A difficult question on representing a square by a sum of 4 (or 5 ) squares

Question

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?

Answer 1

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 $a^2+b^2+c^2$.

Answer 2

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: $$ 6^2 = 2^2 + 4^2 + 4^2 + 0^2 \\ 9^2 = 7^2 + 4^2 + 4^2 + 0^2 $$

Answer 3

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 $$5^2+7^2+11^2 = 22^2-17^2$$

Answer 4

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 $k$.