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What $n$ solves $n^2 + 2n + 2019$ for the expression to be a perfect square?

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    $\begingroup$ What's the source of this problem? The appearance of $2019$ makes this seem like it might be part of an on-going contest. $\endgroup$ – Blue Apr 20 at 18:34
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    $\begingroup$ In standard form, the equation $n^2+2n+2019=0$ results, from the quadratic formula, with a negative discriminant $$n=\frac{-2±\sqrt{4-4*1*2019}}{2*1}$$ so any solution $n$ would be complex. $\endgroup$ – poetasis Apr 20 at 19:03
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$n^2+2n+2019=(n+1)^2+2018=m^2$ implies: $$2018=m^2-(n+1)^2=(m-n-1)(m+n+1)=2\times 1009$$

Can you finish it?

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  • $\begingroup$ No integers solutions? $\endgroup$ – Varaun Ramgoolie Apr 22 at 14:29
  • $\begingroup$ Exactly. Therefore this expression is never a perfect square $\endgroup$ – HAMIDINE SOUMARE Apr 22 at 14:30

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