Proof for PPCG #162106

Assuming n ≡ 1 (mod 4)

f(n) = n

m = n + 1

f(m) = n - m = n - (n + 1) = n - n - 1 = -1

o = m + 1 = n + 2

f(o) = f(m) * o = -1 * o = -o

p = o + 1 = m + 2 = n + 3

f(p) = o // p = (-n - 2) // (n + 3) = |-n - 2| // |n + 3| = (n + 2) // (n + 3) = 0   2 < 3 <=> n + 2 < n + 3

q = p + 1 = o + 2 = m + 3 = n + 4

f(q) = 0 + q = q