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 = x^{2} + 8x + 7 gives

v = u^{2} + 8u + 7(1)andu = v^{2} - 8v + 7(2)

Using (1) in (2):u = (u^{2} + 8u + 7)^{2} - 8(u^{2} + 8u + 7) + 7

which simplifies to:u^{4 }+ 16u^{3} +70u^{2} +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?