In the last post, I hinted at the equality:

d(P(m)) = [tr(M)]^2 - 4*[det(M)] ( ** )

where:

tr(M) is the trace

det(M) is the determinant

and

d(P(m)) is the discriminant of the (associated) characteristic function P(m) (i.e. this is what we used to compute the eigenvalues s and t)

Further, recall that the eigenvalues s and t are rational if and only if the discriminant

d(P(m)) is a (rational) square.

Let us rearrange equation ( ** ) as:

d(P(m)) + 4*[det(M)] = [tr(M)]^2 ( *** )

Thinking Pythagorean, we assume that:

(Assumption #1) det(M) is a (rational) square.

(Assumption #2) D = 0 (that is, we are considering the penultimate equation A + B = C)

Thus, rewriting the equation ( *** ) as follows:

{Alpha}^2 + {Beta}^2 = 1

where:

Alpha = SQRT[d(P(m))]/tr(M)

Beta = 2*SQRT[det(M)]/tr(M)

we will be able to parametrize all rational solutions to Alpha and Beta via:

Alpha = (1 - t^2)/(1 + t^2)

Beta = (2t)/(1 + t^2)

Indeed:

Alpha_X = SQRT[d(P(x))]/tr(X) = (C - A)/(C + A)

Beta_X = 2*SQRT[det(X)]/tr(X) = 2*SQRT[AC]/(C + A)

Alpha_Y = SQRT[d(P(y))]/tr(Y) = (C - B)/(C + B)

Beta_Y = 2*SQRT[det(Y)]/tr(Y) = 2*SQRT[BC]/(C + B)

Alpha_Z = SQRT[d(P(z))]/tr(Z) = 2D/2C = D/C = 0 (since D = 0)

Beta_Z = 2*SQRT[det(Z)]/tr(Z) = 2C/2C = 1

Note the following:

Alpha_X = (1 - [t_X]^2)/(1 + [t_X]^2) = (C - A)(C + A)

Beta_X = 2[t_X]/(1 + [t_X]^2) = 2*SQRT[AC]/(C + A)

Alpha_Y = (1 - [t_Y]^2)/(1 + [t_Y]^2) = (C - B)(C + B)

Beta_Y = 2[t_Y]/(1 + [t_Y]^2) = 2*SQRT[BC]/(C + B)

Alpha_Z = (1 - [t_Z]^2)/(1 + [t_Z]^2) = 0

Beta_Z = 2[t_Z]/(1 + [t_Z]^2) = 1

Solving these equations simultaneously (under the naive assumption that C = 1):

t_X = SQRT(A)

t_Y = SQRT(B)

t_Z = 1 = SQRT(1) = SQRT(C)

Again, if we want all such t_X, t_Y and t_Z to be rational (so that we get rational Pythagorean triples), then we require that A, B and C are also (rational) squares.

(I will stop here for the moment to eat lunch. It's exactly 11:11 AM here in Makati City, Philippines as I type this.)

## 1 comment:

In order to come up with a rational-square discriminant:

d(P(m)) = [tr(M)]^2 - 4*[det(M)]

we can of course choose either:

det(M) = tr(M) - 1

OR

det(M) = -(tr(M) + 1)

This works by completing the square in tr(M).

One eigenvalue then becomes fixed at 1 for one case, and at -1 for the other case. The other eigenvalue has bigger absolute value, albeit arbitrary (because of the OR condition when zero products are decomposed to the respective factors).

It is not obvious to me, at this point, if this gives a complete characterization for rational-square discriminant:

d(P(m)) = [tr(M)]^2 - 4*[det(M)]

Post a Comment