<s>[INST] 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).
 [/INST] Say
$$
x^2 -47x - c = (x-a)(x-b)
$$
for integers $a$ and $b$.
Then
$$
a + b = 47 \\
-ab = C.
$$
So $a$ and $b$ must have only $2,3,5$ in their prime factorization.
Say $b$ is divisible by $5$ (one of them must be). So say $b = 5n$. So then $47 - a$ is divisible by $5$. And $a$ must be $-13 , -8 , -3 , 2, 7, 12, 17, \dots$. So $a = 2 + 5m$. Pick out those that are divisible by only $2$ or $3$.
By trial and error
$$
a = -3, b = 50
$$
would work. 
Another solution is
$$
a = 72, b = - 25.
$$
Note, for example that $a=12$ would give $b = 35$, but then $b$ is divisible by $7$.  
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