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There is no such finite set of pairs $(a_k,b_k)\in\mathbb{Q}_{\geq 0}$. Indeed, because the problem concerns rational representations, we can assume without loss of generality that $(a_k,b_k)\in\mathbb{Z}_{\geq 1}$. If the prime $p$ is represented by some $a_kx^2+b_ky^2$ over $\mathbb{Q}$, then there exists a positive integer $m$ such that $pm^2$ is primitively represented by $a_kx^2+b_ky^2$ over $\mathbb{Z}$. So there is a quadratic form $pm^2x^2+cxy+dy^2$ which is equivalent to $a_kx^2+b_ky^2$. Taking the discriminants of these quadratic forms, we see that $-4a_kb_k$ is a square modulo $p$. Passing to the fundamental discriminants underlying the discriminants $-4a_kb_k$, we would obtain a finite set $\mathcal{D}$ of negative fundamental discriminants such that for every sufficiently large prime $p$, there exists $d\in\mathcal{D}$ with $\chi_d(p)=1$. This is impossible by Lucia's response to MO question 373900.

There is no such finite set of pairs $(a_k,b_k)\in\mathbb{Q}_{\geq 0}^2$. Indeed, because the problem concerns rational representations, we can assume without loss of generality that $(a_k,b_k)\in\mathbb{Z}_{\geq 1}^2$. If the prime $p$ is represented by some $a_kx^2+b_ky^2$ over $\mathbb{Q}$, then there exists a positive integer $m$ such that $pm^2$ is primitively represented by $a_kx^2+b_ky^2$ over $\mathbb{Z}$. So there is a quadratic form $pm^2x^2+cxy+dy^2$ which is equivalent to $a_kx^2+b_ky^2$. Taking the discriminants of these quadratic forms, we see that $-4a_kb_k$ is a square modulo $p$. Passing to the fundamental discriminants underlying the discriminants $-4a_kb_k$, we would obtain a finite set $\mathcal{D}$ of negative fundamental discriminants such that for every sufficiently large prime $p$, there exists $d\in\mathcal{D}$ with $\chi_d(p)=1$. This is impossible by Lucia's response to MO question 373900.

There is no such finite set of pairs $(a_k,b_k)\in\mathbb{Q}_{\geq 0}$. Indeed, because the problem concerns rational representations, we can assume without loss of generality that $(a_k,b_k)\in\mathbb{Z}_{>0}$. If the prime $p$ is represented by some $a_kx^2+b_ky^2$ over $\mathbb{Q}$, then there exists a positive integer $m$ such that $pm^2$ is primitively represented by $a_kx^2+b_ky^2$ over $\mathbb{Z}$. So there is a quadratic form $pm^2x^2+cxy+dy^2$ which is equivalent to $a_kx^2+b_ky^2$. Taking the discriminants of these quadratic forms, we see that $-4a_kb_k$ is a square modulo $p$. Passing to the fundamental discriminants underlying the discriminants $-4a_kb_k$, we would obtain a finite set $\mathcal{D}$ of negative fundamental discriminants such that for every sufficiently large prime $p$, there exists $d\in\mathcal{D}$ with $\chi_d(p)=1$. This is impossible by Lucia's response to MO question 373900.

There is no such finite set of pairs $(a_k,b_k)\in\mathbb{Q}_{\geq 0}$. Indeed, because the problem concerns rational representations, we can assume without loss of generality that $(a_k,b_k)\in\mathbb{Z}_{\geq 1}$. If the prime $p$ is represented by some $a_kx^2+b_ky^2$ over $\mathbb{Q}$, then there exists a positive integer $m$ such that $pm^2$ is primitively represented by $a_kx^2+b_ky^2$ over $\mathbb{Z}$. So there is a quadratic form $pm^2x^2+cxy+dy^2$ which is equivalent to $a_kx^2+b_ky^2$. Taking the discriminants of these quadratic forms, we see that $-4a_kb_k$ is a square modulo $p$. Passing to the fundamental discriminants underlying the discriminants $-4a_kb_k$, we would obtain a finite set $\mathcal{D}$ of negative fundamental discriminants such that for every sufficiently large prime $p$, there exists $d\in\mathcal{D}$ with $\chi_d(p)=1$. This is impossible by Lucia's response to MO question 373900.