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

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

Sep 22, 2017

#2
+27480
+4

Here's a more general solution:

.

Sep 23, 2017

#1
+97581
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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.

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

 n -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 $$\left\lfloor n^2/4 \right\rfloor$$ 25 20 16 12 9 6 4 2 1 0 0 0 1 2 4 6 9 12 16 20 25 $$\lfloor n/2 \rfloor^2$$ 25 25 16 16 9 9 4 4 1 1 0 0 1 1 4 4 9 9 16 16 25 $$\left\lfloor n^2/4 \right\rfloor - \lfloor n/2 \rfloor^2$$ 0 -5 0 -4 0 -3 0 -2 0 -1 0 0 0 1 0 2 0 3 0 4 0

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

Sep 23, 2017
#2
+27480
+4

Here's a more general solution:

.

Alan Sep 23, 2017
#3
+97581
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Thanks Alan, your solution is really neat :)

Melody  Sep 23, 2017