1 | initial version |

Hello, @arneovi! The variables of the form `cN`

, where `N`

is a natural number, are parameters. For example, one of the solutions you obtained is

```
[a_1 == 0, a_2 == 0, b_1 == c2439, b_2 == c2440]
```

This can be rewritten as $a_1=0$, $a_2=0$, $b_1=s$ and $b_2=t$, where $s,t\in\mathbb{R}$. On the other hand, one of the other solutions you obtained is

```
[a_1 == c2443, a_2 == c2443^2, b_1 == 0, b_2 == 0]
```

In this case, you should notice that, unlike the previous example, the two occurrences of the variables of the from `cN`

is the same one. Therefore, this can be rewritten as $a_1=t$, $a_2=t^2$, $b_1=0$ and $b_2=0$.

That should answer your question. However, I should also point out a potential problem that you might or might not find when using `solve()`

. The `solve()`

function accepts an `algorithm`

parameter, which can take the values `maxima`

(the default), `sympy`

, `giac`

and `fricas`

. This indicates the Computer Algebra System that you want to use to solve the equation(s). In the case of `maxima`

, `giac`

and `fricas`

, I obtained exactly the same results as you; however, `sympy`

caused me some trouble. Indeed, when calling

```
solve([eq1, eq2, eq3, eq4], a_1, a_2, b_1, b_2, algorithm='sympy')
```

I obtained some results tat are consistent with the ones you obtained. For example, here is part of the output for this case:

```
[{a_1: 0, a_2: 0},
{a_1: 0, a_2: 0},
{a_2: 0, a_1: 0},
{a_2: 0, b_2: 0},
...]
```

The first three are exactly the same as the first solution I mentioned above. The fourth one is the same as the second result you obtained. However, I also got

```
{a_1: 0, b_1: 1/2*b_2, a_2: 0}
```

which is not part of the solutions given by the other algorithms, which can be proven correct by replacing on your original equations. On the other hand, `Sympy`

also gave me this result:

```
{a_1: -1/6*2^(5/6)*sqrt(3*2^(2/3)*(3*I*sqrt(303) + 1499)^(1/3) + 48*2^(1/3) + 624/(3*I*sqrt(303) + 1499)^(1/3)),
```

b_1: 1/2*b_2,
a_2: 1/3*2^(1/3)*(3*I*sqrt(303) + 1499)^(1/3) + 104/3*2^(2/3)/(3*I*sqrt(303) + 1499)^(1/3) + 16/3}

After replacing in the original equations, it would seem that the whole system is satisfied, except for `eq2`

.

2 | No.2 Revision |

Hello, @arneovi! The variables of the form `cN`

, where `N`

is a natural number, are parameters. For example, one of the solutions you obtained is

```
[a_1 == 0, a_2 == 0, b_1 == c2439, b_2 == c2440]
```

This can be rewritten as $a_1=0$, $a_2=0$, $b_1=s$ and $b_2=t$, where $s,t\in\mathbb{R}$. On the other hand, one of the other solutions you obtained is

```
[a_1 == c2443, a_2 == c2443^2, b_1 == 0, b_2 == 0]
```

In this case, you should notice that, unlike the previous example, the two occurrences of the variables of the from `cN`

is the same one. Therefore, this can be rewritten as $a_1=t$, $a_2=t^2$, $b_1=0$ and $b_2=0$.

That should answer your question. However, I should also point out a potential problem that you might or might not find when using `solve()`

. The `solve()`

function accepts an `algorithm`

parameter, which can take the values `maxima`

(the default), `sympy`

, `giac`

and `fricas`

. This indicates the Computer Algebra System that you want to use to solve the equation(s). In the case of `maxima`

, `giac`

and `fricas`

, I obtained exactly the same results as you; however, `sympy`

caused me some trouble. Indeed, when calling

```
solve([eq1, eq2, eq3, eq4], a_1, a_2, b_1, b_2, algorithm='sympy')
```

I obtained some results tat are consistent with the ones you obtained. For example, here is part of the output for this case:

```
[{a_1: 0, a_2: 0},
{a_1: 0, a_2: 0},
{a_2: 0, a_1: 0},
{a_2: 0, b_2: 0},
...]
```

The first three are exactly the same as the first solution I mentioned above. The fourth one is the same as the second result you obtained. However, I also got

```
{a_1: 0, b_1: 1/2*b_2, a_2: 0}
```

which is not part of the solutions given by the other algorithms, which can be proven correct by replacing on your original equations. On the other hand, `Sympy`

also gave me this result:

```
{a_1: -1/6*2^(5/6)*sqrt(3*2^(2/3)*(3*I*sqrt(303) + 1499)^(1/3) + 48*2^(1/3) + 624/(3*I*sqrt(303) + 1499)^(1/3)),
```

b_1: 1/2*b_2,
a_2: 1/3*2^(1/3)*(3*I*sqrt(303) + 1499)^(1/3) + 104/3*2^(2/3)/(3*I*sqrt(303) + 1499)^(1/3) + 16/3}

After replacing in the original equations, it would seem that the whole system is satisfied, except for `eq2`

~~.~~

`full_simplify()`

.) **My suggestion:** Although very useful, the `solve()`

function not always gives the right and/or complete answer. It is always a good idea to check with different algorithms. Even better, if you can solve the system in a reasonable amount of time and with a reasonable amount of effort, it is best to do it so.

I hope this helps!

3 | No.3 Revision |

Hello, @arneovi! The variables of the form `cN`

, where `N`

is a natural number, are parameters. For example, one of the solutions you obtained is

```
[a_1 == 0, a_2 == 0, b_1 == c2439, b_2 == c2440]
```

This can be rewritten as $a_1=0$, $a_2=0$, $b_1=s$ and $b_2=t$, where $s,t\in\mathbb{R}$. On the other hand, one of the other solutions you obtained is

```
[a_1 == c2443, a_2 == c2443^2, b_1 == 0, b_2 == 0]
```

In this case, you should notice that, unlike the previous example, the two occurrences of the variables of the from `cN`

is the same one. Therefore, this can be rewritten as $a_1=t$, $a_2=t^2$, $b_1=0$ and $b_2=0$.

That should answer your question. However, I should also point out a potential problem that you might or might not find when using `solve()`

. The `solve()`

function accepts an `algorithm`

parameter, which can take the values `maxima`

(the default), `sympy`

, `giac`

and `fricas`

. This indicates the Computer Algebra System that you want to use to solve the equation(s). In the case of `maxima`

, `giac`

and `fricas`

, I obtained exactly the same results as you; however, `sympy`

caused me some trouble. Indeed, when calling

```
solve([eq1, eq2, eq3, eq4], a_1, a_2, b_1, b_2, algorithm='sympy')
```

I obtained some results tat are consistent with the ones you obtained. For example, here is part of the output for this case:

```
[{a_1: 0, a_2: 0},
{a_1: 0, a_2: 0},
{a_2: 0, a_1: 0},
{a_2: 0, b_2: 0},
...]
```

The first three are exactly the same as the first solution I mentioned above. The fourth one is the same as the second result you obtained. However, I also got

```
{a_1: 0, b_1: 1/2*b_2, a_2: 0}
```

which is not part of the solutions given by the other algorithms, which can be proven correct by replacing on your original equations. On the other hand, `Sympy`

also gave me this result:

```
{a_1: -1/6*2^(5/6)*sqrt(3*2^(2/3)*(3*I*sqrt(303) + 1499)^(1/3) + 48*2^(1/3) + 624/(3*I*sqrt(303) + 1499)^(1/3)),
b_1: 1/2*b_2,
a_2: 1/3*2^(1/3)*(3*I*sqrt(303) + 1499)^(1/3) + 104/3*2^(2/3)/(3*I*sqrt(303) + 1499)^(1/3) + 16/3}
```

b_1: 1/2*b_2,
a_2: 1/3*2^(1/3)*(3*I*sqrt(303) + 1499)^(1/3) + 104/3*2^(2/3)/(3*I*sqrt(303) + 1499)^(1/3) + 16/3}

After replacing in the original equations, it would seem that the whole system is satisfied, except for `eq2`

. (Here is a SageCell using this exact solution, just in case you feel curious. However, be warned: I had to use polynomial rings in order to obtain numerical approximations, because the mathematical expressions are too complicated for `full_simplify()`

.)

**My suggestion:** Although very useful, the `solve()`

function not always gives the right and/or complete answer. It is always a good idea to check with different algorithms. Even better, if you can solve the system in a reasonable amount of time and with a reasonable amount of effort, it is best to do it so.

I hope this helps!

4 | No.4 Revision |

`cN`

, where `N`

is a natural number, are parameters. For example, one of the solutions you obtained is

```
[a_1 == 0, a_2 == 0, b_1 == c2439, b_2 == c2440]
```

```
[a_1 == c2443, a_2 == c2443^2, b_1 == 0, b_2 == 0]
```

`cN`

is the same one. Therefore, this can be rewritten as $a_1=t$, $a_2=t^2$, $b_1=0$ and $b_2=0$.

`solve()`

. The `solve()`

function accepts an `algorithm`

parameter, which can take the values `maxima`

(the default), `sympy`

, `giac`

and `fricas`

. This indicates the Computer Algebra System that you want to use to solve the equation(s). In the case of `maxima`

, `giac`

and `fricas`

, I obtained exactly the same results as you; however, `sympy`

caused me some trouble. Indeed, when calling

```
solve([eq1, eq2, eq3, eq4], a_1, a_2, b_1, b_2, algorithm='sympy')
```

```
[{a_1: 0, a_2: 0},
{a_1: 0, a_2: 0},
{a_2: 0, a_1: 0},
{a_2: 0, b_2: 0},
...]
```

```
{a_1: 0, b_1: 1/2*b_2, a_2: 0}
```

`Sympy`

also gave me this result:

```
{a_1: -1/6*2^(5/6)*sqrt(3*2^(2/3)*(3*I*sqrt(303) + 1499)^(1/3) + 48*2^(1/3) + 624/(3*I*sqrt(303) + 1499)^(1/3)),
b_1: 1/2*b_2,
a_2: 1/3*2^(1/3)*(3*I*sqrt(303) + 1499)^(1/3) + 104/3*2^(2/3)/(3*I*sqrt(303) + 1499)^(1/3) + 16/3}
```

After replacing in the original equations, it would seem that the whole system is satisfied, except for `eq2`

. (Here is a SageCell using this exact solution, just in case you feel curious. However, be warned: I had to use polynomial rings in order to obtain numerical approximations, because the mathematical expressions are too complicated for `full_simplify()`

.)

**My suggestion:** Although very useful, the `solve()`

function not always gives the right and/or complete answer. It is always a good idea to check with different algorithms. Even better, if you can solve the system by hand in a reasonable amount of time and with a reasonable amount of effort, it is best to do it so.

I hope this helps!

5 | No.5 Revision |

`cN`

, where `N`

is a natural number, are parameters. For example, one of the solutions you obtained is

```
[a_1 == 0, a_2 == 0, b_1 == c2439, b_2 == c2440]
```

```
[a_1 == c2443, a_2 == c2443^2, b_1 == 0, b_2 == 0]
```

`cN`

is the same one. Therefore, this can be rewritten as $a_1=t$, $a_2=t^2$, $b_1=0$ and $b_2=0$.

`solve()`

. The `solve()`

function accepts an `algorithm`

parameter, which can take the values `maxima`

(the default), `sympy`

, `giac`

and `fricas`

. This indicates the Computer Algebra System that you want to use to solve the equation(s). In the case of `maxima`

, `giac`

and `fricas`

, I obtained exactly the same results as you; however, `sympy`

caused me some trouble. Indeed, when calling

```
solve([eq1, eq2, eq3, eq4], a_1, a_2, b_1, b_2, algorithm='sympy')
```

```
[{a_1: 0, a_2: 0},
{a_1: 0, a_2: 0},
{a_2: 0, a_1: 0},
{a_2: 0, b_2: 0},
...]
```

```
{a_1: 0, b_1: 1/2*b_2, a_2: 0}
```

`Sympy`

also gave me this result:

```
{a_1: -1/6*2^(5/6)*sqrt(3*2^(2/3)*(3*I*sqrt(303) + 1499)^(1/3) + 48*2^(1/3) + 624/(3*I*sqrt(303) + 1499)^(1/3)),
b_1: 1/2*b_2,
a_2: 1/3*2^(1/3)*(3*I*sqrt(303) + 1499)^(1/3) + 104/3*2^(2/3)/(3*I*sqrt(303) + 1499)^(1/3) + 16/3}
```

After replacing in the original equations, it would seem that the whole system is satisfied, except for `eq2`

. (Here is a SageCell using this exact solution, just in case you feel curious. However, be warned: I had to use polynomial rings in order to obtain numerical approximations, because the mathematical expressions are too complicated for `full_simplify()`

.)

**My suggestion:** Although very useful, the `solve()`

function not always gives the right and/or complete answer. It is always a good idea to check with different algorithms. Even better, if you can solve the system by hand in a reasonable amount of time and with a reasonable amount of effort, it is best to do it so.

I hope this helps!

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