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 February 16th, 2019, 11:36 AM #1 Senior Member   Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 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. Thanks from topsquark, idontknow and SDK
 February 16th, 2019, 12:46 PM #3 Senior Member   Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 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: 578 Thanks: 345 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

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