<s>[INST] How to prove a fact about the sum of three squares? How would I go about proving the following?
If $a$, $b$, $c$, $n$ are positive integers, then
$a^2+b^2+c^2 \neq 2^nabc$
I tried doing something similar to the proof for Adrien-Marie Legendre's Three Square theorem:  $a^2+b^2+c^2=n$ iff there are not integers $k$, and $m$ so that $n=4^k(8m+7)$. It didn't quite work out...
$2^nabc$ is always even. So if $a^2+b^2+c^2 = 2^nabc$, then  $a^2+b^2+c^2$ must be even. 
That means there is $a_1$, $b_1$, $c_1$ so that 
$a = 2a_1$, $b = 2b_1$, and $c = 2c_1$ 
So 
$(2a_1)^2+(2b_1)^2+(2c_1)^2 = 2^nabc \rightarrow 2(2a_1^2+2b_1^2+2c_1^2)= 2^nabc$
and we get $2a_1^2+2b_1^2+2c_1^2= 2^{n-1}abc$
We can continue to do this procedure with $a_2$, $b_2$, $c_2$ then $a_3$, $b_3$, $c_3$ then ... $a_n$, $b_n$, $c_n$. 
With $a_n$, $b_n$, $c_n$ we'd get
$2^na_n^2+2^nb_n^2+2^nc_n^2= 2^{n-n}abc=abc$
Since $a_n=2a_{n-1}$ and $a_0=a$, 
$a_n = \frac{a}{2^n}$
and we get 
$2^n(\frac{a}{2^n})^2+2^n(\frac{b}{2^n})^2+2^n(\frac{c}{2^n})^2=abc$
This just becomes the original equation.
$a^2+b^2+c^2 = 2^nabc$
 [/INST] Since the right-hand side is even, either exactly one or all three of $a,b,c$ must be even.
The former case is impossible, as you can easily see by taking both sides mod 4.
In the latter case, let $2^k$ be the greatest power of 2 in the GCD of $a,b,c$. Then
$$\left(\frac{a}{2^k}\right)^2 + \left(\frac{b}{2^k}\right)^2 + \left(\frac{c}{2^k}\right)^2 = 2^{n+k} \frac{a}{2^k} \frac{b}{2^k} \frac{c}{2^k},$$
with at least one of the terms on the left-hand side odd, and we are back in case 1.
EDIT: Note that $n>0$ is essential. When $n=0$,
$$3^2 + 3^2 + 6^2 = 54 = 3\cdot3\cdot6.$$
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