I don't understand what is the intuition/ strategy to obtain $q(u) = \frac{1}{4}(u_1 + u_2 + 2u_3)^2 - \frac{1}{4}(u_1-u_2)^2-u_3^2$ from $q(u) = u_1u_2 + u_1u_3 + u_2u_3.$
I understand the result of course but I don't see how what motivated the choice of grouping $u_1$, $u_2$ and $u_3$ together, and why we have the factor $2$ in front of $u_3$, ...
Actually when we have squares with only two element I understand it very well but when it comes to 3 elements inside the square I become lost.
Thank you for your help !
The idea here is to rewrite $q(u) = u_1u_2 + u_1u_3 + u_2u_3$ as $$q(v) = v_1^2 \pm v_2^2 \pm v_3^3$$ where $v$ is some linear transformation of $q$. It should be obvious that there are multiple solutions. In particular, from the solution you've given, two more are immediately apparent, since $q$ is symmetric in $u_1, u_2, u_3$: $$q(u) = \frac14(2u_1 + u_2 + u_3)^2 - \frac14(u_2-u_3)^2-u_1^2\\q(u)= \frac14(u_1 + 2u_2 + u_3)^2 - \frac14(u_1-u_3)^2-u_2^2$$
So we are not looking for a single solution, which gives us the freedom to make simplifying choices, which is exactly what someone did here.
Start off by letting $v_1 = a_1u_1 +a_2u_2 + a_3u_3$. Then $$v_1^2 = \left[a_1^2u_1^2 + a_2^2u_2^2 + a_3^2u_3^2\right] + (2a_1a_2)u_1u_2 + (2a_1a_3)u_1u_3 + (2a_2a_3)u_2u_3$$ That gives us all the mixed products, but we also pick up the squared terms that we don't want. So we need to subtract them back out with $v_2, v_3$.
We cannot get rid of all three squares with $v_2$ alone. If we set $v_2 = b_1u_1 +b_2u_2 + b_3u_3$, then $$v_1^2 - v_2^2 = (a_1^2 - b_1^2)u_1^2 + (a_2^2 - b_2^2)u_2^2 + (a_3^2 - b_3^2)u_3^2 + \text{ mixed terms}$$ If this were to only consist of mixed terms, we would need $b_i = \pm a_i$ for each $i$. But if we choose the same sign for all three, $v_1^2 = v_2^2$ and we just get $0$. We can assume wlog that $b_1 = a_1$. If $b_2 = a_2$ as well, then the $u_1u_2$ terms in $v_1^2$ and $v_2^2$ will also be the same and $v_1^2 - v_2^2$ would have no $u_1u_2$ term. So instead we take $b_2 = -a_2$. But then $b_3 = a_3$ would give the same problem, leaving no $u_1u_3$ term. And the other choice of $b_3 = -a_3$ leaves $v_1^2 - v_2^2$ with no $u_2u_3$ term. Thus we cannot rewrite $q$ as $q(v) = v_1^2 - v_2^2$ alone. $v_3$ is needed.
So if we cannot get rid of all three squared terms with just $v_2$, can we get rid of two of them? That can be done. Since we can't get rid of $u_3^2$, just ignore it for now and let $b_3 = 0$. Getting rid of $u_1^2$ and $u_2^2$ without also getting rid of $u_1u_2$ requires $b_1 = a_1$ and $b_2 = -a_2$ (or vice versa), so we set $$v_2 = a_1u_1 - a_2u_2$$ which leaves $$v_1^2 - v_2^2 = a_3^2u_3^2 + (4a_1a_2)u_1u_2 + (2a_1a_3)u_1u_3 + (2a_2a_3)u_2u_3$$ Now if we include $u_1$ or $u_2$ in $v_3$, it would bring back $u_1^2$ and $u_2^2$ terms, which we don't want. So $v_3 = cu_3$ for some $c$. And obviously $c = \pm a_3$, if $v_1^2 - v_2^2 - v_3^2$ is to have no $u_3^2$ term. It doesn't matter which sign we pick, so we might as well just use $c=a_3$.
So now $$v_1^2 - v_2^2 - v_3^2 = (4a_1a_2)u_1u_2 + (2a_1a_3)u_1u_3 + (2a_2a_3)u_2u_3$$ which means $$4a_1a_2 = 1\\2a_1a_3 = 1\\2a_2a_3 = 1$$ Thus $$a_1 = a_2 = \frac 1{2a_3}$$ and $$4\frac 1{2a_3}\frac 1{2a_3} = 1\\a_3^2 = 1\\a_3 = \pm 1$$ Arbitrarily choose the positive value, and we get $$a_1 = \frac 12, a_2 = \frac 12, a_3 = 1$$ which is exactly the solution you gave.
We arrived at it by making some arbitrary choices, two of which are trivial, while the other two locked us into this solution.