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MCIV Inequalities 1 Problem 4

 
 
Picture of DL Yu
MCIV Inequalities 1 Problem 4
by DL Yu - Friday, July 5, 2019, 8:33 AM
 

For Problem 4, I can't seem to find a way to approach the problem. If we play arond with equation (iii), we get (I plugged in a=2001 for all equations)

4002m - m^2 <= 4002n - n^2 - 2nm,

4002m - m^2 - mn <= 4002n - n^2 - 3nm,

f <= 4002 - n - 3m,

Which doesn't lead to anything useful, as of my knowledge. Could someone please tell me if this the right approach and I just don't see it, or if not, give me a hint?

Thanks in advance!
(Also can you type LaTeX in here? That might make my writing easier to read.)

 
Picture of Areteem Professor
Re: MCIV Inequalities 1 Problem 4
by Areteem Professor - Sunday, July 7, 2019, 5:57 PM
 

You can try to prove the following:

$f$ is an integer; $f$ is even; $f$ is positive.  So $f\geq 2$.  Then find a pair of values $m$ and $n$ so that $f=2$.  This is the minimum.

The maximum part is more tedious.  Show that $n$ is a factor of $2(a+1)$, and $n\leq 64$, thus $n\leq 52$.   And this gives a maximum of $f$.   I will give more details in class. 

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