# Problem Of The Week #428 August 3rd, 2020

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#### anemone

##### MHB POTW Director
Staff member
Here is this week's POTW:

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Find (without calculus) a fifth degree polynomial $p(x)$ such that $p(x)+1$ is divisible by $(x-1)^3$ and $p(x)-1$ is divisible by $(x+1)^3$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!

#### anemone

##### MHB POTW Director
Staff member
Hi MHB again! I want to apologize for posting the same question I used about a year ago. Please disregard it and here is a new problem of the week #428:

Prove that $\lfloor \sqrt{n}+\sqrt{n+1} \rfloor=\lfloor \sqrt{4n+1} \rfloor =\lfloor \sqrt{4n+2} \rfloor=\lfloor \sqrt{4n+3} \rfloor$ for all positive integers $n$.

#### anemone

##### MHB POTW Director
Staff member
Congratulations to kaliprasad for his correct solution , which you can find below:
Because 4n + 2 and 4n + 3 cannot be squares so we have there exists integer x such that

$x^2 < = 4n + 1 < 4n + 2 < 4n + 3 < (x+1)^2$

Hence $x<= \sqrt{4n+ 1} < \sqrt{4n+2} < \lfloor \sqrt{4n+3} \rfloor < (x+1)$

Hence $x = \lfloor \sqrt{4n+ 1} \rfloor = \lfloor \sqrt{4n+2} \rfloor = \lfloor \sqrt{4n+3} \rfloor \cdots(1)$

we have $n(n+1) = (n+ \frac{1}{2})^2 - \frac{1}{4}$

so $\sqrt{n(n+1)} < n+\frac{1}{2}\cdots(2)$

And $n < \sqrt{n(n+1)}\cdots(3)$
We have
$(\sqrt{n} + \sqrt{n+1})^2 = n + n + 1 + 2\sqrt{n(n+1)}$

$= 2n + 1 + 2\sqrt{n(n+1)} > 2n + 1 + 2n$ (using (3))

or$(\sqrt{n} + \sqrt{n+1})^2 > + 4n + 1$

and $(\sqrt{n} + \sqrt{n+1})^2 = 2n + 1 + 2\sqrt{n(n+1)} < 2n + 1 + 2(n + \frac{1}{2})$ using (2)

or $(\sqrt{n} + \sqrt{n+1})^2 < 4n + 2$

so $4n + 1 < (\sqrt{n} + \sqrt{n+1})^2 < 4n + 2$

so $\lfloor \sqrt{4n+1} \rfloor = \lfloor \sqrt{n} + \sqrt{n+1} \rfloor$

Using (1) and above we get the result

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