MHB Minimum Natural Number for Fraction Expression #328 Aug 24, 2018 POTW

[anemone]

Here is this week's POTW:

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Find the minimum natural number of $\dfrac{a^2+2ab+2017b^2}{a+b}$ where $a$ and $b$ are natural numbers.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[Greg]

Greetings guests and fellow MHB members! :D I'll be filling in for anemone for about four weeks. Your participation in the upcoming POTWs is welcomed, as always!

Two members correctly completed last week's POTW. They are

1. castor28
2. kaliprasad

Here is castor28's solution:

[sp]The expression under consideration can be written as:
$$
f(a,b) = a + b + \frac{2016b^2}{a+b}
$$
We start by assuming that $a$ and $b$ are real numbers subject to the condition $a,b\ge1$. We have:
$$
\frac{\partial f}{\partial a}= 1 - \frac{2016b^2}{(a+b)^2}
$$
This shows that, for fixed $b$, $f(a,b)$ has a single minimum at $a=b(\sqrt{2016}-1)\approx 43.9b$ with value $2b\sqrt{2016}\approx 89.8b$. This is an increasing function of $b$, and the global minimum of $f(a,b)$ in the region $a,b\ge1$ is $89.8$, corresponding to $b=1$. In particular, the smallest integer value that $f(a,b)$ can take is $90$.

We must now use the fact that $a$ and $b$ are integers, and $(a+b)\mid2016b^2$. We start by taking $b=1$ and looking for divisors of $2016$ close to $\sqrt{2016}\approx 44.9$. The closest such divisor is $42$. Taking $a+1=42$ gives $f(41,1)=90$. As we have shown that this is the smallest integer value that $f(a,b)$ can take, there is no need to look at other values of $a$ and $b$.[/sp]