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If \(\left\lfloor n^2/4 \right\rfloor - \lfloor n/2 \rfloor^2 = 2\), then find all integer values of $n$.

michaelcai  Sep 22, 2017

Best Answer 

 #2
avatar+26741 
+4

Here's a more general solution:

 

.

Alan  Sep 23, 2017
 #1
avatar+92751 
+2

The only answer is 5.  I did this by examining a table of values.

There is bound to be a more elegant way of doing it but I do not know what it is.   frown

 

\(\left\lfloor n^2/4 \right\rfloor - \lfloor n/2 \rfloor^2 = 2\)

 

 

n-10-9-8-7-6-5-4-3-2-1012345678910
\(\left\lfloor n^2/4 \right\rfloor \)25201612964210001246912162025
\(\lfloor n/2 \rfloor^2\)2525161699441100114499161625
\(\left\lfloor n^2/4 \right\rfloor - \lfloor n/2 \rfloor^2\)0-50-40-30-20-100010203040

 

So I can see that for all even numbers, positive and negative the answer will be zero.

For the odd numbers a pattern emerges.

It is clear to me that the only integer value of n to make this true is 5

Melody  Sep 23, 2017
 #2
avatar+26741 
+4
Best Answer

Here's a more general solution:

 

.

Alan  Sep 23, 2017
 #3
avatar+92751 
+1

Thanks Alan, your solution is really neat :)

Melody  Sep 23, 2017

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