hanulyeongsam wrote: Find the addition of all positive integers n that follows the following: $ \frac{\sqrt{n}}{2} + \frac{30}{\sqrt{n}} $ is an integer. Let $\sqrt{n}=k,$ we want $\frac{k}{2}+\frac{30}{k}=\frac{k^2+60}{2k}$ to be an integer. Now note that for $\frac{k}{2}+\frac{30}{k},$ to be an integer we can't have $k>30.$ Now we want $\frac{30}{k}$ to either have fractional part $0$ or $1.5$ Simple casework yeilds $k=2, k=6, k=10, k=30$ so we have that $n=4, 36, 100, 900$ are the only solutions that work. (check my work as I speedran this problem in like 30 seconds so I might have sillied a case)

mathmax12 wrote: hanulyeongsam wrote: Find the addition of all positive integers n that follows the following: $ \frac{\sqrt{n}}{2} + \frac{30}{\sqrt{n}} $ is an integer. Let $\sqrt{n}=k,$ we want $\frac{k}{2}+\frac{30}{k}=\frac{k^2+60}{2k}$ to be an integer. Now note that for $\frac{k}{2}+\frac{30}{k},$ to be an integer we can't have $k>30.$ Now we want $\frac{30}{k}$ to either have fractional part $0$ or $1.5$ Simple casework yeilds $k=2, k=6, k=10, k=30$ so we have that $n=4, 36, 100, 900$ are the only solutions that work. (check my work as I speedran this problem in like 30 seconds so I might have sillied a case) The answer (it seems 1040 according to your solution) is correct. Congrats.!

Since$$\frac{n+60}{2\sqrt{n}}=\frac{\sqrt{n}}{2}+\frac{30}{\sqrt{n}}\in \mathbb{Z}$$and $n\in \mathbb{N}$, we get that $\sqrt{n}\in \mathbb{Q}$, which means that $n$ is a perfect square. Let $n=k^2$, with $k\in \mathbb{N}$, then$$\frac{k^2+60}{2k}=\frac{\sqrt{n}}{2}+\frac{30}{\sqrt{n}}\in \mathbb{Z},$$so$$\frac{60-k^2}{2k}=\frac{k^2+60}{2k}-k\in \mathbb{Z},$$which gives$$\frac{60}{k}=\frac{k^2+60}{2k}+\frac{60-k^2}{2k}\in \mathbb{Z}.$$Let $t=\frac{60}{k}\in \mathbb{Z}$. Then$$\frac{k+t}{2}=\frac{k^2+60}{2k}\in \mathbb{Z},$$so $k\equiv t\pmod 2$. Since $kt=60\equiv 0\pmod 2$, we conclude that $k\equiv t\equiv 0\pmod 2$. Further, as $60\not \equiv 0\pmod 8$, we have $k\not \equiv 0\pmod 4$, which gives $k\equiv 2\pmod 4$, because $k\equiv 0\pmod 2$. Since $k\mid 60$ and $k\in \mathbb{N}$, the only possible values of $k$ are $2$, $6$, $10$ and $30$, which work. Hence the $n$ that satisfy the condition are $4$, $36$, $100$ and $900$, which add up to $1040$.

mathmax12 wrote: hanulyeongsam wrote: Find the addition of all positive integers n that follows the following: $ \frac{\sqrt{n}}{2} + \frac{30}{\sqrt{n}} $ is an integer. Let $\sqrt{n}=k,$ we want $\frac{k}{2}+\frac{30}{k}=\frac{k^2+60}{2k}$ to be an integer. Now note that for $\frac{k}{2}+\frac{30}{k},$ to be an integer we can't have $k>30.$ Now we want $\frac{30}{k}$ to either have fractional part $0$ or $1.5$ Simple casework yeilds $k=2, k=6, k=10, k=30$ so we have that $n=4, 36, 100, 900$ are the only solutions that work. (check my work as I speedran this problem in like 30 seconds so I might have sillied a case) I think you have to prove that $k$ is an integer, because for example the solutions to $\frac{k^2+60}{2k}=9$ are $k=9\pm \sqrt{21}$, and then $\left \{\frac{30}{k}\right \}\neq 0,\frac{1}{2}$. Of course this isn't a possible solution because in that case $k^2$ is not an integer, but a priori $\frac{k^2+60}{2k}$ being an integer doesn't imply that $\left \{\frac{30}{k}\right \}$ is either $0$ or $\frac{1}{2}$.

$$n\in\{4,36,100,400,900\}$$

Let's consider the cases. case 1)√n is even: √n is a divisor of 30. n=4,36,100,900 case 2)√n is odd: (√n/2)+(30/√n)=(n+60)/(2√n). n+60 is odd,2√n is even --->Impossible. 4+36+100+900=1040. The answer is 1040.