, 2008

Solution:
We will use the following lemmas.

Lemma 1. If $x \in \mathbb{N}$, then $x^{2} \equiv 0$ or $1 \pmod{3}$.

Proof: Let $x \in \mathbb{N}$, then $x=3k$, $x=3k+1$ or $x=3k+2$, hence
$$\begin{aligned} & x^{2}=9k^{2} \equiv 0 \pmod{3} \\ & x^{2}=9k^{2}+6k+1 \equiv 1 \pmod{3}, \\ & x^{2}=9k^{2}+12k+4 \equiv 1 \pmod{3}, \text{ respectively. } \end{aligned}$$
Hence $x^{2} \equiv 0$ or $1 \pmod{3}$, for every positive integer $x$.

Without proof we will give the following lemma.

Lemma 2. If $a$ is a positive integer then $a \equiv S(a) \pmod{3}$, where $S(a)$ is the sum of the digits of the number $a$.

Further we have
$$\begin{aligned} & (6k+1)^{6k+1}=\left[(6k+1)^{k}\right]^{6} \cdot (6k+1) \equiv 1 \pmod{3} \\ & (6k+2)^{6k+2}=\left[(6k+2)^{3k+1}\right]^{2} \equiv 1 \pmod{3} \\ & (6k+3)^{6k+3} \equiv 0 \pmod{3} \\ & (6k+4)^{6k+4}=\left[(6k+1)^{3k+2}\right]^{2} \equiv 1 \pmod{3} \\ & (6k+5)^{6k+5}=\left[(6k+5)^{3k+2}\right]^{2} \cdot (6k+5) \equiv 2 \pmod{3} \\ & (6k+6)^{6k+6} \equiv 0 \pmod{3} \end{aligned}$$
for every $k=1,2,3, \ldots$.

Let us separate the numbers $1^{1}, 2^{2}, \ldots, 2008^{2008}$ into the following six classes: $(6k+1)^{6k+1}$, $(6k+2)^{6k+2}$, $(6k+3)^{6k+3}$, $(6k+4)^{6k+4}$, $(6k+5)^{6k+5}$, $(6k+6)^{6k+6}$, $k=1,2,\ldots$.

For $k=1,2,3, \ldots$ let us denote by
$s_{k}=(6k+1)^{6k+1}+(6k+2)^{6k+2}+(6k+3)^{6k+3}+(6k+4)^{6k+4}+(6k+5)^{6k+5}+(6k+6)^{6k+6}$.

From (3) we have
$$s_{k} \equiv 1+1+0+1+2+0 \equiv 2 \pmod{3}$$
for every $k=1,2,3, \ldots$.

Let $A$ be the number obtained by writing one after the other (in some order) the numbers $1^{1}, 2^{2}, \ldots, 2008^{2008}$.

The sum of the digits, $S(A)$, of the number $A$ is equal to the sum of the sums of digits, $S\left(i^{i}\right)$, of the numbers $i^{i}, i=1,2, \ldots, 2008$, and so, from Lemma 2, it follows that
$$A \equiv S(A)=S\left(1^{1}\right)+S\left(2^{2}\right)+\ldots+S\left(2008^{2008}\right) \equiv 1^{1}+2^{2}+\ldots+2008^{2008} \pmod{3}$$
Further on $2008=334 \cdot 6+4$ and if we use (3) and (4) we get
$$\begin{aligned} A & \equiv 1^{1}+2^{2}+\ldots+2008^{2008} \\ & \equiv s_{1}+s_{2}+\ldots+s_{334}+2005^{2005}+2006^{2006}+2007^{2007}+2008^{2008} \pmod{3} \\ & \equiv 334 \cdot 2+1+1+0+1=671 \equiv 2 \pmod{3} \end{aligned}$$
Finally, from Lemma 1, it follows that $A$ can not be a perfect square.