Challenge Problem

[jinydu]

:surprised :surprised :surprised

Originally Posted by jinydu

Also, I tried to show that one of the variables must be zero without using the second and third equation.

Instead, I have proven that this is impossible.

Proof by counterexample:

Let A = -1, B = (1+i)/(sqrt 2) = e^(i*pi/4), C = -(1+i)/(sqrt 2) = e^(i*5pi/4), D = 1

Then A = -D and B = -C. Thus, A+D = 0 and B+C = 0. Therefore, A+B+C+D = 0.

Also, A^4 = 1, B^4 = e^(i*pi/4*4) = e^(i*pi) = -1, C^4 = (-B)^4 = B^4 = -1, D^4 = 1

Therefore, A^4 + B^4 + C^4 + D^4 = 1 -1 -1 + 1 = 0

Therefore, we have a nontrivial solution to the system of equations:

A + B + C + D = 0
A^4 + B^4 + C^4 + D^4 = 0

Therefore, any proof that at least one of the variables is 0 must take into account the "intermediate" assumptions (A^2 + B^2 + C^2 + D^2 = 0 and/or A^3 + B^3 + C^3 + D^3 = 0).

[jinydu]

At long last! I have proven the case for n = 4, and along the way, uncovered a new and fascinating problem!

Its not as bad as a completely blind Direct Substitution approach, but it still takes quite a lot of work. Basically, the trick I used was to consider only 3 equations at a time. I thought that doing so could vastly simplify the problem and perhaps identify one or more of the equations are redundant. Instead, I found that every equation is essential to the proof, and that whenever you consider only 3 of the equations, some nontrivial solutions remain. But these nontrivial solutions have a precise parametrization. In this proof, I ignore the 3rd equation until the very end, but keep in mind that if I had ignored a different equation instead, it would have led to different restrictions on the variables.

But beware, this method that I will use is EXTREMELY heavy in algebra. In reality, it is nothing more than a variation of the Direct Substitution method. I'll admit, it isn't the most elegant proof I've ever found. Also note, I don't show every step, because it would be almost impossible to type or read:

(1) A + B + C + D = 0
(2) A^2 + B^2 + C^2 + D^2 = 0
(3) A^3 + B^3 + C^3 + D^3 = 0
(4) A^4 + B^4 + C^4 + D^4 = 0

We will use a proof by contradiction. Assume that there is a solution where at least one of the variables is nonzero. Since nonzero + 0 = nonzero, from equation 1 we can conclude that at least two variables would have to be nonzero. Let's say that A and B are nonzero (it makes no difference, since the equations are symmetrical).

Start by using equation 1:

D = - A - B - C

Substitute into equation 2:

A^2 + B^2 + C^2 + (- A - B - C)^2 = 0

A^2 + B^2 + C^2 + (A^2 + 2AB + B^2 + 2AC + 2BC + C^2) = 0

2A^2 + 2B^2 + 2C^2 + 2AB + 2AC + 2BC = 0

A^2 + B^2 + C^2 + AB + AC + BC = 0

C^2 + (A + B)*C + A^2 + AB + B^2 = 0

This is a quadratic in C. Use the Quadratic Formula:

C = (- A - B)/2 +- sqrt(-3A^2 - 2AB - 3B^2)/2

Substitute this into the equation for D to get:

D = (- A - B)/2 -+ sqrt(-3A^2 - 2AB - 3B^2)/2

Notice that for D, I wrote -+ instead of +-. This is because which sign you choose for D depends on which choice you choose for C (and vice-versa). If you chose + for C, you must choose - for D. If you chose - for C, you must choose + for D.

Next, I use an identity:

(x + sqrt(y)) * (x - sqrt(y)) = x^2 - y

Letting x = (-A - B)/2 and y = -3A^2 - 2AB - 3B^2, we can calculuate C*D. Furthermore, its not hard to show that the value of C*D does not depend on your choice of whether to use (+ for C and - for D) or (- for C and + for D). After some simplifications, you get:

C*D = A^2 + AB + B^2

C^2 * D^2 = (A^2 + AB + B^2)^2

C^2 * D^2 = A^4 + 2(A^3)B + 3(A^2)(B^2) + 2A(B^3) + B^4

Soon, we will make use of the 4th equation. But since the expressions for C and D are so complicated, we derive an identity to simplify it first.

(C^2 + D^2)^2 = C^4 + 2(C^2)(D^2) + D^4 (by the Binomial Theorem)

Rearranging:

C^4 + D^4 = (C^2 + D^2)^2 - 2(C^2)(D^2)

But from equation 2, we can show that C^2 + D^2 = - A^2 - B^2

C^4 + D^4 = (- A^2 - B^2)^2 - 2(C^2)(D^2)

Expanding:

C^4 + D^4 = A^4 + 2(A^2)(B^2) + B^4 - 2(C^2)(D^2)

Now, we substitute it into the fourth equation:

A^4 + B^4 + A^4 + 2(A^2)(B^2) + B^4 - 2(C^2)(D^2) = 0

2A^4 + 2B^4 + 2(A^2)(B^2) - 2(C^2)(D^2) = 0

A^4 + B^4 + (A^2)(B^2) = (C^2)(D^2)

Previously, we calculated the value of C^2 * D^2. We substitute it in now:

A^4 + B^4 + (A^2)(B^2) = A^4 + 2(A^3)B + 3(A^2)(B^2) + 2A(B^3) + B^4

2(A^3)B + 2(A^2)(B^2) + 2A(B^3) = 0

(A^3)B + (A^2)(B^2) + A(B^3) = 0

AB * (A^2 + AB + B^2) = 0

Since we assumed that A and B are nonzero, we are forced to conclude that:

A^2 + AB + B^2 = 0

But earlier on, we proved that

C^2 + (A + B)*C + A^2 + AB + B^2 = 0 (I put it in bold for easy reference)

Now, since A^2 + AB + B^2 = 0, we must have

C^2 + (A + B)*C = 0

Factoring out C, we get:

C * (A + B + C) = 0

But A + B + C = - D (from equation 1):

- C * D = 0

Therefore, either C = 0 or D = 0, or both. Once again, by the symmetry of C and D, it doesn't matter which one we take. We may as well let D = 0.

If we were to continue ignoring Equation 3, we could keep working on the other three equations until we found a parametrization of A, B and C, which I can do in a later post. But since I want to get on with the proof, I will now invoke Equation 3 and finish the proof.

A + B + C = 0
A^2 + B^2 + C^2 = 0
A^3 + B^3 + C^3 = 0
A^4 + B^4 + C^4 = 0

Notice that if we now ignore Equation 4, we have precisely the case of n = 3, which we already determined to have A = B = C = 0 as its only solution. Hence, the only solution for n = 4 is:

A = B = C = D = 0

Q.E.D.

In my next posts, I'll say a bit more about this parametrization, which is an interesting problem in its own right.

[jinydu]

Ok, let me say a bit more about what I mean by parametrization.

Solving all four equations simultaneously is definitely a challenging and interesting problem, but if we ignore one of the equations, the solution set becomes much more "interesting". In fact, I can show that no matter which equation you take away, if you take one away, there are nontrivial solutions. I will discuss what form these solutions must take for each case: Ignoring Equation 1, Ignoring Equation 2, Ignoring Equation 3 and Ignoring Equation 4. Furthermore, I will do it in order of difficulty, from the easiest to the hardest.

It may come as somewhat as a surprise, but the case of ignoring Equation 3 in fact is the simplest one. Without using Equation 3 at all, I already showed in my last post that one of the variables (call it D) is equal to 0. Let's continue and see what happens to A, B and C.

Setting D = 0 in the given equations, we have:

A + B + C = 0
A^2 + B^2 + C^2 = 0
A^4 + B^4 + C^4 = 0

In the previous post, I showed that A^2 + AB + B^2 = 0. We can do something very useful with that.

A^2 = - AB - B^2

Multiply both sides by A

A^3 = - (A^2)*B - A*(B^2)

Now "re-iterate" equation A^2 = - AB - B^2

A^3 = (AB + B^2)*B - A*(B^2)

A^3 = A*(B^2) + B^3 - A*(B^2)

A^3 = B^3

Now that's progress! Multiply both sides by B:

(A^3)*B = B^4

and substitute into Equation 4:

A^4 + (A^3)*B + C^4 = 0

Factor out A^3

(A^3) * (A + B) + C^4 = 0

But from the first equation, A + B = - C

(A^3) * (- C) + C^4 = 0

Factor out C

C * (C^3 - A^3) = 0

Since we don't particularly love trivial solutions, we take C^3 = A^3.

We now have A^3 = B^3 = C^3. Thus, we can set

B = A * r(1)

C = A * r(2)

where r(1) and r(2) are (not necessarily distinct) cube roots of unity.

Substituting this into the first equation:

A + (A * r(1)) + (A * r(2)) = 0

A * (1 + r(1) + r(2)) = 0

Again, since we don't really like trivial solutions, set 1 + r(1) + r(2) = 0.

If we take r(1) = r(2), r(1) = 1 or r(2) = 1, (1 + r(1) + r(2)) can't possibly be 0. This is because if you add a cube root of unity by itself (or equivalently, multiply by 2), you get a number whose absolute value is 2. Adding a third number whose absolute value is 1 can't possibly take you to 0. Thus, the only possible solution is that all three cube roots are different. To summarize, all possible solutions are given by permutations of:

A = A
B = A * e^(i*2pi/3)
C = A * e^(i*4pi/3)
D = 0

Of course, feel free to interchange A, B, C and D.

This concludes the parametrization obtained when ignoring Equation 3.

[jinydu]

The 2nd easiest case comes from ignoring Equation 4. Thus, our equations are:

A + B + C + D = 0
A^2 + B^2 + C^2 + D^2 = 0
A^3 + B^3 + C^3 + D^3 = 0

To save time, we can use some results from previous posts. Any statement that does not rely on Equation 4 is still valid here. In particular, using only equations 1 and 2, we proved that:

C + D = - A - B

C = (- A - B)/2 +- sqrt(-3A^2 - 2AB - 3B^2)/2

D = (- A - B)/2 -+ sqrt(-3A^2 - 2AB - 3B^2)/2

C*D = A^2 + AB + B^2

Again, attempting to cube C and D is extremely tedious, so we use some manipulations to reduce the algebra needed.

(C + D)^3 = C^3 + 3*(C^2)*D + 3*C*(D^2) + D^3

(C + D)^3 = C^3 + D^3 + 3CD*(C + D)

C^3 + D^3 = (C + D)^3 - 3CD*(C + D)

C^3 + D^3 = (C + D)*((C + D)^2 - 3CD)

Using our previously derived results:

C^3 + D^3 = (- A - B)*((- A - B)^2 - 3(A^2 + AB + B^2))

C^3 + D^3 = (- A - B)*(A^2 + 2AB + B^2 - 3A^2 - 3AB - 3B^2)

C^3 + D^3 = (- A - B)*(-2A^2 -AB - 2B^2)

C^3 + D^3 = (A + B)*(2A^2 + AB + 2B^2)

C^3 + D^3 = (2A^3 + (A^2)*B + 2A(B^2)) + (2(A^2)*B + A*(B^2) + 2B^3)

C^3 + D^3 = 2A^3 + 3(A^2)*B + 3A*(B^2) + 2B^3

A^3 + B^3 + C^3 + D^3 = 3A^3 + 3(A^2)*B + 3A*(B^2) + 3B^3

Now, apply Equation 3 by setting the above equation equal to 0

3A^3 + 3(A^2)*B + 3A*(B^2) + 3B^3 = 0

A^3 + (A^2)*B + A*(B^2) + B^3 = 0

A cubic in 2 variables. This may look intimidating, but we can use a trick here. Since we're really only interested in nontrivial solutions, we allow dividing by variables.

Define r = B/A. Therefore, B = Ar

A^3 + (A^2)*Ar + A*(Ar)^2 + (Ar)^3 = 0

A^3 + (A^3)*r + (A^3)*r^2 + (A^3)*r^3 = 0

Factor out A^3

A^3 * (1 + r + r^2 + r^3) = 0

Since we don't generally want A = 0, we have:

1 + r + r^2 + r^3 = 0

The left-hand side is a geometric series. The sum is:

(1 - r^4)/(1 - r)

Set the numerator equal to 0

1 - r^4 = 0

r^4 = 1

r = 1, i, -1, -i

We're almost there! By the symmetry of the original equations, we can now write:

B = A*r(1)
C = A*r(2)
D = A*r(3)

where r(1), r(2) and r(3) are (not necessarily distinct) fourth roots of unity. All we have to do is prove that they are in fact distinct.

By the first equation:

A + A*r(1) + A*r(2) + A*r(3) = 0

Divide through by A to get only nontrivial solutions:

1 + r(1) + r(2) + r(3) = 0

Using similar considerations for Equations 2 and 3, we get:

1 + r(1)^2 + r(2)^2 + r(3)^2 = 0

1 + r(1)^3 + r(2)^3 + r(3)^3 = 0

Immediately, we can knock out 2 possibilities. r(1), r(2) and r(3) can't all be the same, because otherwise:

1 + 3*r(1) = 0

3*r(1) = -1

r(1) = -1/3

But -1/3 is not a fourth root of unity. Hence, by contradiction, r(1), r(2) and r(3) can't all be equal.

Since all three roots can't be the same, either two are the same and 1 is different, or all 3 are different. We can rule out the first possibility by trying all the possible cases.

r(1) = r(2) = 1
1 + 1 + 1 + r(3) = 0
r(3) = -3
Impossible because -3 is not a fourth root of unity

r(1) = r(2) = i
1 + i + i + r(3) = 0
r(3) = -1 -2i
Impossible because -1 -2i is not a fourth root of unity

r(1) = r(2) = -1
1 - 1 - 1 + r(3) = 0
r(3) = 1
This is plausible, but substituting this into the equation 1 + r(1)^2 + r(2)^2 + r(3)^2 = 0 gives the impossible condition:
4 = 0

r(1) = r(2) = -i
1 -i -i + r(3) = 0
r(3) = -1 +2i
Impossible because -1 +2i is not a fourth root of unity

Therefore, all three roots must be different. We can show that none of them can equal 1. A quick proof by contradiction:

Assume r(1) = 1
1 + 1 + r(2) + r(3) = 0
r(2) + r(3) = -2
We are thus forced to conclude that r(2) = r(3) = -1. But we already showed that r(2) and r(3) can't be equal (in fact, we disproved this exact case). This is the needed contradiction.

Therefore, r(1), r(2) and r(3) must all be different and none of them can be 1. The only remaining choice is that one is i, one is -1 and the other is -i. In fact, you can show that:

1 + i - 1 - i = 0
1^2 + i^2 + (-1)^2 + (-i)^2 = 0
1^3 + i^3 + (-1)^3 + (-i)^3 = 0
(Note that if you raise them to their fourth powers and add, you get 4, not 0).

In summary, we have proved that the solution to the case of ignoring Equation 4 is:

A = A
B = Ai
C = -A
D = -Ai

Again, you are free to permute A, B, C and D.

[jinydu]

The next step up in difficulty comes from ignoring Equation 2. Perhaps it shouldn't be too surprising that this set of equations is quite difficult to solve. Equation 2 can always be solved for any one of the variables using the Quadratic Formula, which is easy compared to the other algebra in the problem. Here are the set of equations in this case.

A + B + C + D = 0
A^3 + B^3 + C^3 + D^3 = 0
A^4 + B^4 + C^4 + D^4 = 0

As always, we start with Equation 1. But this time, we rearrange it differently (so as to avoid cubing a trinomial).

A + B = - C - D

Cubing both sides:

A^3 + 3(A^2)*B + 3A*(B^2) + B^3 = - C^3 - 3(C^2)*D - 3C*(D^2) - D^3

A^3 + B^3 + C^3 + D^3 = - 3(A^2)*B - 3A*(B^2) - 3(C^2)*D - 3C*(D^2)

By Equation 3, the left-hand side is equal to 0. Divide the right-hand side by -3.

(A^2)*B + A*(B^2) + (C^2)*D + C*(D^2) = 0

Factor AB from the first two terms and CD from the last two terms:

AB*(A + B) + CD*(C + D) = 0

But C + D = - A - B, so

AB*(A + B) - CD*(A + B) = 0

Factor (A + B) from both terms

(A + B) * (AB - CD) = 0

Therefore, either A = -B or AB = CD. We deal with both of them seperately. First, temporarily assume that A = -B

From the first equation:

-B + B + C + D = 0

C + D = 0

C = -D

If, in the equation A = -B, we raise both sides to the 4th power, we get A^4 = B^4. Similarly, C = -D implies that C^4 = D^4.

Substituting this into the fourth equation:

2A^4 + 2C^4 = 0

A^4 + C^4 = 0

and if we switch B^4 for A^4 and D^4 for C^4, we have:

B^4 + D^4 = 0

In fact, the letters themselves don't matter, since the equations are symmetrical. The important conclusion is that for any possible solution, (A, B, C, D), it is possible to choose two pairs of variables, such that the sum of the fourth powers of each variables within each pair is equal to 0.

Now, let's forget our assumption that A = -B and instead adopt the other assumption, that AB = CD. As you might expect, the analysis of this case is more complicated.

From Equation 1, we have A + B = - C - D. Raising both sides to the fourth power using the Binomial Theorem:

A^4 + 4(A^3)*B + 6(A^2)*(B^2) + 4A*(B^3) + B^4 = C^4 + 4(C^3)*D + 6(C^2)*(D^2) + 4C*(D^3) + D^4

Now, we apply the assumption that AB = CD. Everytime we find CD on the right-hand side, we replace it with AB:

A^4 + 4(A^3)*B + 6(A^2)*(B^2) + 4A*(B^3) + B^4 = C^4 + 4(C^2)*(AB) + 6(A^2)*(B^2) + 4(D^2)*(AB) + D^4

6(A^2)*(B^2) can be cancelled out of both sides.

Factor AB out of both the left and right-hand sides

A^4 + AB*(4A^2 + 4B^2) + B^4 = C^4 + AB*(4C^2 + 4D^2) + D^4

A^4 + 4AB*(A^2 + B^2) + B^4 = C^4 + 4AB*(C^2 + D^2) + D^4

A^4 + 4AB*(A^2 + B^2 - C^2 - D^2) + B^4 = C^4 + D^4

But C^4 + D^4 = - A^4 - B^4 (by Equation 4). We move it to the left-hand side

2A^4 + 4AB*(A^2 + B^2 - C^2 - D^2) + 2B^4 = 0

A^4 + 2AB*(A^2 + B^2 - C^2 - D^2) + B^4 = 0

Let's put that on hold for a second. Go back to the equation A + B = - C - D and square both sides:

A^2 + 2AB + B^2 = C^2 + 2CD + D^2

But since we've assume that CD = AB:

A^2 + 2AB + B^2 = C^2 + 2AB + D^2

Cancelling out 2AB from both sides:

A^2 + B^2 = C^2 + D^2

A^2 + B^2 - C^2 - D^2 = 0

Now we say that the middle term in the bolded equation is equal to 0. Therefore:

A^4 + B^4 = 0

Plugging that into Equation 4, we also find:

C^4 + D^4 = 0

We thus come to the same conclusion that we got when we assumed that A = -B. We can now forget which variable to the fourth power + which other variable to the fourth power = 0, and just say that:

"The important conclusion is that for any possible solution, (A, B, C, D), it is possible to choose two pairs of variables, such that the sum of the fourth powers of each variables within each pair is equal to 0."

You may think that most of the work is now done. Unfortuately, this is far from the case. The next step is to parametrize the solutions. We can use A as our parameter, as usual. Let's take A^4 + B^4 = 0.

Set r = B/A, so B = Ar

A^4 + (Ar)^4 = 0

A^4 * (1 + r^4) = 0

Since we don't like trivial solutions, we set 1 + r^4 = 0

r^4 = - 1

r = e^(i*pi/4), e^(i*3pi/4), e^(i*5pi/4) or e^(i*7pi/4)

Similarly, C^4 + D^4 = 0. By the same reasoning, we have:

k = e^(i*pi/4), e^(i*3pi/4), e^(i*5pi/4) or e^(i*7pi/4)

where D = Ck

Now, we want to find the relationship between A, C and D. Substituting what we know into the Equation 1:

A + Ar = - (C + Ck)

A * (1 + r) = -C * (1 + k)

C = - (A * (1 + r))/(1 + k)

D = Ck = - (Ak * (1 + r))/(1 + k)

We now have parametric equations for B, C and D in terms of A. Unfortunately, we're still not finished because we don't know which combinations of r and k work.

We can reduce the amount of remaining work by expressing k in terms of r. If we calculate k/r, we find that k/r has only four possible values: 1, i, -1, and -i. Thus is far easier to see if graph the possible values of r and k on a complex plane. Notice that they all lie on the unit circle. If each possible value of r or k is represented as a point, the angular distance between any two points is always a multiple of pi/2. Rotating by pi/2 is equivalent to multiplying by i. Doing this rotation again and again is the same as multiplying by i repeatedly, yielding the sequence: 1, i, -1, -i, 1, i, -1, -i, 1, etc. Notice that the sequence repeats and that the only values that appear in the sequence are 1, i, -1 and -i. Hence, the only possible values of k/r are 1, i, -1 and -i. We can therefore write:

k = r, k = ir, k = -r or k = -ir.

Substituting this into our parametric equations gives four possible sets of parametric equations:

A = A
B = Ar
C = -A
D = -Ar

A = A
B = Ar
C = -A * ((1 + r)/(1 + ir))
D = -A * (ir *(1 + r)/(1 + ir))

A = A
B = Ar
C = -A * ((1 + r)/(1 - r))
D = A * (r * (1 + r)/(1 - r))

A = A
B = Ar
C = -A * ((1 + r)/(1 - ir))
D = A * (ir * (1 + r)/(1 - ir))

Since there are four possible values of r and four possible parametric equations, there are 16 different combinations to test. We know that all of these possible solutions satisfy Equation 1, but we still have to test which ones satisfy Equations 2 and Equation 3. This is an exhausting, but fortunately finite task. I'll leave it for the next post...

[Zeta-Flux]

jinydu,

It will be true that you need all of the equations because of the symmetric polynomial argument. If you leave one of your equations out, that means one of the symmetric polynomials does not have to be zero. And so A,B,C,D can satisfy a non-trivial polynomial. :) It shouldn't be too hard to show that leaving out one polynomial gives you a 1-dimensional solution set.

[jinydu]

Thanks for the reply Zeta-Flux. But wouldn't you agree that it is interesting to ask for the precise description of that 1-dimensional set?

Now, I will continue with my promised verification.

------------------------------------------------------------------------

Ok, the first four choices, with k = r, are by far the easiest ones. Again:

A = A
B = Ar
C = -A
D = -Ar

Compute the cubes:

A^3 = A^3
B^3 = A^3 * r^3
C^3 = -A^3
D^3 = -A^3 * r^3

Clearly, A^3 + C^3 = 0 and B^3 + D^3 = 0. Therefore, A^3 + B^3 + C^3 + D^3 = 0

Compute the fourth powers (and remember that r^4 = -1):

A^4 = A^4
B^4 = A^4 * r^4 = -A^4
C^4 = (-A)^4 = A^4
D^4 = (-Ar)^4 = A^4 * r^4 = -A^4

A^4 + B^4 = 0 and C^4 + D^4 = 0. Therefore, A^4 + B^4 + C^4 + D^4 = 0.

Therefore, the first set of parametric equations is good regardless of which of the four values we choose for r. Plugging in the specific values of r, we get four parametric equations:

For r = e^(i*pi/4)
A = A
B = A * e^(i*pi/4)
C = -A
D = -A * e^(i*pi/4) = A * e^(i*5pi/4)

For r = e^(i*3pi/4)
A = A
B = A * e^(i*3pi/4)
C = -A
D = -A * e^(i*3pi/4) = A * e^(i*7pi/4)

For r = e^(i*5pi/4)
A = A
B = A * e^(i*5pi/4)
C = -A
D = -A * e^(i*5pi/4) = A * (i*pi/4)

For r = e^(i*7pi/4)
A = A
B = A * e^(i*7pi/4)
C = -A
D = -A * e^(i*7pi/4) = A * (i*3pi/4)

Notice that the last two solutions are just permutations of the first two. Since the variables are symmetric then, they are essentially the same, and we can forget about them. Thus, we in fact have only two parametric solutions so far:

A = A
B = A * e^(i*pi/4)
C = -A
D = -A * e^(i*pi/4) = A * e^(i*5pi/4)

A = A
B = A * e^(i*3pi/4)
C = -A
D = -A * e^(i*3pi/4) = A * e^(i*7pi/4)

--------------------------------------------------------------------------

Next, we let k = ir

A = A
B = Ar
C = -A * ((1 + r)/(1 + ir))
D = -A * (ir *(1 + r)/(1 + ir)) = C*ir

In this case, let's check for the 4th Equation first:

A^4 = A^4
B^4 = A^4 * r^4 = -A^4
C^4 = C^4 (I could write this in terms of A, but that would be tiresome to compute)
D^4 = (C*ir)^4 = C^4 * i^4 * r^4 = C^4 * 1 * -1 = -C^4

So, A^4 + B^4 = 0 and C^4 + D^4 = 0. Therefore, A^4 + B^4 + C^4 + D^4 = 0.

Now for the hard part, checking which values of r are consistent with the 3rd Equation. As you can see, the formula for C in terms of A is quite complicated, so trying to calculate C^3 directly, using that expression, would be horrendously complicated, and actually performing the division would be even worse. Fortunately, I found a much more efficient method.

Start from Equation 1, and use the basic substitutions:

A + Ar + C + iCr = 0

We can solve for r

r*(A + iC) = - A - C

r = (- A - C)/(A + iC)

So far, nothing exciting there, since it is (currently) difficult to write C in terms of A. But we're going to do something unexpected.

Back to our Equation 1, with the substitutions.

A + Ar + C + iCr = 0

This time, let's try something different. Move the last two terms to the other side

A + Ar = - C - iCr

Group terms:

A*(1 + r) = -C(1 + ir)

And cube both sides

(A^3)*(1 + 3r + 3r^2 + r^3) = (-C^3)*(1 + 3ir - 3r^2 - ir^3)

Now distribute again (grouping was just a trick to make cubing a little less mistake-prone):

A^3 + 3(A^3)r + 3(A^3)(r^2) + (A^3)(r^3) = -C^3 - 3(C^3)ir + 3(C^3)r + (C^3)ir

Rearrange terms:

A^3 + (A^3)(r^3) + C^3 - (C^3)ir = -3(A^3)r - 3(A^3)(r^2) - 3(C^3)ir + 3(C^3)(r^2)

By Equation 3, the left-hand side equals 0. Therefore:

-3(A^3)r - 3(A^3)(r^2) - 3(C^3)ir + 3(C^3)(r^2) = 0

Divide through by -3

(A^3)r + (A^3)(r^2) + (C^3)ir - (C^3)(r^2) = 0

Factor r and r^2

(r^2)*(A^3 - C^3) + r*(A^3 + iC^3) = 0

Divide through by r, and solve for r:

r*(A^3 - C^3) + (A^3 + iC^3) = 0

r*(A^3 - C^3) = -A^3 - iC^3

r = (-A^3 - iC^3)/(A^3 - C^3)

But earlier on, we already derived an expression for r (I put it in bold for easier reference). We can now equate this expression with that one:

(-A^3 - iC^3)/(A^3 - C^3) = (- A - C)/(A + iC)

Multiply both sides by -1 to get rid of the negative signs

(A^3 + iC^3)/(A^3 - C^3) = (A + C)/(A + iC)

Cross-multiply, and then expand

(A^3 + iC^3)*(A + iC) = (A + C)*(A^3 - C^3)

A^4 + iA(C^3) + i(A^3)C - C^4 = A^4 + (A^3)C - A(C^3) - C^4

Cancel A^4 - C^4 out from both sides

iA(C^3) + i(A^3)C = (A^3)C - A(C^3)

Dividing by AC

iC^2 + iA^2 = A^2 - C^2

At last! A simple equation that matches the (apparent) simplicity of the original problem. Now, just divide by A^2

i*((C^2)/(A^2)) + i = 1 - ((C^2)/(A^2))

(1 + i)*((C^2)/(A^2)) = 1 - i

(C/A)^2 = (1 - i)/(1 + i)

(C/A)^2 = -i = e^(i*3pi/2)

C/A = e^(i*3pi/4) or e^(i*7pi/4)

C = Ae^(i*3pi/4) or Ae^(i*7pi/4)

Earlier, we derived this equation for r:

r = (- A - C)/(A + iC)

Substituting in the values for C:

r = (- A - Ae^(i*3pi/4))/(A + i * Ae^(i*3pi/4)) or (- A - Ae^(i*7pi/4))/(A + i * Ae^(i*7pi/4))

Dividing the numerator and denominator by A:

r = (-1 - e^(i*3pi/4))/(1 + i * e^(i*3pi/4)) or (-1 - e^(i*7pi/4))/(1 + i * e^(i*7pi/4))

r = (-1 - e^(i*3pi/4))/(1 + e^(i*5pi/4)) or (-1 - e^(i*7pi/4))/(1 + e^(i*pi/4))

Let’s look at the first possibility (with i*3pi/4). Expand using Euler’s Equation:

r = (-1 - cos 3pi/4 - i*sin 3pi/4))/(1 + cos 5pi/4 + i*sin 5pi/4))

Now, we use the exact values for these cosines and sines:

cos 3pi/4 = -1/sqrt(2), sin 3pi/4 = 1/sqrt(2)

cos 5pi/4 = -1/sqrt(2), sin 5pi/4 = -1/sqrt(2)

r = (-1 + 1/sqrt(2) - i/sqrt(2))/(1 –1/sqrt(2) - i/sqrt(2))

Multiplying numerator and denominator by sqrt(2):

r = (1 - sqrt(2) - i)/(-1 + sqrt(2) - i)

Now, multiply numerator and denominator by (-1 + sqrt(2) + i):

r = (2sqrt(2) - 2 + i*(2 - 2sqrt(2))/(4 - 2sqrt(2))

Divide through by 2:
r = (sqrt(2) - 1 + i*(1 - sqrt(2))/(2 - sqrt(2))

r = (1/sqrt(2)) * ((2 - sqrt(2))/(2 – sqrt(2)) + (1/sqrt(2)) * i((sqrt(2) - 2)/(2 - sqrt(2))

r = (1/sqrt(2)) * 1 + (1/sqrt(2)) * -i

r = 1/sqrt(2) - i/sqrt(2)

r = cos (7pi/4) + i*sin (7pi/4)

r = e^(i*7pi/4)

Thus:

B = Ar = Ae^(i*7pi/4)

D = C*ir = C * i * e^(i*7pi/4) = C * e^(i*pi/4)

D = (Ae^(i*3pi/4)) * e^(i*pi/4)

D = Ae^(i*pi) = -A

Now, we look at the other possibility: r = (-1 - e^(i*7pi/4))/(1 + e^(i*pi/4)). Again, expanding using Euler’s Formula:

r = (-1 - (cos (7pi/4) + i*sin (7pi/4)))/(1 + (cos (pi/4) + i*sin (pi/4)))

r = (-1 - cos (7pi/4) - i*sin (7pi/4))/(1 + cos (pi/4) + i*sin (pi/4))

Using the exact values for the cosine and sine of 7pi/4 and pi/4:

r = (-1 - 1/sqrt(2) + i/sqrt(2))/(1 + 1/sqrt(2) + i/sqrt(2))

Multiplying numerator and denominator by sqrt(2):

r = (-sqrt(2) - 1 + i)/(sqrt(2) + 1 + i)

Now, multiply numerator and denominator by (sqrt(2) + 1 - i):

r = (-2 - 2sqrt(2) + i*(2sqrt(2) + 2))/(4 + 2sqrt(2))

Dividing by numerator and denominator by 2:

r = (-1 - sqrt(2) + i*(sqrt(2) + 1))/(2 + sqrt(2))

r = (-1/sqrt(2)) * ((sqrt(2) + 2)/(2 + sqrt(2)) + (-1/sqrt(2)) * i(-2 -sqrt(2))/(2 + sqrt(2))

r = (-1/sqrt(2)) * 1 + (-1/sqrt(2)) * -i

r = -1/sqrt(2) + i/sqrt(2)

r = cos (3pi/4) + i*sin (3pi/4)

r = e^(i*3pi/4)

Thus:

B = Ar = Ae^(i*3pi/4)

D = C * i * r = C * i * e^(i*3pi/4)

D = C * e^(i*5pi/4) = Ae^(i*7pi/4) * e^(i*5pi/4)

D = Ae^(i*3pi) = -A

Thus, our two parametric solutions for k = ir are:

A = A
B = Ae^(i*7pi/4)
C = Ae^(i*3pi/4)
D = -A

and

A = A
B = Ae^(i*3pi/4)
C = Ae^(i*7pi/4)
D = -A

But alas, these two solutions are just permutations of each other! Even worse, you’ll find that they are both permutations of a solution we derived for the case of k = r, which I’ll repeat below:

A = A
B = A * e^(i*3pi/4)
C = -A
D = A * e^(i*7pi/4)

This means that all the work since the last dotted line has failed to generate any new solutions!

Oh well, better luck with k = -r and k = -ir, in the next post.

[jinydu]

Before I continue, let me point out an error in one of my previous posts.

Originally Posted by jinydu

but we still have to test which ones satisfy Equations 2 and Equation 3.

I should have instead said Equation 3 and Equation 4. Remember, we're ignoring Equation 2. With that in mind, I'll continue:

-------------------------------------------------------------------

Next, we take k = -r. Thus, we have the equations:

A = A
B = Ar
C = -A * ((1 + r)/(1 - r))
D = A * (r * (1 + r)/(1 - r)) = -C * r

Start by checking consistency with equation 4:

A^4 = A^4
B^4 = A^4 * r^4 = -A^4
C^4 = C^4
D^4 = (-C)^4 * r^4 = C^4 * -1 = -C^4

Evidently, A^4 + B^4 = 0 and C^4 + D^4 = 0. Therefore, A^4 + B^4 + C^4 + D^4 = 0. Thus, all possible values of r are consistent with equation 4.

In order for Equation 3 to be satisfied, we must have:

A^3 + (Ar)^3 + C^3 + (-Cr)^3 = 0

Expanding, and doing further manipulations:

A^3 + (A^3)(r^3) + C^3 - (C^3)(r^3) = 0

(A^3) * (1 + r^3) + (C^3) * (1 - r^3) = 0

(A^3) * (1 + r^3) = (C^3) * (r^3 - 1)

(C^3)/(A^3) = (r^3 + 1)/(r^3 - 1)

But we already know that:

C/A = - (1 + r)/(1 - r)

or

(C^3)/(A^3) = - (1 + 3r + 3r^2 + r^3)/(1 - 3r + 3r^2 - r^3)

(C^3)/(A^3) = (r^3 + 3r^2 + 3r + 1)/(r^3 - 3r^2 + 3r - 1)

Equating, these two expressions for (C^3)/(A^3):

(r^3 + 1)/(r^3 - 1) = (r^3 + 3r^2 + 3r + 1)/(r^3 - 3r^2 + 3r - 1)

Cross-multiplying:

(r^3 + 1)*(r^3 - 3r^2 + 3r - 1) = (r^3 - 1)*(r^3 + 3r^2 + 3r + 1)

r^6 - 3r^5 + 3r^4 - r^3 + r^3 - 3r^2 + 3r - 1 = r^6 + 3r^5 + 3r^4 + r^3 - r^3 - 3r^2 - 3r -1

-3r^5 + 3r^4 - 3r^2 + 3r - 1 = 3r^5 + 3r^4 - 3r^2 - 3r - 1

6r^5 - 6r = 0

Dividing both sides by 6r

r^4 - 1 = 0

r^4 = 1

But earlier, we already deduced that r^4 = -1. This is a contradiction. Hence, there are no solutions to the case where k = -r.

---------------------------------------------------------------------

This leaves us with the last possibility, k = -ir. Our equations are:

A = A
B = Ar
C = -A * ((1 + r)/(1 - ir))
D = A * (ir * (1 + r)/(1 - ir)) = C * -ir

Its not hard to show that this is always consistent with Equation 4:

A^4 = A^4
B^4 = A^4 * r^4 = -A^4
C^4 = C^4
D^4 = C^4 * (-ir)^4 = -C^4

Hence, A^4 + B^4 = 0 and C^4 + D^4 = 0, so A^4 + B^4 + C^4 + D^4 = 0 and Equation 4 is satisfied.
We now need to check which values of r are consistent with Equation 3, but as you’ve probably realized by now, this is where most of the work is. This looks so tantalizingly similar to the case of k = -ir (not surprising, since i and -i are both square roots of -1, which is how i is defined), that it seems that there should be a way to use our knowledge of the solutions to that case to help us with this one. Unfortunately, I haven't been able to find a way to do this; they're close, but not quite close enough. In any case:

Using our expressions for C and A, calculate C/A:

C/A = - ((1 + r)/(1 - ir))

Cubing both sides:

(C^3)/(A^3) = - ((1 + 3r + 3r^2 + r^3)/(1 - 3ir - 3r^2 + ir^3))

And by Equation 3:

A^3 + (A^3)*(r^3) + C^3 + i*(C^3)*(r^3) = 0

(A^3) * (1 + r^3) + (C^3) * (1 + ir^3) = 0

(C^3) * (1 + ir^3) = - (A^3) * (1 + r^3)

(C^3)/(A^3) = - ((1 + r^3)/(1 + ir^3))

Equating both expressions for (C^3)/(A^3):

- ((1 + r^3)/(1 + ir^3)) = - ((1 + 3r + 3r^2 + r^3)/(1 - 3ir - 3r^2 + ir^3))

Cross-multiplying and getting rid of the negative signs:

(r^3 + 1) * (ir^3 - 3r^2 - 3ir + 1) = (ir^3 + 1) * (r^3 + 3r^2 + 3r + 1)

ir^6 - 3r^5 - 3ir^4 + r^3 + ir^3 - 3r^2 - 3ir + 1 = ir^6 + 3ir^5 + 3ir^4 + ir^3 + r^3 + 3r^2 + 3r + 1

(3 + 3i)r^5 + (6i)r^4 + 6r^2 + (3 + 3i)r = 0

But r^4 = -1 and r^5 = -r, so

6r^2 - 6i = 0

r^2 - i = 0

r^2 = i

r = e^(i*pi/4) or e^(i*5pi/4)

Therefore:

C/A = - (1 + e^(i*pi/4))/(1 - i*e^(i*pi/4)) or - (1 + e^(i*5pi/4))/(1 - i*e^(i*5pi/4))

We take a closer look at the first one (involving pi/4):

C/A = - (1 + e^(i*pi/4))/(1 - i*e^(i*pi/4))

C/A = - (1 + e^(i*pi/4))/(1 - e^(i*3pi/4))

Expanding using Euler's Equation:

C/A = (-1 - cos pi/4 - i*sin pi/4))/(1 - cos (3pi/4) - i*sin (3pi/4))

Using the exact values for the cosine and sine of pi/4 and 3pi/4:

C/A = (-1 - 1/sqrt(2) - i/sqrt(2))/(1 + 1/sqrt(2) - i/sqrt(2))

Multiplying numerator and denominator by sqrt(2):

C/A = (-sqrt(2) - 1 - i)/(sqrt(2) + 1 - i)

Multiplying numerator and denominator by (sqrt(2) + 1 + i):

C/A = (-2 - 2sqrt(2) + i*(-2 - 2sqrt(2)))/(4 + 2sqrt(2))

Dividing numerator and denominator by 2:

C/A = (-1 - sqrt(2) + i*(-1 - sqrt(2)))/(2 + sqrt(2))

C/A = (-1 - sqrt(2))/(2 + sqrt(2)) + i*(-1 - sqrt(2))/(2 + sqrt(2))

To find the value of (-1 - sqrt(2))/(2 + sqrt(2)), multiply numerator and denominator by (2 - sqrt(2))

(-1 - sqrt(2))/(2 + sqrt(2)) = (- sqrt(2))/2 = -1/sqrt(2).

Therefore:

C/A = -1/sqrt(2) - i/sqrt(2)

C/A = e^(i*5pi/4)

C = Ae^(i*5pi/4)

B = Ar = Ae^(i*pi/4)

D = -iCr = Ae^(i*12pi/4) = Ae^(i*pi) = -A

Now, we look at the second possibility (5pi/4):

C/A = - (1 + e^(i*5pi/4))/(1 - i*e^(i*5pi/4))

C/A = (-1 - e^(i*5pi/4))/(1 - e^(i*7pi/4))

C/A = (-1 - (cos (5pi/4) + i*sin (5pi/4)))/(1 - (cos (7pi/4) + i*sin (7pi/4)))

C/A = (-1 - cos (5pi/4) - i*sin (5pi/4))/(1 - cos (7pi/4) - i*sin (7pi/4))

Using the exact values for the cosine and sine of 5pi/4 and 7pi/4:

C/A = (-1 + 1/sqrt(2) + i/sqrt(2))/(1 - 1/sqrt(2) + i/sqrt(2))

Multiply numerator and denominator by sqrt(2):

C/A = (-sqrt(2) + 1 + i)/(sqrt(2) - 1 + i)

Multiply numerator and denominator by (sqrt(2) - 1 - i):

C/A = (-2 + 2sqrt(2) + i*(-2 + 2sqrt(2))/(4 - 2sqrt(2))

Divide numerator and denominator by 2:

C/A = (-1 + sqrt(2) + i*(-1 + sqrt(2))/(2 - sqrt(2))

C/A = (-1 + sqrt(2))/(2 - sqrt(2)) + i*(-1 + sqrt(2))/(2 - sqrt(2))

To find the value of (-1 + sqrt(2))/(2 - sqrt(2)), multiply numerator and denominator by (2 + sqrt(2))

(-1 + sqrt(2))/(2 - sqrt(2)) = sqrt(2)/2 = 1/sqrt(2)

C/A = 1/sqrt(2) + i/sqrt(2)

C/A = e^(i*pi/4)

C = Ae^(i*pi/4)

B = Ar = Ae^(i*5pi/4)

D = -iCr = Ae^(i*12pi/4) = Ae^(i*pi) = -A

Therefore, our two parametric solutions are:

A = A
B = Ae^(i*pi/4)
C = Ae^(i*5pi/4)
D = -A

and

A = A
B = Ae^(i*5pi/4)
C = Ae^(i*pi/4)
D = -A

But once again, you'll see that these two solutions are, once again, just permutations of each other! And not only that, they are both permutations of a solution we derived in the case k = r (repeated below again);

A = A
B = A * e^(i*pi/4)
C = -A
D = -A * e^(i*pi/4) = A * e^(i*5pi/4)

Thus, it can be seen that all of our efforts after k = r have completely failed to generate any new solutions at all!

---------------------------------------------------------------------

In summary, we have two parametric sets of equations that completely describe all possible solutions obtained by ignoring Equation 1:

A = A
B = Ae^(i*pi/4)
C = -A
D = Ae^(i*5pi/4)

A = A
B = Ae^(i*3pi/4)
C = -A
D = Ae^(i*7pi/4)

In fact, the set of solutions is even more unified than these two sets of equations would lead one to believe. If you were to draw a graph of the first solution set on a complex plane, then rotate all four points pi/4 radians clockwise about the origin, you would get the graph of the second solution set! Furthermore, it is possible to describe the solutions even more concisely:

A = A
B = Aiz
C = -A
D = -Aiz
where z = e^(i*pi/4) or e^(i*-pi/4) and you must choose the same value of z in both instances.

Of couse, you're allowed to permute A, B, C and D.

[jinydu]

Ok, I'll confess, the final, and most difficult case has me stumped, at least for now: Ignoring Equation 1. This gives the system of equations:

(2) A^2 + B^2 + C^2 + D^2 = 0
(3) A^3 + B^3 + C^3 + D^3 = 0
(4) A^4 + B^4 + C^4 + D^4 = 0

The method I used to attack this problem is essentially the same as that used for the other 3 cases: Eliminate C and D by expressing them in terms of A and B. But this time, the amount of nonrepetitive calculations is greater than all three other cases combined.

From Equation 2:

D^2 = -A^2 - B^2 - C^2

D = +- sqrt(-A^2 - B^2 - C^2)

Plugging this into Equation 4 (so as to cancel out the square root, thus simplifying the problem) gives:

A^4 + B^4 + C^4 + (A^2 + B^2 + C^2)^2 = 0

Expanding that out and simplifying ultimately gives:

A^4 + (A^2)(B^2) + (A^2)(C^2) + B^4 + (B^2)(C^2) + C^4 = 0

I can think of this as a quadratic in C^2, which I can solve with the quadratic formula to get:

C^2 = (-A^2 - B^2 +- sqrt(-3A^2 - 2(A^2)(B^2) - 3B^4))/2

C = +- sqrt((-A^2 - B^2 +- sqrt(-3A^2 - 2(A^2)(B^2) - 3B^4))/2)

And plugging that into the equation for D gives:

D = +- sqrt((-A^2 - B^2 -+ sqrt(-3A^2 - 2(A^2)(B^2) - 3B^4))/2)

This is almost exactly the same, except -+ instead of +- inside the first square root. The reason for this is that the choice of +- for C and D is not independent. If + is chosen for C, - must be chosen for D, and vice-versa. Thus, for each fixed value of C, there are only two possible values for D, not 4.

Now comes the most tiresome part yet, substituting these values into the third equation. Computing C^3 + D^3 would be almost impossible in practice (at least by hand), so I used the identities:

C^3 + D^3 = (C + D) * (C^2 - CD + D^2) = (C + D)*(C^2 + D^2) - (C + D)*(CD)

and the fact that C^2 + D^2 = -A^2 - B^2 (by Equation 2).

After lots of simplifying, and twice squaring both sides, I got the stunning equation:

3A^12 + 6(A^10)(B^2) + 4(A^9)(B^3) + 9(A^8)(B^4) + 12(A^7)(B^5) + 4(A^6)(B^6) + 12(A^5)(B^7) + 9(A^4)(B^8) + 4(A^3)(B^9) + 6(A^2)(B^10) + 3 = 0

Then, I use the substitution r = B/A, divide through by A^12, to get an equation that I have nicknamed: The Grand Polynomial

3r^12 + 6r^10 + 4r^9 + 9r^8 + 12r^7 + 4r^6 + 12r^5 + 9r^4 + 4r^3 + 6r^2 + 3 = 0

So far, this polynomial has steadfastly resisted every attack I have thrown at it, although I have made some progress.

When confronted with such a large polynomial, my first reaction was to look for simple factors (or at least, rule them out). Since all the coefficients are positive, I can immediately rule out any positive, real roots. Not surprisingly, the Rational Roots Theorem also fails to uncover any roots. Less obviously, I can rule out all nth roots of unity up to n = 4 (I could extend that further, but I doubt it would be worth the effort). I have considered using the derivative to prove that no real roots exist (which is what Mathematica indicates) and that no repeated roots exist either (which would be a corrolary to a proof that no real roots exist, since a + bi cannot equal a - bi if b is nonzero). Unfortunately, the derivative is an 11th degree polynomial (ok, I can trivially factor out r to reduce it to a 10th degree, but that's still an imposing challenge), and proving anything about the roots of the derivative is likely to be as difficult as proving anything about the roots of the Grand Polynomial itself.

I have posed this problem to various people in the mathematics department at university, but so far, nobody has managed to give a complete solution. By exploiting the palindromic symmetry, my teaching assistant found a substitution to reduce the degree of the polynomial.

Divide the Grand Polynomial through by r^6 to give an equation that looks even more symmetrical:

3r^6 + 6r^4 + 4r^3 + 9r^2 + 12r + 4 + 12(1/r) + 9(1/r^2) + 4(1/r^3) + 6(1/r^4) + 3 = 0

Grouping terms:

3(r^6 + 1/r^6) + 6(r^4 + 1/r^4) + 4(r^3 + 1/r^3) + 9(r^2 + 1/r^2) + 12(r + 1/r) + 4 = 0

Now, using the Binomial Theorem, we can derive the identities:

(r^2 + 1/r^2) = (r + 1/r)^2 - 2

(r^3 + 1/r^3) = (r + 1/r)^3 - 3(r + 1/r)

(r^4 + 1/r^4) = (r + 1/r)^4 - 4(r + 1/r)^2 + 2

(r^6 + 1/r^6) = (r + 1/r)^6 - 6(r + 1/r)^4 + 9(r + 1/r)^2 - 2

Substituting this into the previous equation, and using the substitution z = r + 1/r gives:

3z^6 - 12z^4 + 4z^3 + 12z^2 - 8 = 0

This equation seems tantalizingly close to being solved. If I could just eliminate the constant term, I would have z^2 multiplied by a quartic, which could be solved using Ferrari's Quartic Formula. If I could eliminate the cubic term, I would have a cubic in terms of z^2, which could be solved using the Cubic Formula. Unfortunately, I have not found a way to do either of these.

Of course, this is a sextic (6th degree) polynomial, which means that, by Abel and Galois' theorems, a solution by radicals may not exist. For this reason, one of the professors I posed to question to was originally skeptical that a solution by radicals existed. However, he used Maple to compute the Galois group of the polynomial; it came out as 6T4. Under the reference listed in Maple's Help article, I found that 6T4 was listed as A4. Quoting from the reference:

"An is the alternating group of degree n"

When I told this to the professor, he said, to my delight, that A4 is a solvable group, which means a solution by radicals to the sextic must exist! Unfortunately, he doesn't know of any way to exploit knowledge about the Galois group to explicitly find the solution by radicals. Nevertheless, the guarantee that an exact solution does exist (assuming that Maple's calculation is correct) has increased my desire to find such a solution. Unfortunately, both the 12th degree polynomial and the 6th degree polynomial have resisted all further attempts at reduction...

[Crook]

I have a solution of the problem. If someone is interested I can post it.

[jinydu]

Thank you for your interest. I would certainly be interested in the solution.