The Square Problem!

$\textbf{Case 1: } n=1$

By mere observation, First term of the sequence $14^1 + 11=25=5^2$ is a perfect square.

$\textbf{Case 2: } n \ge 2$

Every term of the sequence is of the form $4k+3$ where $k \in \mathbb N$

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Every number is either of the form $4m$ or $4m+1$ or $4m+2$ or $4m+3$ where $m \in \mathbb Z$

$(4m)^2 = 16m^2 = \boxed{4k} \text{ where }k=4m^2$

$(4m+1)^2 = 16m^2 + 8m+1 = \boxed{4k+1} \text{ where } k=4m^2+2m$

$(4m+2)^2 = 16m^2 + 16m + 4 = \boxed{4k} \text{ where } k=4m^2+4m+1$

$(4m+3)^2 = 16m^2+24m+9 = \boxed{4k+1} \text{ where }k=4m^2+6m+2$

$\therefore \textbf{No perfect square is of the form 4k+3}$

Hence the only solution is $\large \boxed{n=1}$

Observe that for n=1, the first term of the sequence is a perfect square. For n>=2, all the terms of the sequence are of the form 4k+3 So they cant be perfect squares.

Ans= 1