The rotational symmetry suggests that the square will have its centre at the origin, so that its corners will have coordinates of the form: (u, v), (-v, u), (-u, -v), (v, -u).
Taking the first two of these points to lie on y = x2 + 8x + 7 gives
v = u2 + 8u + 7(1)andu = v2 - 8v + 7(2)
Using (1) in (2):u = (u2 + 8u + 7)2 - 8(u2 + 8u + 7) + 7
which simplifies to:u4 + 16u3 +70u2 +47u = 0
This quartic has four real roots:
u = -7.756981174, -7.7427228955, -0.8157898708, 0
The first three of these give the squares found in previous postings. The fourth gives a square with corners at (0, 7), (-7, 0), (0, -7) and (7, 0).
I’ve been trying to find a way of avoiding the need to solve a cubic to find the three irrational roots, but so far with no success. Any ideas?
Four squares by analysis (spoiler)
