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I need all the values for $n$, where $n^2 + 2019$ is exact square.
I've found all up to $999,999,999$.

Should I keep going or am I not going to find any higher?

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    $\begingroup$ Hint: $n^2+2019=m^2\implies 2019=(m+n)(m-n)$. $\endgroup$
    – lulu
    Mar 28, 2019 at 9:02
  • $\begingroup$ and $2019=(673)(3)$ $\endgroup$
    – E.H.E
    Mar 28, 2019 at 9:06
  • $\begingroup$ I get 8 solutions. If we assume $n,a>0$ then there are only two solutions. $\endgroup$
    – Lozenges
    Mar 28, 2019 at 10:24
  • $\begingroup$ How come 8 solutions ? $\endgroup$ Mar 30, 2019 at 21:57

2 Answers 2

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Hint: If $n^2+2019=a^2$ for some $a$, then $a$ needs to be at least $n+1$. If $n$ is big enough, then $(n+1)^2>n^2+2019$ so there can be no more solutions after that. Can you compute whether checking until $999,999,999$ is enough?

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    $\begingroup$ Yep, found out I should have stopped much before that :D $\endgroup$ Mar 28, 2019 at 9:22
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We are looking for the solutions for $a$ in the equation: $a^2 = n^2 + 2019$ where $a, n \in \mathbb{Z}$. Rewriting,

\begin{align*} a^2 - n^2 = 2019 \\ (a + n)(a - n) =2019 \end{align*}

Does this provide enough of a hint to continue?

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  • $\begingroup$ Yep, absolutely! $\endgroup$ Mar 28, 2019 at 9:14

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