\over f(a,b,2n)} = 2^t {  \prod p^{ [4n:\lambda_p] (w_p+\{4n,p\})} 
                 \over  
             \prod p^{ [2n:\lambda_p] (w_p+\{ 2n , p \} ) } 
            }
$$
Here for all odd primes p we have $ \small \{ 4n , p \} = \{ 2n, p \} = \{ n,p \} $,
$$ \small { f(a,b,4n)\over f(a,b,2n)} = 2^t \prod p^{ ([4n:\lambda_p]- [2n:\lambda_p]) (w_p+\{ n, p \} ) } $$
Conclusion: (updated)
In the rhs we get only that primefactors p, whose order divide 4n but not 2n and - having $ \small n = 2^m \cdot o, o \text{ odd } $ thus must be exactly $ \small 4 \cdot 2^m r $ where r is any odd divisor of n, while on the lhs we get all primes in the factorization whose order equal any divisor of n. But the sets of primes must be equal to allow a solution for the original problem.
A: Writing the equation as
$$
c^n+(ab)^n = (a^n+b^n)^2,
$$
one has, via results of Darmon and Merel, and Poonen, that there
are no solutions for $n \geq 4$, provided the three terms are coprime.
If $n=2$, I think Pythagorean triply stuff shown that there are again no
solutions. In the case $n=3$, I guess one could again look at the
parametrizations for $X^3+Y^3=Z^2$, say in Cohen's GTM. I'm not sure how 
hard this case is....
The condition on coprimality might be a problem, but it's too early
in the morning for me to be sure!
