A difficult question on representing a square by a sum of 4 (or 5 ) squares The question above is not about the decomposition of a square $$m^2 = a^2 + b^2 + c^2 + d^2 $$ as a sum of 4 or more squares. We know it is always possible and we also have algorithms to do that.  
My question is about going backward, that is if we only have a partial decomposition of $$m^2$$ as a sum of 4 (or 5) squares , can we rebuild $$m^2$$ If we have say 3 out of 4 or 4 out of 5 squares, can we reconstruct the square?
 A: Suppose you have any three integers $a,b,c > 0$, and denote by sake of simplicity $k = a^2+b^2+c^2$. Then the equation
$$a^2+b^2+c^2 + y^2 = x^2$$
(where $x,y$ are the unknowns) may have more than one solution. Indeed this is equivalent to
$$(x-y)(x+y) = k$$
whose solutions are $$x= \frac{1}{2} \left( d+ \frac{k}{d} \right), \ \ y= \frac{1}{2} \left| d- \frac{k}{d} \right|$$
where $d$ is any divisor of $k$.
A: Not uniquely.
Since for instance, given $a=3, b=5, c=11$
$18^2= 3^2+5^2+11^2+13^2$
$78^2= 3^2+5^2+11^2+77^2$
As we can write $(m+d)(m-d) = a^2+b^2+c^2$
there are as many solutions for $m,d$ as there factorisations of $a^2+b^2+c^2$.
A: If we have 
$$
m^2 = a^2 + b^2 + c^2 + d^2
$$
where we know $b$, $c$ and $d$ and that there is some unknown $a$, then you ask if $m$ can be uniquely determined.
Counter example:
$$
6^2 = 2^2 + 4^2 + 4^2 + 0^2 \\
9^2 = 7^2 + 4^2 + 4^2 + 0^2
$$
A: You are asking if, given $a,b,c$, we can find $x,y$ in,
$$a^2+b^2+c^2+y^2=x^2$$
The answer is yes, but it may not be unique since,
$$a^2+b^2+c^2=(x-y)(x+y)$$
hence depends on the factorization of the LHS. For example, given $5,7,11$, then,
$$5^2+7^2+11^2 = 14^2-1^2$$
or
$$5^2+7^2+11^2 = 22^2-17^2$$
