integer harmonic means

Let $u$ and $v$ be positive integers.  As is seen in the parent entry (http://planetmath.org/IntegerContraharmonicMeans), there exist nontrivial cases($u\ne v$) where their contraharmonic mean

$c:={\displaystyle \frac{u^2+v^2}{u+v}}=u+v-{\displaystyle \frac{2uv}{u+v}}$(1)

is an integer.  Because the subtrahend of the last is the harmonic mean of $u$ and $v$, the equation means that the contraharmonic mean $c$ and the harmonic mean

$h:={\displaystyle \frac{2uv}{u+v}}$(2)

of $u$ and $v$ are simultaneously integers.

The integer contraharmonic mean of two distinct positiveintegers ranges exactly the set of hypotenuses of Pythagoreantriples (see contraharmonic integers), but the integer harmonicmean of two distinct positive integers the wider set $\{3,\mathrm{\hspace{0.17em}4},\mathrm{\hspace{0.17em}5},\mathrm{\hspace{0.17em}6},\mathrm{\dots }\}$.  As a matter of fact, onecathetus of a right triangle is the harmonic mean of the samepositive integers $u$ and $v$ the contraharmonic mean of whichis the hypotenuse of the triangle (seePythagorean triangle (http://planetmath.org/PythagoreanTriangle)).

The following table allows to compare the values of $u$, $v$,$c$, $h$ when  .

Some of the propositions concerning the integer contraharmonic means directly imply corresponding propositions of the integer harmonic means:

Proposition 1.  For any value of $u>2$, there are at least two greater values

$v_1:=(u-1)u,v_2:=(2u-1)u$(3)

of $v$ such that $h$ in (2) is an integer.

Proposition 2.  For all  $u>1$, a necessary condition for $h\in \mathbb{Z}$  is that

$$\gcd (u,v)>\mathrm{\hspace{0.33em}1}.$$

Proposition 3.  If $u$ is an odd prime number, then the values (3) are the only possibilities for  $v>u$  enabling integer harmonic means with $u$.

Proposition 5.  When the harmonic mean of two different positive integers $u$ and $v$ is an integer, their sum is never squarefree.

Proposition 6.  For each integer $u>0$ there are only a finite number of solutions  $(u,v,h)$  of the Diophantine equation (2).

Proposition 6 follows also from the inequality

$$\frac{1}{h}=\frac{1}{2}\left(\frac{1}{u}+\frac{1}{v}\right)>\frac{1}{2u}$$

which yields the estimation

(4)

(cf. the above table).  This is of course true for any harmonic means $h$ of positive numbers $u$ and $v$.  The difference of $2u$ and $h$ is $\frac{2u^2}{u+v}$.

The estimation (4) implies that the number of solutions is less than $2u$.  From the proof of the corresponding proposition in the http://planetmath.org/node/11241parent entry one can see that the number in fact does not exceed $u-1$.