If $u_1=1$ and $u_{n+1} = n+\sum_{k=1}^n u_k^2$, then $u_n$ is never a square. Cute problem I saw on quora:
If
$$u_1 = 1,\qquad
u_{n+1} = n+\sum_{k=1}^n u_k^2
$$
show that,
for $n \ge 2$,
$u_n$ is never a square.
\begin{align}
n&=1:& u_2 &= 1+1 = 2\\
n&=2:& u_3 &= 2+1+4 = 7\\
n&=3:& u_4 &= 3+1+4+49 = 57\\
\end{align}
And, as usual,
I have a generalization:
If
$$a \ge 1,\quad b \ge 0,\quad u_1 = 1,\quad
u_{n+1} = an+b+\sum_{k=1}^n u_k^2,$$
where $a, b \in \mathbb{Z}$, then
for $n \ge 3$,
$u_n$ is never a square.
Note:
My solutions to these
do not involve
any explicit expressions
for the $u_n$.
 A: The recurrence can be rewritten as
\begin{eqnarray*}
u_{n+1}=u_n^2+u_n+1.
\end{eqnarray*}
It is easy to show that 
\begin{eqnarray*}
u_n^2 < u_{n+1} <(u_n+1)^2.
\end{eqnarray*}
Now observe that there is no whole numbers between $u_n$ & $u_n+1$.
A: The problem, indeed, is very cute. 
We just need to check the condition $\textrm{mod } 5$:
$$u_1 \equiv 1 \textrm{ mod } 5$$
$$u_2 \equiv 2 \textrm{ mod } 5$$
$$u_3 \equiv 2 \textrm{ mod } 5$$
etc.
Therefore, one can show by induction that for any $n \in \mathbb{N}$:
$$u_{n+1} \equiv \left[ n+1+\sum_{j=2}^n (-1) \right] \textrm{ mod } 5 \equiv 2 \textrm{ mod } 5.$$
The latter is never true for squares, since squares are equal to either 0,1 or 4 mod 5.
To prove the generalization, we use the recurrence relation of type 
$$u_{n+1} = u_n^2+u_n+a.$$
Notice that $u_{n+1}>u_n^2$ since $a>0$. 
Therefore, if $u_{n+1}$ is a perfect square, then $a\geq u_n+1 > u_n$.
But $a>u_n$ is impossible since for any $n \geq 3$: 
$$u_{n} = u_{n-1}^2+u_{n-1}+a >a.$$
This gives us a contradiction, so $u_n$ is never a perfect square for any 
$n\geq 3$.
