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Let $F$ be a field, $n$ be a positive integer. Denote by $h_{F}(n)$ the maximal dimension of a subspace $X\subset F^n$ such that $(x,y)=0$ for any two (not necessary distinct) vectors $x,y\in F^n$, where $(x,y)=x_1y_1+\dots+x_ny_n$ for $x=(x_1,\dots,x_n)$, $y_1=(y_1,\dots,y_n)$. y=(y_1,\dots,y_n)$. For example,$h_{\mathbb{R}}(n)=0$for any$n$. It is clear that$h_F(n)$is non-decreasing by$n$, that$h_F(n)\leq n/2$(such$X$is contained in$X^{\perp}$, hence$\dim(X)\leq \dim(X^{\perp})=n-\dim(X)$) and that always$h_{F}(n+k)\geq h_F(n)+h_F(k)$. It allows to get$h_{F}(n)=[n/2]$for$F=\mathbb{C}$or$F=\mathbb{F}_p$with prime$p=4k+1$or$p=2$. Now take$p=4k+3$and$F=\mathbb{F}_p$. It is easy to get$h_{F}(2)=0$,$h_{F}(3)=1$,$h_{F}(4)=2$(take span of$(a,b,c,0)$and$(0,-c,b,a)$with$a^2+b^2+c^2=0$). Hence$h_{F}(4n)=2n$,$h_{F}(4n+1)=2n$,$h_{F}(4n+3)=2n+1$. But what about$h_F(4n+2)$? If it equals$2n+1$for some$n$, then also for all greater$n$. But does there always exist such$n$and if it does exist, how to find it as a function of$p$? I managed only to observe by hands that$h_{F}(6)=2$for$p=3$. 2 added 7 characters in body Let$F$be a field,$n$be a positive integer. Denote by$h_{F}(n)$the maximal dimension of a subspace$X\subset F^n$such that$(x,y)=0$for any two (not necessary distinct) vectors$x,y\in F^n$, where$(x,y)=x_1y_1+\dots+x_ny_n$for$x=(x_1,\dots,x_n)$,$y_1=(y_1,\dots,y_n)$. For example,$h_{\mathbb{R}}(n)=0$for any$n$. It is clear that$h_F(n)$is non-decreasing by$n$, that$h_F(n)\leq n/2$(such$X$is contained in$X^{\perp}$, hence$\dim(X)\leq \dim(X^{\perp})=n-\dim(X)$) and that always$h_{F}(n+k)\geq h_F(n)+h_F(k)$. It allows to get$h_{F}(n)=[n/2]$for$F=\mathbb{C}$or$F=\mathbb{F}_p$with prime$p=4k+1$or$p=2$. For prime Now take$p=4k+3$and$F=\mathbb{F}p$it F=\mathbb{F}_p$.