a_507_bc wrote: Find all positive integers $n$, for which there exist positive integers $a>b$, satisfying $n=\frac{4ab}{a-b}$. Define $c:=a-b>0$. We have $n=4b+\frac{4b^2}{c}$. In particular $n>4$. By choosing $c=4b^2, c=2b^2$ or $c=b^2$ we see that every number $n>4$ with $n\not\equiv3\pmod{4}$ works. If $n=\frac{4ab}{a-b}$ then $mn=\frac{4(ma)(mb)}{ma-mb}$. So every number $n>4$ that is not a prime $\equiv3\pmod{4}$ works. If $n=p$ for a prime $p\equiv3\pmod{4}$ we have $cp=4b(b+c)$. We can not have $p\mid 4$ and $p\mid b$ (this would imply $cp<4p(b+c)$). So $p\mid b+c$ and $4b\mid c$. Let $c=4bk$. We have $kp=b(1+4k)$. Since $p\nmid b$ we have $p\mid 1+4k$ and $b\mid k$. Taking $\pmod{k}$ gives $k\mid b$. So $k=b$ and $p=1+4k$ contrading $p\equiv 3\pmod{4}$. So exactly the numbers $n>4$ that are not primes $\equiv3\pmod{4}$ work.

We uploaded our solution https://calimath.org/pdf/MEMO2023-T7.pdf on youtube https://youtu.be/aVE9TX08oTk.

Let $gcd(a, b) = d$ and $a = xd$, $b = yd$. Then $n = \frac{4xyd}{x-y}$ which means $x | n$, $y | n$. If $ n = 4k$ $\Rightarrow$ set $x = 2, y = 1, d = k$ If $ n = 4k + 1$ $\Rightarrow$ set $x = 4k + 1, y = 1, d = k$ If $ n = 4k + 2$ $\Rightarrow$ set $x = 2k + 1, y = 1, d = 1$ If $ n = 4k + 3$ then we have two cases: 1) $n$ has a divisor $\equiv 1 (mod 4)$ $\Rightarrow$ call that divisor $4s + 1$ and let $n = (4s + 1)r$ where $s, r \in N$. Then, set $x = 4s + 1, y = 1, d = sr$ 2) $n$ doesn't have any divisor $\equiv 1 (mod 4)$ $\Rightarrow$ n must be a prime in form $4k + 3$. Then, we can only set $x = p$ and $y = 1$ which doesn't work.

Let $k = a - b$. Hence, we need to find all such $n$ that $n = \frac{4(b+k)b}{k}$, where $b$ and $k$ are positive integers. \[\frac{4(b+k)b}{k} = \frac{4b^2}{k} + 4b.\]Case 1: $n \equiv 0, 1, 2 \pmod{4}$. Taking $k$ equal to $4b^2$, $2b^2$, or $b^2$ yields any positive integer $n$ equal to $1$, $2$, or $0$ modulo 4, respectively. Case 2: $n \equiv 3 \pmod{4}$. $\gcd \left(\frac{4b^2}{k}, 4b\right) \neq 1$, hence $n$ is not prime. We prove that any composite $n \equiv 3 \pmod{4}$ can be represented as such a fraction. $n \equiv 3 \pmod{4}$, hence $n$ is divisible by a prime $p \equiv 3 \pmod{4}$. So, $\frac{n}{p} = 4l + 1$, where $l$ is a positive prime number. Let $b = pl$, $k = \frac{4b^2}{p}$, then $\frac{4b^2}{k} + 4b = p(4l + 1) = n$.