Consider the equation

A + B = C + D (where, for now, we assume all of A, B and C are positive, D is unrestricted)

Assuming a nonzero sum C + D:

(A + B)/(C + D) = 1 (from which it also follows that the sum A + B is nonzero)

Assuming C is not equal to D:

[(A + B)(C - D)]/[(C + D)(C - D)] = 1

[ (AC - BD) + (BC - AD) ] / [ C^2 - D^2 ] = 1

[ (AC - BD)/(C^2 - D^2) ] + [ (BC -AD)/(C^2 - D^2) ] = 1

Passing on to determinant notation:

det X + det Y = det Z

where the 2x2 matrices X, Y and Z are defined as follows:

X

A B

D C

Y

B A

D C

Z

C D

D C

(The numbers A, B, C and D were laid out in their respective positions by "basket-weaving".)

Some linear algebra:

The traces of these matrices are:

tr(X) = A + C

tr(Y) = B + C

tr(Z) = 2C

Solving for the eigenvalues of each of these 3 matrices:

det(X - xI) = 0 = (A - x)(C - x) - BD = x^2 - (A + C)x + (AC - BD)

det(Y - yI) = 0 = (B - y)(C - y) - AD = y^2 - (B + C)y + (BC - AD)

det(Z - zI) = 0 = (C - z)(C - z) - D^2 = z^2 - (2C)z + (C^2 - D^2)

It is evident that the characteristic polynomial P(m) for a 2x2 matrix M then takes the form:

P(m) = m^2 - [tr(M)]m + [det(M)]

and its discriminant d(P(m)) is then:

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

In particular, if s and t are the eigenvalues of a particular 2x2 matrix M, then:

det(M) = st

tr(M) = s + t

Consequently (assuming s <= t):

s = (1/2)*([tr(M)] - SQRT[d(P(m)])

t = (1/2)*([tr(M)] + SQRT[d(P(m)])

For the matrices X, Y and Z above, the respective discriminants are:

d(P(x)) = [tr(X)]^2 - 4[det(X)] = (A + C)^2 - 4(AC - BD) = (A^2 + 2AC + C^2) - 4AC + 4BD

= (A - C)^2 + 4BD

d(P(y)) = [tr(Y)]^2 - 4[det(Y)] = (B + C)^2 - 4(BC - AD) = (B^2 + 2BC + C^2) - 4BC + 4AD

= (B - C)^2 + 4AD

d(P(z)) = [tr(Z)]^2 - 4[det(Z)] = (2C)^2 - 4(C^2 - D^2) = 4C^2 - 4C^2 + 4D^2 = (2D)^2

(Note that, if we want all three discriminants to be perfect squares, we can take D = 0. The original equation under consideration then becomes A + B = C. Indeed, I had the ABC conjecture in mind while trying to do these things the other day -- but it turned out these were closely related to another open problem.)

If the discriminant d(P(m)) is a (rational) square, it follows that the eigenvalues s and t are also rational, and conversely.

We now compute the eigenvalues (under the assumption s <= t):

s_X = (1/2)*(A + C - SQRT((A - C)^2 + 4BD))

t_X = (1/2)*(A + C + SQRT((A - C)^2 + 4BD))

s_Y = (1/2)*(B + C - SQRT((B - C)^2 + 4AD))

t_Y = (1/2)*(B + C + SQRT((B - C)^2 + 4AD))

s_Z = (1/2)*(2C - SQRT((2D)^2)) = (1/2)*(2C - 2D) = C - D

t_Z = (1/2)*(2C + SQRT((2D)^2)) = (1/2)*(2C + 2D) = C + D

Under the additional assumption that D = 0, these expressions simplify to:

s_X = A

t_X = C

s_Y = B

t_Y = C

s_Z = C

t_Z = C

With this assumption about the vanishing of D, the original matrices X, Y and Z become:

X'

A B

0 C

Y'

B A

0 C

Z'

C 0

0 C

Thus, we have two upper-triangular matrices X' and Y', and Z' which is a scalar multiple of the 2x2 identity matrix. But that is not really the crucial point.

Note that we have shown that:

A = s_X <= t_X = C

B = s_Y <= t_Y = C

C = s_Z <= t_Z = C

without knowing the actual values of A, B and C. How did this happen?

Hint: SQRT(x^2) = ABS(x), where ABS is the absolute value function.

In summary: Note how the value of D (which was initially unrestricted, and which we later take to be zero for purposes of "simplifying" the computations for the eigenvalues) imposed A LOT of STRUCTURAL RESTRICTIONS on the three matrices.

Later during the day, I will present the details of my

**(dis-)proof**for the inequality "m < p^k", which was the last stumbling block for my current approach in an attempt to resolve the OPN conjecture.