# Number of possible integer values of $x$ for which $\frac{x^3+2x^2+9}{x^2+4x+5}$ is integer

How many integer numbers, $$x$$, verify that the following

$$\begin{equation*} \frac{x^3+2x^2+9}{x^2+4x+5} \end{equation*}$$

is an integer?

I managed to do:

$$\begin{equation*} \frac{x^3+2x^2+9}{x^2+4x+5} = x-2 + \frac{3x+19}{x^2+4x+5} \end{equation*}$$

but I cannot go forward.

• The denominator is larger than the numerator for most values. Either solve $3x+19\geq x^2+4x+5$ or find a bound for its solutions. Then test the integer values in that range. – logarithm May 9 at 18:45
• You only need to check few values of $x$: from $-4$ to $+3$. – Math Lover May 9 at 18:48

Let $$t=x+2$$, then $$t^2+1\mid 3t+13$$ and thus $$t^2+1\mid t(3t+13)-3(t^2+1)= 13t-3$$ so $$t^2+1\mid 13(3t+13)-3(13t-3) = 178$$

So $$t^2+1\in\{1,2,89,178\}\implies t=\pm 1,0 \implies x\in\{-1,-2,-3\}$$

You're off to a good start. Now note that the denominator $$x^2+4x+5$$ is quickly larger than the numerator $$3x+19$$; you can quickly reduce the problem to only finitely many values for $$x$$ to check.

More details: (Hover to show)

The fraction is certainly not an integer if the denominator is greater than the numerator, i.e. if $$x^2+4x+5>3x+19,$$ unless perhaps the numerator is zero, but that is not possible in this case. The quadratic formula shows that the inequality above holds if $$x\leq5$$ or $$x\geq4$$. Then it remains to check whether the fraction is an integer for $$x$$ in the range $$-4\leq x\leq3$$.

Here is another method that works well when Diophantine quadratics $$ax^2+bx+c=0$$ are involved, it consists in saying that since a solution should exists then $$\exists \delta\in\mathbb Z\mid b^2-4ac=\delta^2$$.

In this problem we'd like $$\quad\dfrac{3x+19}{x^2+4x+5}=n\quad$$ to be an integer.

Thus applying the method to $$nx^2+(4n-3)x+(5n-19)=0$$

gives $$\quad(4n-3)^2-4n(5n-19)=p^2\iff 4n^2-52n+(p^2-9)=0$$

Which by itself should have solutions in $$n$$ : $$\quad(-52)^2-4(4)(p^2-9)=q^2\iff 2848-16p^2=q^2$$

We can divide by $$16$$ (setting $$q=4r$$) to get $$p^2+r^2=178$$

This has solutions $$(\pm 3,\pm 13)$$.

• case 1 : $$p^2=3^2=9$$

Leads to $$4n^2-52n=0\iff 4n(n-13)=0\iff n=0,\ 13$$

$$n=0\implies 3x+19=0$$ impossible

$$n=13\implies 13x^2+49x+46=(x+2)(13x+23)=0$$ and we verify that $$x=-2$$ is solution.

• case 2: $$p^2=13^2=169$$

Leads to $$4n^2-52n+160=4(n-5)(n-8)=0\iff n=5,\ 8$$

$$n=5\implies 5x^2+17x+6=(x+3)(5x+2)=0$$ and we verify that $$x=-3$$ is solution.

$$n=8\implies 8x^2+29x+21=(x+1)(8x+21)=0$$ and we verify that $$x=-1$$ is solution.

Finally we have found all solutions $$x\in\{-3,-2,-1\}$$