February 16th, 2019, 11:36 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 511 Thanks: 79  Square of a number
Find all values of $\displaystyle n$ such that $\displaystyle n^2 +64n +646 \; $ is a square of a natural number . $\displaystyle n\in \mathbb{N}$ . 
February 16th, 2019, 12:34 PM  #2 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry 
If $n^2 + 64n + 646 = m^2$ with $n$ and $m$ natural, then $n = \dfrac{64 \pm \sqrt{64^2  4 \times (646  m^2)}}{2}$. So $64^2  4 \times (646  m^2) = 4 \times (378 + m^2)$ is a square number. This implies $378 + m^2$ is itself a square, say $378 + m^2 = k^2$. But $378 \equiv 2 \bmod 4$, while the difference of two squares must be $0$, $1$ or $3$ $\bmod 4$. This is a contradiction. Hence there are no solutions.

February 16th, 2019, 12:46 PM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 511 Thanks: 79 
Saw it in a old page , seems like it is wrong.

February 16th, 2019, 01:04 PM  #4  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,656 Thanks: 2634 Math Focus: Mainly analysis and algebra  Quote:
\begin{align}n^2 + 64n + 646 &= m^2 \\ n^2 + 64n + 1024 &= m^2 + 378 \\ (n+32)^2 &= m^2 + 378\end{align}  
February 16th, 2019, 01:36 PM  #5 
Senior Member Joined: Sep 2016 From: USA Posts: 609 Thanks: 378 Math Focus: Dynamical systems, analytic function theory, numerics 
It's worth pointing out that taking the original expression mod 4 gives the result immediately, since $n^2 + 64n + 646$ can only be $2$ or $3$ mod 4 while any square can only be $0$ or $1$ mod 4. Of course, I noticed this after having the benefit of reading cjem's solution, so I doubt I would have initially solved it differently than cjem. Last edited by skipjack; February 17th, 2019 at 09:04 AM. 
February 16th, 2019, 01:36 PM  #6  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,414 Thanks: 1024  Quote:
If the 646 is changed to 640, then 3 solutions: n = 3, 18, 65  

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