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I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{4}:=\langle k^4\vert k\in K\backslash 0 \rangle$, I want to find (all?) $a$ and $b$ in $F$ such that p1==1 modulo $K^{4}$ and p2==-1 modulo $K^{*4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{4}:=\langle k^4\vert k\in K\backslash 0 \rangle$, \rangle$ i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1==1 p1$\equiv$1 modulo $K^{4}$ and p2==-1 p2$\equiv$-1 modulo $K^{*4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{4}:=\langle k^4\vert k\in K\backslash 0 \rangle$ i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{4}$ and p2$\equiv$-1 modulo $K^{*4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{4}:=\langle k^4\vert k\in K\backslash 0 \rangle$ \rangle$. i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{4}$ and p2$\equiv$-1 modulo $K^{*4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{4}:=\langle 4}:=( k^4\vert k\in K\backslash 0 \rangle$. )$. i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{4}$ and p2$\equiv$-1 modulo $K^{*4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{4}:=( k^4\vert k\in K\backslash K\setminus 0 )$. i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{4}$ and p2$\equiv$-1 modulo $K^{*4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{4}:=( k^4\vert k\in K\setminus 0 )$. 4}$, i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{4}$ and p2$\equiv$-1 modulo $K^{*4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{$K4}$, ^{4}$, i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{4}$ and p2$\equiv$-1 modulo $K^{*4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{4}$, $K^{4}$

i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{4}$ and p2$\equiv$-1 modulo $K^{*4}$.$K^{4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{4}$

i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{4}$ $K^{*4}$ and p2$\equiv$-1 modulo $K^{4}$.$K^{\ast4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{4}$$K^{\ast4}$

i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{*4}$ $K^{4}$ and p2$\equiv$-1 modulo $K^{\ast4}$.$K^{4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{\ast4}$

i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{4}$ $K^{\ast4}$ and p2$\equiv$-1 modulo $K^{4}$.$K^{\ast4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{\ast4}$$K^{\ast4}:=( k^4\vert k\in K\setminus 0 )$

i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{\ast4}$ and p2$\equiv$-1 modulo $K^{\ast4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{\ast4}:=( $K^{\ast4}:=\langle k^4\vert k\in K\setminus 0 )$\rangle$

i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{\ast4}$ and p2$\equiv$-1 modulo $K^{\ast4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{\ast4}:=\langle k^4\vert k\in K\setminus 0 \rangle$

\rangle$ i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{\ast4}$ and p2$\equiv$-1 modulo $K^{\ast4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{\ast4}:=\langle k^4\vert k\in K\setminus 0 \rangle$ i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{\ast4}$ and p2$\equiv$-1 modulo $K^{\ast4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{\ast4}:=\langle k^4\vert k\in K\setminus 0 \rangle$ i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{\ast4}$ and p2$\equiv$-1 modulo $K^{\ast4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.

I am trying to solve two, 2-variable polynomial equations over $F:=\mathbb{Q}(i)$ modulo $K:=F(\sqrt{2})$. Specifically, if p1 = $a^2+6b^2$, p2 = $3a^2+2b^2$, and $K^{\ast4}:=\langle k^4\vert k\in K\setminus 0 \rangle$ i.e. the group of 4th powers of nonzero elements of $K$. I want to find (all?) $a$ and $b$ in $F$ such that p1$\equiv$1 modulo $K^{\ast4}$ and p2$\equiv$-1 modulo $K^{\ast4}$.

Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.