Let $u$ and $v$ are positive integers. There exist nontrivial cases where their contraharmonic mean

(1)

The nontrivial integer contraharmonic means are in Sloane's sequence A146984.

Proposition 1. For any value of $u > 2$ , there are at least two greater values of $v$ such that $c \in \mathbb{Z}$ .

Proof. One has the identities

(2)

(3)

In (2) and (3), the value of $v$ is a multiple of $u$ , but it needs not be always so in order to $c$ be an integer, e.g. we have $u = 12,\; v = 20,\; c = 17$ .

Proposition 2. For all $u > 1$ , a necessary condition for $c \in \mathbb{Z}$ is that $$\gcd(u,\,v) > 1.$$

Proof. Suppose that we have positive integers $u,\,v$ such that $\gcd(u,\,v) = 1$ . Then as well, $\gcd(u\!+\!v,\,uv) = 1$ , since otherwise both $u\!+\!v$ and $uv$ would be divisible by a prime $p$ , and thus also one of the factors $u$ and $v$ of $uv$ would be divisible by $p$ ; then however $p \mid u\!+\!v$ would imply that $p \mid u$ and $p \mid v$ , whence we would have $\gcd(u,\,v) \geqq p$ . Consequently, we must have $\gcd(u\!+\!v,\,uv) = 1$ .

We make the additional supposition that $\displaystyle\frac{u^2\!+\!v^2}{u\!+\!v}$ is an integer, i.e. that $$u^2\!+\!v^2 = (u\!+\!v)^2-2uv$$ is divisible by $u\!+\!v$ . Therefore also $2uv$ is divisible by this sum. But because $\gcd(u\!+\!v,\,uv) = 1$ , the factor 2 must be divisible by $u\!+\!v$ , which is at least 2. Thus $u = v = 1$ .

The conclusion is, that only the ``most trivial case'' $u = v = 1$ allows that $\gcd(u,\,v) = 1$ . This settles the proof.

Proposition 3. If $u$ is an odd prime number, then (2) and (3) are the only possibilities enabling integer contraharmonic means.

Proof. Let $u$ be a positive odd prime. The values $v = (u-1)u$ and $v = (2u-1)u$ do always. As for other possible values of $v$ , according to the Proposition 2, they must be multiples of the prime number $u$ : $$v = nu, \quad n \in \mathbb{Z}$$ Now $$\mathbb{Z} \ni \frac{u^2+v^2}{u\!+\!v} = \frac{(n^2\!+\!1)u}{n\!+\!1},$$ and since $u$ is prime, either $u \mid n\!+\!1$ or $n\!+\!1 \mid n^2\!+\!1$ .

In the former case $n+1 = ku$ , one obtains $$c = \frac{(n^2\!+\!1)u}{n\!+\!1} = \frac{(k^2u^2-2ku+2)u}{ku} = ku^2-2+\frac{2}{k},$$ which is an integer only for $k = 1$ and $k = 2$ , corresponding (2) and (3).

In the latter case, there must be a prime number $p$ dividing both $n\!+\!1$ and $n^2\!+\!1$ , whence $p \nmid n$ . The equation $$n^2+1 \;=\; (n+1)^2-2n$$ then implies that $p \mid 2n$ . So we must have $p \mid 2$ , i.e. necessarily $p = 2$ . Moreover, if we had $4 \mid n\!+\!1$ and $4 \mid n^2\!+\!1$ , then we could write $n\!+\!1 = 4m$ , and thus $$n^2\!+\!1 = (4m\!-\!1)^2\!+\!1 = 16m^2\!-\!8m\!+\!2 \not\equiv 0 \pmod{4},$$ which is impossible. We infer, that now $\gcd(n\!+\!1,\,n^2\!+\!1) = 2$ , and in any case $$\gcd(n\!+\!1,\,n^2\!+\!1) \;\leqq\; 2.$$ Nevertheless, since $n\!+\!1 \geqq 3$ and $n\!+\!1 \mid n^2\!+\!1$ , we should have $\gcd(n\!+\!1,\,n^2\!+\!1) \geqq 3$ . The contradiction means that the latter case is not possible, and the Proposition 3 has been proved.

Proposition 4. If $(u_1,\,v,\,c)$ is a nontrivial solution of (1) with $u_1 < c < v$ , then there is always another nontrivial solution $(u_2,\,v,\,c)$ with $u_2 < v$ . On the contrary, if $(u,\,v_1,\,c)$ is a nontrivial solution of (1) with $u < c < v_1$ , there exists no different solution $(u,\,v_2,\,c)$ .

For example, there are the solutions $(2,\,6,\,5)$ and $(3,\,6,\,5)$ ; $(5,\,20,\,17)$ and $(12,\,20,\,17)$ .

Proof. The Diophantine equation (1) may be written

(4)

(5)

If one solves (1) for $v$ , the smaller root $$\frac{c-\sqrt{c^2+4cu-4u^2}}{2} \;=\; \frac{2(u-c)u}{c+\sqrt{c^2+4cu-4u^2}}$$ is negative. Thus there cannot be any $(u,\,v_2,\,c)$ .