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Difference between revisions of "1991 OIM Problems/Problem 5"
Line 28: Line 28:
 
Let <math>K=2n</math>, <math>y=2m</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>
 
Let <math>K=2n</math>, <math>y=2m</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>
  
 +
Now we try the possible combinations of <math>n</math> and <math>m</math>:
  
 +
<math>\begin{cases}
 +
m=0\text{; }n=0\text{; }P=(0^2+0^2)/2=0; & \text{NO}\\
 +
m=0\text{; }n=1\text{; }P=(0^2+1^2)/2=2; & \text{YES}\\
 +
m=0\text{; }n=2\text{; }P=(0^2+2^2)/2=8; & \text{YES}\\
 +
m=0\text{; }n=3\text{; }P=(0^2+3^2)/2=18; & \text{YES}\\
 +
m=0\text{; }n=4\text{; }P=(0^2+4^2)/2=32; & \text{YES}\\
 +
m=0\text{; }n=5\text{; }P=(0^2+5^2)/2=50; & \text{YES}\\
 +
m=0\text{; }n=6\text{; }P=(0^2+6^2)/2=72; & \text{YES}\\
 +
m=0\text{; }n=7\text{; }P=(0^2+7^2)/2=98; & \text{YES}\\
 +
m=1\text{; }n=1\text{; }P=(1^2+1^2)/2=4; & \text{YES}\\
 +
m=1\text{; }n=2\text{; }P=(1^2+2^2)/2=10; & \text{YES}\\
 +
m=1\text{; }n=3\text{; }P=(1^2+3^2)/2=20; & \text{YES}\\
 +
m=1\text{; }n=4\text{; }P=(1^2+4^2)/2=34; & \text{YES}\\
 +
m=1\text{; }n=5\text{; }P=(1^2+5^2)/2=52; & \text{YES}\\
 +
m=1\text{; }n=6\text{; }P=(1^2+6^2)/2=74; & \text{YES}\\
 +
m=1\text{; }n=7\text{; }P=(1^2+7^2)/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+3^2)/2=26; & \text{YES}\\
 +
m=2\text{; }n=4\text{; }P=(2^2+4^2)/2=40; & \text{YES}\\
 +
m=2\text{; }n=5\text{; }P=(2^2+5^2)/2=58; & \text{YES}\\
 +
m=2\text{; }n=6\text{; }P=(2^2+6^2)/2=80; & \text{YES}\\
 +
m=2\text{; }n=7\text{; }P=(2^2+7^2)/2=106; & \text{NO}\\
 +
m=3\text{; }n=3\text{; }P=(3^2+3^2)/2=36; & \text{YES}\\
 +
m=3\text{; }n=4\text{; }P=(3^2+4^2)/2=50; & \text{YES}\\
 +
m=3\text{; }n=5\text{; }P=(3^2+5^2)/2=68; & \text{YES}\\
 +
m=3\text{; }n=6\text{; }P=(3^2+6^2)/2=90; & \text{YES}\\
 +
m=3\text{; }n=7\text{; }P=(3^2+7^2)/2=116; & \text{NO}\\
 +
m=4\text{; }n=4\text{; }P=(4^2+4^2)/2=64; & \text{YES}\\
 +
m=4\text{; }n=5\text{; }P=(4^2+5^2)/2=82; & \text{YES}\\
 +
m=4\text{; }n=6\text{; }P=(4^2+6^2)/2=104; & \text{NO}\\
 +
m=4\text{; }n=7\text{; }P=(4^2+7^2)/2=130; & \text{NO}\\
 +
m=5\text{; }n=5\text{; }P=(5^2+5^2)/2=100; & \text{YES}\\
 +
m=5\text{; }n=6\text{; }P=(5^2+6^2)/2=122; & \text{NO}\\
 +
m=5\text{; }n=7\text{; }P=(5^2+7^2)/2=148; & \text{NO}\\
 +
m=6\text{; }n=6\text{; }P=(6^2+6^2)/2=144; & \text{NO}\\
 +
m=6\text{; }n=7\text{; }P=(6^2+7^2)/2=170; & \text{NO}\\
 +
m=7\text{; }n=7\text{; }P=(7^2+7^2)/2=196; & \text{NO}
 +
\end{cases}</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 20:21, 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 $0 \le m \le 7$

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

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

  • 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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