Difference between revisions of "1991 OIM Problems/Problem 5"

Revision as of 21:01, 22 December 2023

Let $P(x,y) = 2x^2 - 6xy + 5y^2$ . We will say that an integer $a$ is a value of $P$ if there exist integers $b$ and $c$ such that $a=P(b,c)$ .

i. Determine how many elements of {1, 2, 3, ... ,100} are values of $P$ .

ii. Prove that the product of values of $P$ is a value of $P$ .

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Part i.

Let $x$ , $y$ , $P$ be integers

$2x^2 - 6xy + 5y^2-P=0$ , then solving for $x$ using the quadratic equation we have:

$x=\frac{3y \pm \sqrt{2P-y^2}}{2}$

Let $K$ be an integer and $K^2=2P-y^2$ . Therefore, $P=\frac{K^2+y^2}{2}$ Since $1 \le P \le 100$ , then $0 \le K,y \le 14$ because $15^2/2>100$

Note. I actually competed at this event in Argentina when I was in High School representing Puerto Rico. I have no idea what I did on this one nor how many points they gave me. Probably close to zero on this one.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

@@ Line 18: / Line 18: @@
 <math>x=\frac{3y \pm \sqrt{2P-y^2}}{2}</math>
-Let <math>K</math> be an integer and <math>K^2=2P-y^2</math>.  Since <math>1 \le P \le 100</math>, then
+Let <math>K</math> be an integer and <math>K^2=2P-y^2</math>. Therefore, <math>P=\frac{K^2+y^2}{2}</math> Since <math>1 \le P \le 100</math>, then <math>0 \le K,y \le 14</math> because <math>15^2/2>100</math>
 * Note.  I actually competed at this event in Argentina when I was in High School representing Puerto Rico.  I have no idea what I did on this one nor how many points they gave me.  Probably close to zero on this one.