MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$\forall i,j \in$ $\{$$1, \cdots,n$$\},$ let $x_{i},y_{i}$ be unknowns and $n_{ij} \in \mathbb{Z}$ with $i \le j$ be the knowns.

Consider the following $\frac{n(n+1)}{2}$ with $n > 2$ overdetermined bilinear equations:

$\sum_{\substack{j=1,}{j \ne i}}^{n} x_{j}y_{j} = -n_{ii} + x_{i}y_{i} \in \mathbb{Z}$.

$x_{i}y_{j} + x_{j}y_{i} = n_{ij} \in \mathbb{Z}$ when $i < j$.

When is the system solvable and when is it solvable over $\mathbb{Z}$?

share|cite|improve this question

Let's first focus on the equations: $$\sum_{j=1, j\ne i}^n x_j y_j = -n_{ii} + x_i y_i.$$ Denoting $s = \sum_{j=1}^n x_j y_j$, we have $$2x_iy_i = n_{ii} + s.$$ Summing up over $i=1,2,\dots,n$, we get $$2s = \sum_{i=1}^n n_{ii} + n\cdot s.$$ implying that (for $n\ne 2$) $$s = \frac{-1}{n-2} \sum_{i=1}^n n_{ii},$$ Therefore, $$x_i y_i = n_{ii} - \frac{1}{n-2} \sum_{i=1}^n n_{ii}.$$

Now, the equation $$x_i y_j + x_j y_i = n_{ij}$$ multiplied by $2 x_i x_j$ turns into $$(n_{jj} + s) x_i^2 + (n_{ii} + s) x_j^2 = 2 n_{ij} x_i x_j.$$ Plugging in $j=1$ and dividing by $x_1^2$, we further have $$(n_{11} + s) z_i^2 - 2 n_{i1} z_i + (n_{ii} + s) = 0$$ which is a quadratic equation w.r.t. $z_i = x_i / x_1$ and can be easily solved.

Values of $y_i$ can be found similarly.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.