Find $C$ such that $x^2 - 47x - C = 0$ has integer roots, and further conditions Have been working on this for years. Need a system which proves that there exists a number $C$ which has certain properties. I will give a specific example, but am looking for a system which could possibly be generalized.
Find (or prove there exists) $C$, such that the quadratic $x^2 - 47x - C = 0$ has integer roots, and furthermore, $C$ must have ALL OF $2$, $3$ and $5$ as its only prime factors (though each of these can be to any positive integer power).

Well if $ax^2+bx+c$ has integer roots, then $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ is an integer. Since in this case $a=1$ and $b=-47$ and we want to find $C$, $$\frac{47\pm \sqrt{(-47)^2+4C}}{2}=\frac{47\pm \sqrt{2209+4C}}{2}$$ is an integer. $47$ is odd. Since the fraction is an integer, the numerator must be divisible by $2$. Since $47\pm \text{odd_number}$ is even (and even means divisible by two), $\sqrt{2209+4C}$ must be an odd number. That in turn means $2209+4C$ must be an odd square.
We also know that $C=2^a\times 5^b\times 3^c$. So now we need to solve the equation $2209+4\times 2^a5^b3^c=x^2$ for integers. You can use a computer to solve it from here if you like. 
$$2209+4\times 2^a5^b3^c=\text{a square bigger than 2209}\\ \Longrightarrow \text{a square bigger than 2209}-2209=4\times 2^a5^b3^c\\ \Longrightarrow (47+l)^2-47^2=4\times 2^a5^b3^c$$
For some integer $l$. See if you take it from here...
