Find all integer $t$ such that $\frac{t^2+9}{t+9}$ is an integer Find all integer $$ t \geq 0 $$ such that $$ \frac{t^2+9}{t+9} =k $$ k is integer.
How to solve this?
I've tried to solve this as an equation with parameter 
$$ 
t^2-kt-9k+9=0 \\
D= k^2-36k+36\\ 
t_1=\frac{k-\sqrt{k^2-36k+36}}{2} \\
t_2=\frac{k+\sqrt{k^2-36k+36}}{2} 
$$
so we need $$\sqrt{k^2-36k+36}$$ to be an int and also $$k-\sqrt{k^2-36k+36} \\
k+\sqrt{k^2-36k+36}$$ must be even... So I'm kinda stuck
 A: Your discriminant is wrong. You should get:
$$D=k^2-4(9-9k)=k^2\color{red}+36k\color{red}-36$$

The easier approach is to note that $\frac{t^2-81}{t+9}=t-9$ is an integer, so $$\frac{t^2+9}{t+9}=\frac{t^2-81}{t+9}+\frac{90}{t+9}$$ is an integer if and only if $\frac{90}{t+9}$ is an integer.

If you must use the discriminant method...
You only really need the discriminant to be a square, since if the polynomial has rational roots, then the roots are integers, since the polynomial has integer coefficients and lead coefficient $1.$
Completing the square, you need $D=(k+18)^2-(18^2+36)=(k+18)^2-360=n^2$ for some $n$, or $(k+18-n)(k+18+n)=360$.
Any factorization $90=AB$ gives us $k+18-n=2A, k+18+n=2B$, or $k=A+B-18$ and $n=B-A$  and $t=\frac{-k\pm n}{2}$ and this gives $t=A-9$ or $t=B-9.$ 
(Why did we use factorizations of $90$ rather than $360?$)

More geneally, if $f$ is a polynomial with integer coefficients, and $a$ is an integer such that $f(a)\neq 0$ then the set of $t$ such that:
$$\frac{f(t)}{t-a}$$
can be determined as follows. Let $u=t-a$ then:
$$\frac{f(t)}{t-a}=\frac{f(u+a)}{u}.$$ And $u$ divides $f(u+a)$ if and only if $u$ divides $f(a).$ So $t+a\mid f(t)$ if any only if $t+a\mid f(a).$
So write all the integer factors of $f(a)$ and then add $a$ to each of them.
In your example, $f(t)=t^2+9, a=-9,$ and $f(a)=90.$
A: Write $s=t+9$ then:
$$ {t^2+9\over t+9} = {s^2-18s+90\over s} = s-18+{90\over s}$$
so $t+9\mid 90$...
A: As an alternative note that
$$t^2+9=(t+9)^2-18t-72=(t+9)^2-18(t+9)+90\equiv 90\mod{(t+9)}$$
