How many integers $n$ are there such that $3 \leq n \leq 10$ and $121_n$ (the number written as $121$ in base $n$) is a perfect square?
To solve this problem, I need to think about what it means for $121_n$ to be a perfect square. This means that there is some integer $m$ such that $121_n = m^2$. I also need to remember how to convert numbers from different bases to base 10. For example, $121_n = 1 \cdot n^2 + 2 \cdot n + 1$. I can start with $n = 10$, since that is the easiest case. In base 10, $121_{10} = 121$, which is a perfect square, since $121 = 11^2$. So, $n = 10$ is one possible value. Next, I can try $n = 9$. In base 9, $121_9 = 1 \cdot 9^2 + 2 \cdot 9 + 1 = 100$, which is also a perfect square, since $100 = 10^2$. So, $n = 9$ is another possible value. Next, I can try $n = 8$. In base 8, $121_8 = 1 \cdot 8^2 + 2 \cdot 8 + 1 = 81$, which is again a perfect square, since $81 = 9^2$. So, $n = 8$ is another possible value. Next, I can try $n = 7$. In base 7, $121_7 = 1 \cdot 7^2 + 2 \cdot 7 + 1 = 64$, which is still a perfect square, since $64 = 8^2$. So, $n = 7$ is another possible value. Next, I can try $n = 6$. In base 6, $121_6 = 1 \cdot 6^2 + 2 \cdot 6 + 1 = 49$, which is also a perfect square, since $49 = 7^2$. So, $n = 6$ is another possible value. Next, I can try $n = 5$. In base 5, $121_5 = 1 \cdot 5^2 + 2 \cdot 5 + 1 = 36$, which is again a perfect square, since $36 = 6^2$. So, $n = 5$ is another possible value. Next, I can try $n = 4$. In base 4, $121_4 = 1 \cdot 4^2 + 2 \cdot 4 + 1 = 25$, which is still a perfect square, since $25 = 5^2$. So, $n = 4$ is another possible value. Finally, I can try $n = 3$. In base 3, $121_3 = 1 \cdot 3^2 + 2 \cdot 3 + 1 = 16$, which is also a perfect square, since $16 = 4^2$. So, $n = 3$ is another possible value. I have tried all the possible values of $n$ between 3 and 10, and they all work. Therefore, there are 8 integers $n$ that satisfy the given conditions. 