Maximum possible integer square root polynomial 2: Much harder

We are working with the most general form of this equation except that we're working with a^2 (if working with a ...things gets real)

Let $y=\sqrt{a^2x^2+bc+c}$

Since a,b,c,x are all greater than 0, then $y>ax$ . Thus we can represent the LHS as \(ax+n).

Squaring both sides, we get \((ax+n)^2-a^2x^2=bx+c\)

Solving for X we get $x=\dfrac{n^2-c}{b-2an}$ .

Rather than plugging this in for x and bashing for y, because the graph of this function will look like a "V". We can simply maximize x.

Induction leads to the following conclusions:

N must be maximized

for n to be maximized and x to be positive, a must be minimized.

for x to be positive and n to be maximized, b must be maximized.

b=2an+1 because if b=2an, then the equation can be represented as $ax+\sqrt{c}$ . Thus it had no bound and b is EVEN. It is specified below.

c must be minimized for x to be maximized.

Thus our equation becomes

$x=\dfrac{1499^2-1}{2999-2(1)(1499)}\Rightarrow x=1499^2-1$

Plugging back in x, we get

$y=\sqrt{1(1499^2-1)^2+2999(1499^2-1)-1}$

$\therefore \boxed{y=2,248,499}$