3. Given two sets of numbers, set $A$ is: $1,2, \cdots, 100$; set $B$ is: $1^{2}, 2^{2}, \cdots, 100^{2}$. For a number $x$ in set $A$, if there is a number $y$ in set $B$ such that $x+y$ is also a number in set $B$, then $x$ is called an "associated number". Therefore, the number of such associated numbers in set $A$ is $\qquad$.
3.73.

Let $x+y=a^{2}, y=b^{2}$, then $1 \leqslant b<a \leqslant 100$.
And $x=a^{2}-b^{2}=(a+b)(a-b) \leqslant 100$, since $a+b$ and $a-b$ have the same parity, hence $a+b \geqslant(a-b)+2$.
(1) If $a-b=1$, then $a+b$ is odd, and $3 \leqslant a+b \leqslant$ 99. Thus, $a+b$ can take the values $3,5,7, \cdots, 99$, a total of 49 values, at this time, the corresponding $x$ can also take these 49 values.
(2) If $a-b=2$, then $a+b$ is even, and $4 \leqslant a+b \leqslant$ 50. Thus, $a+b$ can take the values $4,6,8, \cdots, 50$, a total of 24 values, at this time, the corresponding $x$ can take the values $8,12,16, \cdots, 100$ for 24 values.
In other cases, the values of $x$ obtained all fall into the above scenarios.
If $a-b=$ is odd, then $a+b=$ is odd.
And $x=a^{2}-b^{2} \geqslant a+b \geqslant 3$, falling into (1).
If $a-b=$ is even, then $a+b=$ is even.
And $x=(a-b)(a+b)$ is a multiple of 4, and $a-b \geqslant 2$, $a+b \geqslant 4$, hence $x \geqslant 8$, falling into (2).

Therefore, there are a total of $49+24=73$ such $x$.