1991 OIM Problems/Problem 5

Problem

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

Solution

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 $0 \le K \le 14$ , $-14 \le y \le 14$ because $15^2/2>100$

Since $(-y)^2=y^2$ we can look at the combinations of $y$ with $K$ for non-negative values. So, we can use: $0 \le y \le 14$ to find the values of $P$

Since $x=\frac{3y \pm K}{2}$ , $P=\frac{K^2+y^2}{2}$ , then to get integers $x$ and $P$ , both expressions $K^2+y^2$ and $3y \pm K$ need to be even. This happens when either $K$ and $y$ are both odd, or both even.

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.

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1991 OIM Problems/Problem 5

Problem

Solution

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