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Difference between revisions of "1991 OIM Problems/Problem 5"
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Since <math>(-y)^2=y^2</math> we can look at the combinations of <math>y</math> with <math>K</math> for non-negative values.  So, we can use: <math>0 \le y \le 14</math> to find the values of <math>P</math>
 
Since <math>(-y)^2=y^2</math> we can look at the combinations of <math>y</math> with <math>K</math> for non-negative values.  So, we can use: <math>0 \le y \le 14</math> to find the values of <math>P</math>
  
−
Since  
+
Since <math>x=\frac{3y \pm K}{2}</math>, <math>P=\frac{K^2+y^2}{2}</math>, then to get integers <math>x</math> and <math>P</math>, both expressions <math>K^2+y^2</math> and <math>3y \pm K</math> need to be even.  This happens when either <math>K</math> and <math>y</math> 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.  
 
* 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.  

Revision as of 20:08, 22 December 2023

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.

See also

https://www.oma.org.ar/enunciados/ibe6.htm

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