Making the sum of 5th power of integers, a perfect square. Yesterday this question was posed in a contest. It contains pretty easy questions like asking range of $ab+bc+ca$ when $a^2+b^2+c^2=1$, etc.  
But this question is something else. I haven't been able to solve this after $4$ hours.  
Question:
Find, $n$ if
$133^5+27^5+84^5+110^5=n^2$.  
I checked answer on calculator, it is $248832$.
I tried all sort of things I can do like trying to factor the expression, converting two of the odd nos. into $(m+n)^5+(m-n)^5$ form. But all in vain.
I hope anyone can help me here. 

This is less than a complete solution but too long for a comment.
Suppose $n^2 = a^5 + b^5 + c^5 + d^5$.  Then:
$$n^2 = (a+b)m + c^5 + d^5\quad\quad(1)$$
where:
$$m = a^4 -a^3b + a^2b^2 -ab^3 + b^4$$
Applying $(1)$ with $(a,b,c,d) = (133,27,110,84)$:
$$n^2 = 160m + 110^5 + 84^5 = 2^5(5m + 55^5 + 42^5)$$
Hence $2^5 | n^2$ and so $2^3 | n$.
Also, applying $(1)$ with $(a,b,c,d) = (133,110,84,27)$:
$$n^2 = 243m + 84^5 + 27^5 = 3^5(m + 28^5 + 9^5)$$
Hence $3^5 | n^2$ and so $3^3 | n$.
Combining the above results:
$$2^33^3 | n$$
