Integral value of $n$ that makes $n^2+n+1$ a perfect square. Find all integers $n$ for which $n^2+n+1$ is a perfect square.
By hit and trial we get $n=-1,0$ but could someone suggest any genuine approach as how to approach this problem?

We have: $n^2 + n + 1 = m^2 \implies n^2 + n + (1-m^2) = 0 \implies \triangle = 1^2 - 4(1)(1-m^2) = k^2 \implies 1-4(1-m^2) = k^2 \implies 4m^2 - k^2 = 3 \implies (2m+k)(2m-k) = 3\implies 2m+k = 3, 2m-k = 1$ or $2m+k = -1, 2m-k = -3$. Either case gives $n^2 + n = 0 \implies n = -1, 0$. 
