Find all integers $k$ such that $\frac{k^4+k^3-5k^2+k}{k^2+k-6}$ is an integer.
I tried factorization and ended up with $\frac{k(k^3+k^2-5k+1)}{(k-2)(k+3)}$. Please give me some hints. Thank you!
Find all integers $k$ such that $\frac{k^4+k^3-5k^2+k}{k^2+k-6}$ is an integer.
1
$\begingroup$
elementary-number-theory
integers
-
0See k =-3 satisfies the $k^3 +k^2 - 5k + 1$ . So $k+3$ is a factor. – 2017-02-16
-
0@ChirantanChowdhury Not really $(-3)^3+(-3)^2-5(-3)+1=-2$ – 2017-02-16
-
1You want first to do polynomial division so that you are left with a numerator of degree less than $2$. Then, if it is not obvious, you can split what remains into partial fractions (i.e. fractions with scalar numerators and denominators $k-2$ and $k+3$). If $|k|$ is too large, these fractions will be too small to give an integer, and there will be a small number of cases left to consider. – 2017-02-16
1 Answers
6
HINT: $$\frac{k^4+k^3-5k^2+k}{k^2+k-6}=\frac{(k^2+1)(k^2+k-6)+6}{k^2+k-6}$$