90. If the first digit of a four-digit number, which is a perfect square, is decreased by 3, and the last digit is increased by 3, the result is also a perfect square. Find this number.
S o l u t i o n. Let the given number be $a^{2}$, and the modified number $b^{2}$, then $a^{2}-3000+3=b^{2}$, or $a^{2}-b^{2}=2997$, from which $(a-b) \times$ $\times(a+b)=2997=3 \cdot 3 \cdot 3 \cdot 3 \cdot 37$. Clearly, $a+b$ is greater than 37, and from the problem statement, it follows that $a+b$ is less than 200 ( $a$ and $b$ are two-digit numbers). Therefore, $a+b=3 \cdot 37=111, a-b=3 \cdot 3 \cdot 3=27$. Then $2 a=138, a=69, a^{2}=4761$.

A n s w e r: 4761 .