Find x,y,N s.t. x³ - y³ - 2xy - N = 0

# Find natural numbers x, y, N that satisfy x³ - y³ - 2xy - N = 0
# See https://www.youtube.com/watch?v=2Zov0KXPswk for more

def f(n):
    # Find all x,y ∈ ℕ s.t. x³ - y³ - 2xy - N = 0
    # Limits:
    # y < x because y³ + 2xy + N = x³, and x,y,N are all non-negative
    # x < N + 1, 0therwise there can be no y that satisfies this relation
    #   Rationale: if x,y = N+1,N,
    #   then x³ - y³ - 2xy - N = (N+1)³ - N³ - 2N(N+1) - N = (N+1)² > 0
    #   any smaller value of y will also give a result greater than zero
    #   so there is no value of y for which x³ - y³ - 2xy - N = 0
    result = []
    for x in range(1,n+1):
        x_cubed_minus_n = x**3 - n
        # Find corresponding y by binary search (if it exists)
        ymin, ymax = 0, x-1
        while ymin <= ymax:
            y = (ymin + ymax) // 2
            z = x_cubed_minus_n - y**3 - 2*x*y
            if z == 0:
                result.append((x,y))
                break
            elif z > 0:
                ymin = y + 1
            else:
                ymax = y - 1
    return result


for n in range(1000):
    result = f(n)
    # If there are one or more solutions for this value of n,
    # print them out
    if len(result) > 0:
        print(n,result)