Bilinear system of Diophantine Equations - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T03:37:48Z http://mathoverflow.net/feeds/question/70406 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70406/bilinear-system-of-diophantine-equations Bilinear system of Diophantine Equations unknown (yahoo) 2011-07-15T07:06:53Z 2011-08-27T11:57:44Z <p>$\forall i,j \in$ $\{$$1, \cdots,n$$\},$ let $x_{i},y_{i}$ be unknowns and $n_{ij} \in \mathbb{Z}$ with $i \le j$ be the knowns.</p> <p>Consider the following $\frac{n(n+1)}{2}$ with $n > 2$ overdetermined bilinear equations:</p> <p>$\sum_{\substack{j=1,}{j \ne i}}^{n} x_{j}y_{j} = -n_{ii} + x_{i}y_{i} \in \mathbb{Z}$.</p> <p>$x_{i}y_{j} + x_{j}y_{i} = n_{ij} \in \mathbb{Z}$ when $i &lt; j$.</p> <p>When is the system solvable and when is it solvable over $\mathbb{Z}$?</p> http://mathoverflow.net/questions/70406/bilinear-system-of-diophantine-equations/73839#73839 Answer by Max Alekseyev for Bilinear system of Diophantine Equations Max Alekseyev 2011-08-27T11:57:44Z 2011-08-27T11:57:44Z <p>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, $$x_i y_i = n_{ii} - \frac{1}{n-2} \sum_{i=1}^n n_{ii}.$$</p> <p>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.</p> <p>Values of $y_i$ can be found similarly.</p>