I found this old math competition question and I am completely at sea with this one. The only thing I can honestly say I've tried is plug in numbers and work from there, but it hasn't worked and I'm sure there's a more formal, direct approach to this problem. This is something I'm really interested in knowing how you solve, though!
So how would you solve this?
By the quadratic formula, we need $a^2-4b$ and $a^2-4(b+1)$ both to be perfect squares.
The only squares that differ by $4$ are...
Okay. You will be knowing the formula for x which we get by completing the squares. $$x={-b\pm\sqrt(b^2-4ac)\over 2a}$$ Hence the basic thing is that $\sqrt(b^2-4ac)$ must be an integer. Which implies $b^2-4ac$ must be a whole square. Doing this for the two quadratic equations we get: $$a^2-4b=m^2$$ $$a^2-4b-4=n^2$$ For m,n in integers. Notice carefully that the two squares differ by 4.This doesn't happen for any integer. You can easily check this. The first few squares are 1,4,9,16......for The difference between the squares is is 3,5,7 and this keeps on increasing.The difference is 4 only for $0^2$ and $2^2$. Hence $b^2-4ac=4$ $$x={-a\pm\sqrt(a^2-4b)\over2}$$ Now we know that $\sqrt(a^2-4b)=2$. Hence if x has to be an integer then a is even. What about b? We know that $a^2-4b=4$, hence $$a=2\sqrt(b+1)$$ So b has to be of the form $n^2-1$ for n<1000. Which means a=2n but a<1000, this further restricts n to n<500. Hence we have 499 ordered pairs.
\sqrt{a + b +c} gives $\sqrt{a+b+c}$.
$\endgroup$
If $0=x_1^2+ax_1+b=x_2^2+ax_2+b+1,$ then by subtracting, $0=x_1^2-x_2^2+a(x_1-x_2)-1,$ so $1=(x_1-x_2)(x_1+x_2+a).$
Therefore $x_1-x_2=\pm 1=x_1+x_2+a.$ Solving this pair of linear equations, we have $$2x_1+a=\pm 2 \text { and } 2x_2+a=0.$$ So $x_2=-a/2$ which implies $$0=x_2^2+ax_2+b+1=a^2/4-a^2/2+b+1=b+1-a^2/4.$$ Since $8\leq 4(b+1)\leq 3996$ and $a^2=4(b+1)$ this gives $3\leq a\leq 63.$ But $a$ is even because $2x_1+a=\pm 2.$ So $a=2c$ where $2\leq c\leq 31.$ And $b=a^2/4-1=c^2-1.$
