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 February 16th, 2019, 11:36 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 647 Thanks: 94 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: 313 Thanks: 112 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. Thanks from topsquark, idontknow and SDK February 16th, 2019, 12:46 PM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 647 Thanks: 94 Saw it in a old page , seems like it is wrong. February 16th, 2019, 01:04 PM   #4
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Quote:
 Originally Posted by cjem 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 $378 + m^2 = k^2$.
Perhaps a tidier way of expressing this is by completing the square:
\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: 648 Thanks: 412 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. Thanks from topsquark and cjem Last edited by skipjack; February 17th, 2019 at 09:04 AM. February 16th, 2019, 01:36 PM   #6
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Quote:
 Originally Posted by idontknow Find all values of $\displaystyle n$ such that $\displaystyle n^2 +64n +646 \;$ is a square of a natural number . $\displaystyle n\in \mathbb{N}$ .
n^2 + 64n + 646 = u^2 : NO solutions if u = natural number

If the 646 is changed to 640, then 3 solutions: n = 3, 18, 65 Tags number, square Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post idontknow Number Theory 2 March 27th, 2018 06:19 PM jiasyuen Number Theory 5 March 10th, 2015 02:57 PM M_B_S Algebra 23 November 20th, 2013 12:01 AM Dougy Number Theory 6 June 17th, 2012 07:00 PM Cruella_de_Vil Number Theory 5 March 25th, 2012 06:03 PM

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