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### solving system of equations over number field modulo another number field

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}$. Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it. ### solving system of equations over number field modulo another number field I am trying to solve two, 2-variable polynomial equations over$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}$. Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it. ### solving system of equations over number field modulo another number field I am trying to solve two, 2-variable polynomial equations over$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}$. Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it. ### solving system of equations over number field modulo another number field I am trying to solve two, 2-variable polynomial equations over$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}$. Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it. ### solving system of equations over number field modulo another number field I am trying to solve two, 2-variable polynomial equations over$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.

### solving system of equations over number field modulo another number field

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.

### solving system of equations over number field modulo another number field

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.

### solving system of equations over number field modulo another number field

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}$$K^{\ast4} i.e. the group of 4th powers of nonzero elements of K. I want to find (all?) a and b in F such that p1\equiv1 modulo K^{*4} K^{4} and p2\equiv-1 modulo K^{\ast4}.K^{4}. Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it. ### solving system of equations over number field modulo another number field I am trying to solve two, 2-variable polynomial equations over F:=\mathbb{Q}(i) modulo K:=F(\sqrt{2}). Specifically, if p1 = a^2+6b^2, p2 = 3a^2+2b^2, and K^{\ast4} i.e. the group of 4th powers of nonzero elements of K. I want to find (all?) a and b in F such that p1\equiv1 modulo K^{4} K^{\ast4} and p2\equiv-1 modulo K^{4}.K^{\ast4}. Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it. ### solving system of equations over number field modulo another number field I am trying to solve two, 2-variable polynomial equations over F:=\mathbb{Q}(i) modulo K:=F(\sqrt{2}). Specifically, if p1 = a^2+6b^2, p2 = 3a^2+2b^2, and K^{\ast4}$$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.

### solving system of equations over number field modulo another number field

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.

### solving system of equations over number field modulo another number field

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}$. Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it. ### solving system of equations over number field modulo another number field I am trying to solve two, 2-variable polynomial equations over$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}$. Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.  17 retagged FrédéricC 4314 ●3 ●37 ●90 ### solving system of equations over number field modulo another number field I am trying to solve two, 2-variable polynomial equations over$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}$. Any amount of walk through or pointing in the right direction, or telling me this might not be doable would be great! I am relatively new to sage, or at least it has been years since I've used it.  18 retagged FrédéricC 4314 ●3 ●37 ●90 ### solving system of equations over number field modulo another number field I am trying to solve two, 2-variable polynomial equations over$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.