## T-7

Find all positive integers $n$ for which there exist positive integers $a>b$ satisfying

$$
n=\frac{4 a b}{a-b}
$$

Answer. Any composite $n \neq 4$.


Solution. We say that $n$ is good if there exist such positive integers $a$ and $b$ (and bad otherwise).

First, we show that $n=4$ is bad. Suppose otherwise. Then there exist positive integers $a, b$ such that

$$
4=\frac{4 a b}{a-b}
$$

But this rewrites as $(a+1)(b-1)=-1$ and the left-hand side is non-negative for any two positive integers $a, b$, a contradiction.

Now consider $n \geq 6$. We make three observations.

First, note that if $n$ is good then any its multiple $n^{\prime}=k \cdot n$ is also good - it suffices to take $a^{\prime}=k \cdot a$ and $b^{\prime}=k \cdot b$. In particular, for $a=2, b=1$ we have $4 a b /(a-b)=8$, hence all multiples of 8 are good.

Second, take any positive integer $t$ of the form $t=4 k+1$. By setting $a=t, b=1$ we get

$$
\frac{4 a b}{a-b}=\frac{4 \cdot t}{4 k}=\frac{t}{k}
$$

hence by setting $a=t \cdot k, b=k$ we obtain that any multiple of any number of the form $t=4 k+1$ is good.

Third, likewise, take any positive integer $t$ of the form $t=4 k-1$. By setting $a=t, b=1$ we get

$$
\frac{4 a b}{a-b}=\frac{4 \cdot t}{4 k-2}=\frac{2 t}{2 k-1}
$$

hence every multiple of double of any number of the form $t=4 k-1$ is good.

Now we combine the observations. Consider any composite number $n \geq 6$. If the prime factorization of $n$ contains a prime of the form $4 k+1$ then $n$ is good (by the second observation). Similarly, if $n$ contains at least two (not necessarily distinct) primes of the form $4 k-1$ then their product is of the form $4 k+1$, and thus $n$ is good (by the second observation). If $n$ contains a prime of the form $4 k-1$ and it is even, then $n$ is also good (by the third observation). Thus,
it remains to consider the powers of 2 . But we know that all multiples of 8 are good (by the first observation) and 4 is bad, so we are done.

