700*. In what number system is the number $11111_{d}$ a perfect square?
$\triangle$ Let's represent this number as a sum, decomposing it by powers of $d$:

$$
11111_{d}=d^{4}+d^{3}+d^{2}+d+1
$$

We need to find a natural number $d>1$ such that the resulting sum is a square of a quadratic trinomial with argument $d$ and integer coefficients.

We will bound this sum between two squares:

$$
\left(d^{2}+\frac{d}{2}\right)^{2}<d^{4}+d^{3}+d^{2}+d+1<\left(d^{2}+\frac{d}{2}+1\right)^{2}
$$

(verify the inequality!). Multiply the last inequality by 4 to avoid fractions:

$$
\left(2 d^{2}+d\right)^{2}<4 d^{4}+4 d^{3}+4 d^{2}+4 d+4<\left(2 d^{2}+d+2\right)^{2}
$$

Since the left and right parts of this double inequality are squares of natural numbers differing by 2, and the middle part must also be a square of a natural number, the base of the latter square can only be $2 d^{2}+d+1$. Solve the corresponding equation:

$$
\begin{gathered}
\left(2 d^{2}+d+1\right)^{2}=4 d^{4}+4 d^{3}+4 d^{2}+4 d+4 \\
4 d^{4}+d^{2}+1+4 d^{3}+4 d^{2}+2 d=4 d^{4}+4 d^{3}+4 d^{2}+4 d+4 \\
d^{2}-2 d-3=0, \quad d=3
\end{gathered}
$$

Answer: in the ternary system.