Universal sums of three quadratic polynomials

Let a, b, c, d, e and f be integers with a > c > e > 0, b > -a and b equivalent to a (mod 2), d > -c and d equivalent to c (mod 2), f > -e and f equivalent to e (mod 2). Suppose that b > d if a = c, and d > f if c = e. When b(a-b), d(c-d) and f(e-f) are not all zero, we prove that if each n is an element of N = {0, 1, 2,...} can be written as x(ax + b)/2 + y(cy + d)/2 + z(ez + f)/2 with x, y, z is an element of N then the tuple (a, b, c, d, e, f) must be on our list of 473 candidates, and show that 56 of them meet our purpose. When b is an element of [0, a), d is an element of [0, c) and f is an element of [0, e), we investigate the universal tuples (a, b, c, d, e, f) over DOUBLE-STRUCK CAPITAL Z for which any n is an element of N can be written as x(ax+b)/2+y(cy+d)/2+z(ez+f)/2 with x, y, z is an element of DOUBLE-STRUCK CAPITAL Z, and show that there are totally 12,082 such candidates some of which are proved to be universal tuples over DOUBLE-STRUCK CAPITAL Z. For example, we show that any n is an element of N can be written as x(x + 1)/2 + y(3y + 1)/2 + z(5z + 1)/2 with x, y, z is an element of DOUBLE-STRUCK CAPITAL Z, and conjecture that each n is an element of N can be written as x(x + 1)/2 + y(3y + 1)/2 + z(5z + 1)/2 with x, y, z is an element of N.