719. Find all numbers of the form $222 \ldots 2$ that can be represented as the sum of two perfect squares.
$\triangle$ Let

$$
222 \ldots 2=a^{2}+b^{2}
$$

where $a$ and $b$ are integers. Then the numbers $a$ and $b$ can only be odd:

$$
a=2 k+1, b=2 l+1 \quad(k \in Z, l \in Z)
$$

Therefore, the sum $a^{2}+b^{2}$ when divided by 8 gives a remainder of 2:

$$
a^{2}+b^{2}=(2 k+1)^{2}+(2 l+1)^{2}=4 k(k+1)+4 l(l+1)+2
$$

On the other hand, of the numbers of the form $222 \ldots 2$, only the number 2 gives a remainder of 2 when divided by 8, since if the number of twos in this number is greater than 1, then

$$
22 \ldots 2=22 \ldots 2 \cdot 100+22
$$

where the first term of the obtained sum is divisible by 8, and the second term gives a remainder of 6 when divided by 8.

Answer: 2.