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A sum of two squares below the bar, beside a repeated factor, is split over its conjugates (#1804)
Under u = tan(x) and t = sqrt(c + d u), the 1 + u^2 of
(c + d tan(x))^(3/2)/(a + b tan(x))^2 is (t^2 - c)^2 + d^2, a quartic beside the
square of a d + b (t^2 - c). The split by residues reads factors of the first
and second degree only, so the quotient went to the Hermite reduction, which
solved for its numerators with the symbols in every entry and ran past two
minutes. A written sum of two squares P^2 + Q^2, the greater of their degrees
two, beside a repeated factor, is written (P - i Q)(P + i Q) now and the
quotient asked again: four rows of 4.3.2.1 with one root over a power of
a + b tan are answered in one to two seconds, and two with two roots, where the
sum is (b - d t^2)^2 + (c t^2 - a)^2 under the root of their quotient.
The constants in that split hold i, and it is short only with them in lowest
terms (#1788's first part).
Part of #718.
Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
Co-authored-by: Claude Opus 5.5 (1M context) <noreply@anthropic.com>1 parent 54c1d10 commit ba77f7a
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