Let $a$ be a positive integer such that there are exactly 10 perfect squares greater than $a$ and less than $2a$.

a) Find the smallest possible value of $a$.

b) Find the largest possible value of $a$.

#
Solution

a) Since there are exactly ten perfect squares greater than $a$ and less than $2a$, these ten perfect squares must be consecutive. Therefore, there must exist a positive integer $x$ such that

$$
(x-1)^{2} \leq a\frac{961}{2}>480
$$

which gives us

$$
a \geq 481
$$

Now, we observe that between 481 and $2(481)=962$, there are exactly ten perfect squares. Indeed, since $22^{2}=484$ and $31^{2}=961$, we have that

$$
481 < 22^{2} < 23^{2} < 24^{2} < 25^{2} < 26^{2} < 27^{2} < 28^{2} < 29^{2} < 30^{2} < 31^{2} < 962
$$

Next, we need to find the largest possible value of $a$ such that there are exactly ten perfect squares between $a$ and $2a$. We consider the range from 684 to $2(684)=1368$, so:

$$
684 < 27^{2} < 28^{2} < 29^{2} < 30^{2} < 31^{2} < 32^{2} < 33^{2} < 34^{2} < 35^{2} < 36^{2} < 1368
$$

From equation (.4) we deduce that it is not possible to choose a value greater than 684 for $a$. Therefore, we conclude that the largest possible value for $a$ is 684.