[@Singla2010 Lemma 4.11], we prove that, if $q$ is an odd prime power, then the number of solutions of quadratic-[`IP1S`]{}in ${{\mathbb}{F}}_q^{n\times n}$ is bounded from above by: $$q^{{\left(\sum_{1\leq i\leq r}\sum_{1\leq j\leq d_i}(2\,d_i-2\,j+1)s_{i,j}\right)}}-1.$$

Open Question: The Irregular Case {#open-question-the-irregular-case .unnumbered}
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Given a quadratic instance of [`IP1S`]{}, a nondegenerate instance is an instance wherein the matrix whose rows are all the rows of $H_1,\ldots,H_m$ has rank $n$. In paragraph \[par:variables\], we see how to transform some degenerate instances into nondegenerate instances. However, nondegenerate instances are not always regular instances. There are cases, the so-called *irregular* cases, such that the vector space of matrices spanned by $H_1,\ldots,H_m$ does not contain a nondegenerate matrix. This situation is well illustrated by the following example $f_1=x_1x_3,f_2=x_2x_3$. Any linear combination of $f_1,f_2$ is degenerate, while ${{\mathbf}{f}}=(f_1,f_2)$ is not. Note that we can decide in randomized polynomial time if an instance of quadratic-[`IP1S`]{}is irregular since it is equivalent to checking if a determinant is identically equal to zero; thus it is a particular instance of polynomial identity testing. In the irregular case, it is clear that our algorithm fails. In fact, it seems that most known algorithms dedicated to quadratic-[`IP1S`]{} ([@DBLP:conf/pkc/BouillaguetFFP11; @MacariotRatPG2013]) will fail on these instances; 