Mathematical and Physical Journal
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Problem C. 1189. (November 2013)

C. 1189. Find all integers n such that \frac{n^2+2n-8}{n^2+n-12} is also an integer.

(5 pont)

Deadline expired on December 10, 2013.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. \(\displaystyle \frac{n^2+2n-8}{n^2+n-12}=\frac{n^2+n-12+n+4}{n^2+n-12}=1+\frac{n+4}{n^2+n-12}\). Vegyük észre, hogy \(\displaystyle n^2+n-12=(n+4)(n-3)\), így a kifejezés tovább alakítható: \(\displaystyle 1+\frac{n+4}{n^2+n-12}=1+\frac{n+4}{(n+4)(n-3)}=1+\frac{1}{n-3}\), ahol \(\displaystyle n\neq3\) és \(\displaystyle n\neq-4\). Ez pontosan akkor egész szám, ha \(\displaystyle n-3=-1\) vagy \(\displaystyle n-3=1\). Ekkor \(\displaystyle n=2\) vagy \(\displaystyle n=4\).

Megjegyzés: Ha valaki a számlálót rögtön átalakítja: \(\displaystyle n^2+2n-8=(n+4)(n-2)\), akkor a törtet így írhatja fel: \(\displaystyle \frac{n^2+2n-8}{n^2+n-12}=\frac{(n+4)(n-2)}{(n+4)(n-3)}=\frac{n-2}{n-3}=1+\frac{1}{n-3}\).


196 students sent a solution.
5 points:82 students.
4 points:69 students.
3 points:26 students.
2 points:6 students.
1 point:4 students.
0 point:6 students.
Unfair, not evaluated:3 solutions.

Problems in Mathematics of KöMaL, November 2013