# Finding $n$ such that $n^2 + 2n + 2019$ is a perfect square

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

• What's the source of this problem? The appearance of $2019$ makes this seem like it might be part of an on-going contest. – Blue Apr 20 at 18:34
• 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. – poetasis Apr 20 at 19:03

$$n^2+2n+2019=(n+1)^2+2018=m^2$$ implies: $$2018=m^2-(n+1)^2=(m-n-1)(m+n+1)=2\times 1009$$