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.)