Proof of Adamson's conjecture on palindromic continued fractions, re: A179238

Proof of Adamson's conjecture: Gary W. Adamson, re: A179238 in OEIS stating that :

a. given t = irrational Tan A, then if t = the convergent to an infinitely periodic palindromic continued fraction, then Tan 2A is rational; and the converse.

b. Tan 2A is rational when Tan A is rational iff, Tan A = t = the convergent to an infinitely periodic palindromic continued fraction.

Example: barover[a], with a = 1 getting [1,1,1,1,...] = .618....; with the inverse 1.618.... = t (either case) = Tan A.; where 1.618... is irrational.. If t = .618...then Tan 2A = 2.0. If t = 1.618..., then Tan 2A = -2.
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We will present the sections of the proof then put the separate components together.
Part A: Given t = Tan A, then the double angle 2A triangle hs lets of 2t and t^2-1, with hyptenuse t^2+1.
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The key step here is the relationship Q = t - 1/t = rational; example 1.618... - .618...= 1.
In order for t to be irrational but Tan 2A to be rational, t - 1/t = Q must be rational (to be shown); but for now continuing with this algrbra,; we have
Q = 1 - 1/t, then Qt = t^2 - 1, and
2t * (QT) = 2t * (t^2 - 1); with 2t / (t^2-1) = 2t / Qt = 2/Q; (and Q is rational).
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Alternatively, using Tan 2A = 2TanA / (1 - Tan^2 A,) , let t = Tan A, say phi^(-3).
Then Tan 2A = 2*(phi^-3) / (1 - phi^-3) =
2 / (phi^3 - phi^(-3); fulfilling our required identity such that phi^3 - phi^(-3) = 4, since
phi^3 = 4.236...; where .236...= [4,4,4,4,4,4,....] ; palindromic
...Thus, a specific case. of Tan 2A = 2 Tan A / (1 - Tan^2 A), where Tan A = phi^3 or it's inverse.
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Next, we investigate Part B:
States that t - 1/t must be rational, and looking at the "tail" of an infinitely periodic continued fraction (refer to Wolfram's Mathworld); the "tail" using the formula, and applying to barover [1,2,3] = [1,2,3,1,2,3,1,2,3,...] this is not palindromic and is thus a counter example. The "tail" becomes (2x + 7 / (3x + 10) setting this = to x, getting 3x^2 + 8x - 7, the roots to this being the convergent to barover [1,2,3].
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We can straightaway obtain the formula 3x^2 + 8x - 7 by writing the partial quotients of [1,2,3] underneath as follows:
1,.......2,........3
1/1 2/3 7/10.
We put 2/3 and 7/10 into a 2x2 format as:
2, 7
3, 10 and perform the following operation such that lower left term (3) = a, upper right = c;
and (lower right - upper left) = b, or f(x), (a,b,c) = ax^2 + bx - c = 0 or
3x^2 + 8x - 7 = 0, and we note that a is not equal to c or 3 is not equal to 7.
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Next, it follows that in constants of the form used in our conjecture c - 1/c = rational; the corresponding quadratic equation must be such that a must = c.; this being the case, dividing through by (a, c) we obtain an equation of the form x^2 - bx - 1 = 0, where b is rational.
This equation is the characteristic polynomial for an infinitely periodic, palindromic continued fraction, equavalent to the statement in the above 2x2 format that the lower left term must equal the upper right term. In this case 3 is not equal to 7, and the roots are not of the form t - 1/t = rational.
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Part C. In order for the lower left term to equal the upper right term, we again refer to the "tail" of an infinitely periodic continued fraction which in the case of barover[1,2,3] =
2x + 7 / (3x+10), then setting this to x, we get 3x^2 + 8x - 7 = 0; again 3 is not equal to 7.
We need the tail of the form where upper right and lower left terms are equal, in other words,
in the convergents to [1, 2, 3] we have the partial quotients underneath:
1,.......2,.......7 and
1,.......3,......10. since [1,2,3] = 7/10 and the partial quotient to the left = 2/3.
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Part D. We refer to Mathworld "Continued fraction", theorem 30 and 31 which state that:
pn / p(n-1) = [an, a(n-1), ....ao] = (30) and
qn / (q(n-1) = [an....a1]................=(31)
...In other words, (30) applies to the reversal of [a,b,c] saying we start with an, then procede to a0. For example, given [1,2,3,4] , theorem 30 uses [4, 3, 2, 1]. Where [1,2,3,4] = 30/43 as follows:
1,.......2,.......7,........30
1,.......3,......10,......43
...then reverse using [4, 3, 2, 1] = obtaining
1,.......3,.......7,.......10
4,......13,.....30,.....43.; so in essence, theorem 30 switches the position of the upper left and lower right terms, with the 7 and 43 terms invariant.
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Now if the reversal of [a,b,c] remains invariant and thus the upper right and lower left terms are unchanged, this could only occur if the reversal is palindromic, such as [1, 2, 2, 1] where our convergents are
1,....2,......5,......7
1,....3,......7,.....10.
Thus only in a palindromic continued fraction is the lower left term = to the upper right.
Part E.
It follows from the formula of the "tail" say given (using different a,b,c,...):
we set x =( ax + b) / (cx + d); but again changing the usage of our (a,b,c,...) the quadratic formula for the tail if the (upper right = lower left) condition holds must be such that in
ax^2 + bx - c, a must = c. where a,b,c are all rational.
...But only when in this quadratic, a = c does the result hold (dividing through by a = c), we obtain
x^2 - bx - 1 = 0 (with an allowance for changing signs); and if b is rational, then we have our necessary equation with roots t and 1/t such that for example 1.618... - .618 = 1, rational, or
t - 1/t is rational, but t is irrational.
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Combining all of the steps, Q.E.D. especially noting the key step "upper right = lower left term";
we have shown that in given t irrational, t must be the convergent of an infinitely periodic, palindromic continued fraction in order that angle 2A is rational. Of course, if Tan 2A is rational, t can be rational; but our conjecture only applies to the case in which Tan 2A is rational and t (Tan A) is irrational. The converse of the conjecture is also true.