Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1991 OIM Problems/Problem 5"
Line 70: Line 70:
 
m=7\text{; }n=7\text{; }P=2(7^2+7^2)=196; & \text{NO}
 
m=7\text{; }n=7\text{; }P=2(7^2+7^2)=196; & \text{NO}
 
\end{cases}</math>
 
\end{cases}</math>
 +
 +
'''Case 2:''' Both <math>K</math> and <math>y</math> are odd.
 +
 +
Let <math>K=2n+1</math>, <math>y=2m+1</math> where integers <math>n</math> and <math>m</math> with <math>0 \le n \le 7</math> and <math>0 \le m \le 7</math>
 +
 +
<math>P=\frac{K^2+y^2}{2}=\frac{4n^2+4m^2+4n+4n+2}{2}=2(n^2+m^2)+2(n+m)+1</math>
 +
 +
Now we try the possible combinations of <math>n</math> and <math>m</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.  
 
* 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 21:25, 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. Thus we will try both cases:

Case 1: Both $K$ and $y$ are even.

Let $K=2n$, $y=2m$ where integers $n$ and $m$ with $0 \le n \le 7$ and $0 \le m \le 7$

$P=\frac{K^2+y^2}{2}=\frac{4n^2+4m^2}{2}=2(n^2+m^2)$

Now we try the possible combinations of $n$ and $m$:

$\begin{cases} m=0\text{; }n=0\text{; }P=2(0^2+0^2)=0; & \text{NO}\\ m=0\text{; }n=1\text{; }P=2(0^2+1^2)=2; & \text{YES}\\ m=0\text{; }n=2\text{; }P=2(0^2+2^2)=8; & \text{YES}\\ m=0\text{; }n=3\text{; }P=2(0^2+3^2)=18; & \text{YES}\\ m=0\text{; }n=4\text{; }P=2(0^2+4^2)=32; & \text{YES}\\ m=0\text{; }n=5\text{; }P=2(0^2+5^2)=50; & \text{YES}\\ m=0\text{; }n=6\text{; }P=2(0^2+6^2)=72; & \text{YES}\\ m=0\text{; }n=7\text{; }P=2(0^2+7^2)=98; & \text{YES}\\ m=1\text{; }n=1\text{; }P=2(1^2+1^2)=4; & \text{YES}\\ m=1\text{; }n=2\text{; }P=2(1^2+2^2)=10; & \text{YES}\\ m=1\text{; }n=3\text{; }P=2(1^2+3^2)=20; & \text{YES}\\ m=1\text{; }n=4\text{; }P=2(1^2+4^2)=34; & \text{YES}\\ m=1\text{; }n=5\text{; }P=2(1^2+5^2)=52; & \text{YES}\\ m=1\text{; }n=6\text{; }P=2(1^2+6^2)=74; & \text{YES}\\ m=1\text{; }n=7\text{; }P=2(1^2+7^2)=100; & \text{YES}\\ m=2\text{; }n=2\text{; }P=2(2^2+2^2)=16; & \text{YES}\\ m=2\text{; }n=3\text{; }P=2(2^2+3^2)=26; & \text{YES}\\ m=2\text{; }n=4\text{; }P=2(2^2+4^2)=40; & \text{YES}\\ m=2\text{; }n=5\text{; }P=2(2^2+5^2)=58; & \text{YES}\\ m=2\text{; }n=6\text{; }P=2(2^2+6^2)=80; & \text{YES}\\ m=2\text{; }n=7\text{; }P=2(2^2+7^2)=106; & \text{NO}\\ m=3\text{; }n=3\text{; }P=2(3^2+3^2)=36; & \text{YES}\\ m=3\text{; }n=4\text{; }P=2(3^2+4^2)=50; & \text{YES}\\ m=3\text{; }n=5\text{; }P=2(3^2+5^2)=68; & \text{YES}\\ m=3\text{; }n=6\text{; }P=2(3^2+6^2)=90; & \text{YES}\\ m=3\text{; }n=7\text{; }P=2(3^2+7^2)=116; & \text{NO}\\ m=4\text{; }n=4\text{; }P=2(4^2+4^2)=64; & \text{YES}\\ m=4\text{; }n=5\text{; }P=2(4^2+5^2)=82; & \text{YES}\\ m=4\text{; }n=6\text{; }P=2(4^2+6^2)=104; & \text{NO}\\ m=4\text{; }n=7\text{; }P=2(4^2+7^2)=130; & \text{NO}\\ m=5\text{; }n=5\text{; }P=2(5^2+5^2)=100; & \text{YES}\\ m=5\text{; }n=6\text{; }P=2(5^2+6^2)=122; & \text{NO}\\ m=5\text{; }n=7\text{; }P=2(5^2+7^2)=148; & \text{NO}\\ m=6\text{; }n=6\text{; }P=2(6^2+6^2)=144; & \text{NO}\\ m=6\text{; }n=7\text{; }P=2(6^2+7^2)=170; & \text{NO}\\ m=7\text{; }n=7\text{; }P=2(7^2+7^2)=196; & \text{NO} \end{cases}$

Case 2: Both $K$ and $y$ are odd.

Let $K=2n+1$, $y=2m+1$ where integers $n$ and $m$ with $0 \le n \le 7$ and $0 \le m \le 7$

$P=\frac{K^2+y^2}{2}=\frac{4n^2+4m^2+4n+4n+2}{2}=2(n^2+m^2)+2(n+m)+1$

Now we try the possible combinations of $n$ and $m$:


  • 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_OIM_Problems/Problem_5&oldid=208151"