Flooring Square-Roots

For discussing Olympiad Level Algebra (and Inequality) problems
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sowmitra
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Flooring Square-Roots

Unread post by sowmitra » Tue Jun 12, 2012 5:26 pm

Prove that, \[\displaystyle \left \lfloor \sqrt{n}+\sqrt{n+1} \right \rfloor = \left \lfloor \sqrt{4n+2} \right \rfloor\]\[\forall n \epsilon \mathbb{N}\]
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sowmitra
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Re: Flooring Square-Roots

Unread post by sowmitra » Tue Jun 12, 2012 10:00 pm

Partial Solution:
Let, $a^2\leq{n}<(a+1)^2$
Then, Case 1:$n=a^2$.
So, $\sqrt{n}=a$, and, $\sqrt{n+1}=a+f$, where, $0<f<1$.
Therefore, \[\left \lfloor \sqrt{n}+\sqrt{n+1} \right \rfloor=\left \lfloor a+a+f \right \rfloor=2a\]
On the other hand, ${4n+2}=4a^2+2=(2a)^2+2$. Here, $(2a)^2+2$ is surely not a square, because, the distance between $(2a)^2$ and $(2a+1)^2$ is at least $5$. So,\[\left \lfloor \sqrt{4n+2} \right \rfloor=\left \lfloor \sqrt{(2a)^2+2} \right \rfloor=2a\]
A similar argument shows that the equality holds if $(n+1)$ is also a square. But, unfortunately I have not been able to prove it if neither $n$ nor $(n+1)$ is a square.
Really need some help. :?
Last edited by sowmitra on Tue Jun 12, 2012 10:12 pm, edited 1 time in total.
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shehab ahmed
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Re: Flooring Square-Roots

Unread post by shehab ahmed » Tue Jun 12, 2012 10:07 pm

Hint:
you can suppose that n=a^2 + b where 0<b<2a+1 and consider some other cases

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sowmitra
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Re: Flooring Square-Roots

Unread post by sowmitra » Thu Jun 14, 2012 6:51 pm

Finally, a breakthrough!...... :D (after a hideous amount of struggle :P ):
First, let, $a^2 \leq n<(a+1)^2$, as before.
Now, suppose, $n=a^2+z$, and furthermore, $\sqrt{n}=a+f_1$ and $\sqrt{n+1}=a+f_2$, where, $f_1$ and $f_2$ are the fractional parts of $\sqrt{n}$ and $\sqrt{n+1}$ respectively.
Case 1:$0<z \leq (a-1)$
Case 2:$(a+1) \leq z<(2a+1)$
Case 3:$z=a$

Case 1:
If $0<z \leq (a-1)$, then, $\sqrt{n}=\sqrt{a^2+z} \leq \sqrt{a^2+a-1} \leq\sqrt{(a+ \frac{1}{2})^2- \frac{5}{4}}<\sqrt{(a+\frac{1}{2})^2}=a+\frac{1}{2}$
Similarly, $\sqrt{n+1} \leq \sqrt{a^2+a}\leq\sqrt{(a+ \frac{1}{2})^2-\frac{1}{4}}<a+\frac{1}{2}$
So, $\left \lfloor \sqrt{n}+\sqrt{n+1} \right \rfloor<a+\frac{1}{2}+a+\frac{1}{2}=2a+1$

Now,\[\left \lfloor \sqrt{n}+\sqrt{n+1} \right \rfloor=\left \lfloor \sqrt{(a^2+z)}+\sqrt{(a^2+z+1)} \right \rfloor=\left \lfloor a+f_1+a+f_2 \right \rfloor=2a+\left \lfloor f_1+f_2 \right \rfloor\]
Therefore, $\left \lfloor \sqrt{n}+\sqrt{n+1} \right \rfloor \geq 2a$
Combining these two, we have,\[2a\leq\left \lfloor \sqrt{n}+\sqrt{n+1} \right \rfloor<2a+1\]
which implies that, $\left \lfloor \sqrt{n}+\sqrt{n+1} \right \rfloor=2a$
On the other hand, $\sqrt{4n+2}=\sqrt{4a^2+4z+2}\leq\sqrt{4a^2+4a-2}\leq\sqrt{(2a+1)^2-3}<\sqrt{(2a+1)^2}=2a+1$
Again, \[\left \lfloor \sqrt{4n+2} \right \rfloor=\left \lfloor \sqrt{(2a)^2+4z+2} \right \rfloor>\sqrt{(2a)^2}=2a\]
So, on the right hand side, we have,\[2a<\left \lfloor \sqrt{4n+2} \right \rfloor<2a+1\]
Therefore, $\left \lfloor \sqrt{4n+2} \right \rfloor=2a$, and, we have equality. :)
Oh! Finally finished Latexing. :mrgreen:
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