Homer Simpson's Fermat equation Disprove the following equation

$3987^{12} + 4365^{12} = 4472^{12}$

First, since both the two numbers on the LHS were odd and the RHS was even, I tried dividing by 3
and found 

$3 \mid  3987$
  and $3 \mid 4365$

so

$3 \mid  3987^{12}$
  and $3 \mid 4365^{12}$

and by Divisibility of integer combinations we have

$3 \mid  (3987^{12}x + 4365^{12}y)$

For the RHS, $4472$ being even will obviously not be divisible by 3
4472/3 = 1490 with remainder 2

So, $3 \nmid 4472$ and thus $3 \nmid 4472^{12}$

In particular
let  $x = 1$ and $y = 1$
we will get

$3 \mid (3987^{12} + 4365^{12})$

But we have $3987^{12} + 4365^{12} = 4472^{12}$
so

$3 \mid 4472^{12}$ # a contradiction

thus the original statement is false
Is this a valid proof?
Is it safe to assume this step? $3 \nmid 4472 \Rightarrow 3 \nmid 4472^{12}$
 A: The statement '$4472$ being even will obviously not be divisible by $3$' is false (consider $6$). Aside from that, the reasoning is correct. If a prime $p$ does not divide $n$ it divides no power of $n$.
A: Is it safe to assume if $n\not |m$ then $n\not |m^k$?
If $n $ has any prime factors that $m $ doesn't, then absolutely!
But if every prime factor of $n $ is a prime factor of $m $ then.... it depends on if the power of $k$ will make the prime factors of $m^k$ higher than the power of $n $.
Example:
Let $n=12=2^2*3$ and $m=18=2*3^2$.  $n \not | m $.  They have the same prime factors $2,3$ but $n $ has a higher power of $2$ than $m$ does.
But $m^3$ multiplies those powers three fold: $m^3=2^3*3^6$ so now $n|m^3$.
But if $n$ has ANY prime factor, $p $ that $m$ does not, $n\not |m$ because $m$ has  no $p $ factor.  Raising $m $ to any power isn't going to add any more new prime factors.
Let $4472 =p^aq^b $ be the prime factorisation.  We don't actually care what it is (it's something like $4472=8*559$ or something.... I don't care).  The only thing we care is that $3$ isn't in it.  So $4472^{12}=p^{12a}q^{12b} $ and $3$ didn't slip in.  Nothing slipped in. So we know $3\not |4472^{12}.$
A: $\gcd(3987,13)=\gcd(4365,13)=1$
$\gcd(4472,13)=13$
$3987^{12} + 4365^{12} \equiv 1 + 1 \equiv 2 \pmod{13}$
$4472^{12} \equiv 0 \pmod{13}$
