> Hence M31 should appear...

I don't think square degrees should be used here. People don't visually approximate that way when fitting round objects into a space.

Andromeda 3.167° across its longest length. Moon 0.518°. That's about 6 moons across the longest length.

Andromeda is 1° across the shorter axis. Approximately 2 moons.

So, appearing about 10-12 moons in area maximum, shaving some off at the corners.

I should have mentioned that area = π r² is only an approximation for the moon’s apparent size in square degrees. What we really want to calculate is the surface area of the spherical cap (on a unit sphere) with the same angular diameter as the moon. A formula for that is 2π(1 – cos r) [1], where r again is the radius of the moon in degrees. In this case, the two formulas give the same result to four significant digits (e.g. [2]).

[1] https://en.m.wikipedia.org/wiki/Spherical_cap

[2] http://www.wolframalpha.com/input/?i=2pi%281-cos%5B%2834.1%2...