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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: 1/2b_2, a_2: 1/32^(1/3)(3Isqrt(303) + 1499)^(1/3) + 104/32^(2/3)/(3Isqrt(303) + 1499)^(1/3) + 16/3}

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

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: 1/2b_2, a_2: 1/32^(1/3)(3Isqrt(303) + 1499)^(1/3) + 104/32^(2/3)/(3Isqrt(303) + 1499)^(1/3) + 16/3}

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

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. (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!

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. (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!

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. (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!