Question

find all positive integers n for which
\[n^2+n+7\]
is a complete square

Answer 1

I believe, the only solution will be \(n=1,6\). Give me a minute to type up why.

Answer 2

First, notice that for it to be a positive square, we require the following relation to be true: \[n^2+n+7\geq n^2+2n+1\]Since \(n^2+2n+1=(n+1)^2\). If we solve that relation, we find that \(n\leq 6\).

Use trial and error on \(n=1,2,3,4,5,6\) and you find \(n=1,6\) work.

Answer 3

\[n^2< n^2+n+7<(n+3)^2 \\ n^2+n+7=(n+1)^2 \ \ or \ \ (n+2)^2 \\ n=6,1\]

Answer 4

@KingGeorge
nice work

Answer 5

Thanks.