Find $C$ such that $x^2 - 47x - C = 0$ has integer roots, and further conditions

Question

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).

Answer 1

$$C=150=2\cdot3\cdot5^2\implies x\in\{-3,50\}$$ $$C=-90=-2\cdot3^2\cdot5\implies x\in\{2,45\}$$ $$C=-480=-2^5\cdot3\cdot5\implies x\in\{15,32\}$$ $$C=-540=-2^2\cdot3\cdot5^2\implies x\in\{20,27\}$$

Answer 2

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 $7$.

Answer 3

The quadratic formula implies that this is equivalent to getting $47^2+4C$ to be the square of an odd integer.

$$47^2-(2n+1)^2=-4C\\ 4(23-n)(24+n)=-4C\\ (n-23)(n+24)=C$$

So we need to find $n$ such that $(n-23)(n+24)$ is a multiple of $30$ with no prime factors other than $2,3,5$ and $47^2+4(n-23)(n+24)\geq 0$.

Inspection reveals $n=3$ as a solution with $C=-20\cdot 27=-2^23^35=-540$. Then $47^2+4C=49$ and the quadratic formula gives roots $$\frac{47+7}{2}=27\text{ and }\frac{47-7}{2}=20$$ to the equation $$x^2-47x+540=0$$

Edit:

This was "too long for a comment."

I speculate that the method I give above can be pushed a little further to get which values of $n$ work, hence giving an integral parametrization of the solutions ($C$) of the problem.

The same approach can be applied to $x^2-397x-C=0$ with $2,3,5,7,11,13,17,19$ as prime factors. You are right that "inspection" only gets so far. I don't want to spend the time on the case you put in your comment, but my initial approach would be use the method above to get a parametrization of $C$ in terms of $n$, use congruences and the chinese remainder theorem to restrict the possible values of $n$, and to write a short algorithm finding the first few $n$ that work.

Answer 4

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...