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Difference between revisions of "1991 OIM Problems/Problem 5"
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<math>x=\frac{3y \pm \sqrt{2P-y^2}}{2}</math>
 
<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>. 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>
+
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 \le 14</math>, <math>-14 \le y \le 14</math> because <math>15^2/2>100</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
  
 
* 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 20:05, 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

  • 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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