Tessellate S.T.E.M.S. (2019) - Mathematics - Category B - Set 2 - Objective Problem 1

Since $n^2 = 3k^2+3k+1,$ we rearrange terms to get $(2n)^2-3(2k+1)^2=1$ and using your favorite Pell equation solution method, we get a sequence of solutions $n=1,13,181,2521,\ldots.$ Here it is in OEIS . There is a remark in the OEIS entry that each such $n$ is a sum of consecutive squares, which means that $2n-1$ is a square. (The sequence $1,5,19,71,\ldots$ of square roots also appears in the solutions to the same Pell equation, so that should give a road map to the proof, although I have not written it out.)

Anyway, assuming this fact, if $2n+79$ is also a square, say $a^2,$ then we get $a^2-b^2 = 80$ for some $b.$ So $a-b$ and $a+b$ are factors of $80$ with the same parity. The possible choices are $(2,40), (4,20), (8,10),$ which lead to $(a,b) = (21,19), (12,8), (9,1)$ respectively. This leads to $n=181, 32.5, 1$ respectively. Two of those actually appear in the sequence of solutions, and the larger of the two is $\fbox{181}.$