<s>[INST] Is it possible that $a^2+b^2+c^2 = d^2+e^2+f^2$? 
Let $a,b,c$ be nonnegative integers such that $a \leq b \leq c, 2b \neq a+c$ and $\frac{a+b+c}{3}$ is an integer. Is it possible to find three nonnegative integers $d,e,$ and $f$ such that $d \leq e \leq f, f \neq c,$ and such that $a^2+b^2+c^2 = d^2+e^2+f^2$?

I have tried it for examples such as $a=1,b=2,c=6$, we have $1^2+2^2+6^2 = 41 = 0^2+4^2+5^2$ and it always seems to work. I am wondering how to use the fact that $\frac{a+b+c}{3}$ is an integer and wondering if it may be related to the AM-GM inequality. How should I use this information to solve the question?
 [/INST] Short answer (explained in the comments above):
$$ (4x)^2+(4x^2-1)^2    =  (4x^2+1)^2 $$
can be "interlaced" with
$$ (4x^2+1)^2 + (8x^4+4x^2)^2 =  (8x^4+4x^2+1)^2 $$
to get:
$$ (4x)^2 + (4x^2-1)^2 + (8x^4+4x^2)^2 = 0^2 + 0^2 + (8x^4+4x^2+1)^2 $$
or:

$$ (12x)^2 + (12x^2-3)^2 + (24x^4+12x^2)^2 = 0^2 + 0^2 + (24x^4+12x^2+3)^2 $$

i.e. an infinite number of solutions.
Easier approach: we may take three positive integers $N=4u+1$, $M=4v-1$, $ N+M=4(u+v)$
and write them as a difference of two squares, 
$$N=a^2-d^2,\quad  M=b^2-e^2,\quad  N+M=f^2-c^2.$$
We get $a^2+b^2+c^2=d^2+e^2+f^2$ with $f> c$ and we just need very few work to ensure the other constraints.
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