Problem of the Week # 247 - Jan 02, 2017

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The discussion centers around Problem of the Week #247, which challenges participants to prove that for any integers \(a\), \(b\), and \(c\), there exists a positive integer \(n\) such that \(\sqrt{n^3 + an^2 + bn + c}\) is not an integer. This problem is derived from Problem B-6 of the 1998 William Lowell Putnam Mathematical Competition. Participants kaliprasad and vidyarth provided correct solutions, showcasing their mathematical reasoning and problem-solving skills.

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Here is this week's POTW, the first of the (prime) new year!

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Prove that, for any integers $a, b, c$, there exists a positive integer $n$ such that $\sqrt{n^3+an^2+bn+c}$ is not an integer.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Re: Problem Of The Week # 247 - Jan 02, 2017

This was Problem B-6 in the 1998 William Lowell Putnam Mathematical Competition.

Congratulations to kaliprasad and vidyarth for their correct solutions. vidyarth's solution follows:

Let us assume, for the sake of contradiction, that, $P(n)=n^3+an^2+bn+c$ is a perfect square. Now, $P(1)$ and $P(3)$; $P(2)$ and $P(4)$ are squares of the same parity. Hence, their differences must be divisible by $4$. Thus, $P(1)-P(3)=26+8a+2b$ and $P(2)-P(4)=56+12a+2b$ each must be divisible by $4$ which in the first condition gives us $b$ to be odd and, in the second case, $b$ to be even, thus giving us a contradiction.
 

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