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 (uv) where their contraharmonic mean

c:=u2+v2u+v=u+v-2uvu+v (1)

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

h:=2uvu+v (2)

of u and v are simultaneously integers.

The integer contraharmonic mean of two distinct positive integers ranges exactly the set of hypotenusesMathworldPlanetmath of Pythagorean triplesMathworldPlanetmath (see contraharmonic integersPlanetmathPlanetmath), but the integer harmonic mean of two distinct positive integers the wider set  {3, 4, 5, 6,}.  As a matter of fact, one cathetusMathworldPlanetmath of a right triangleMathworldPlanetmath is the harmonic mean of the same positive integers u and v the contraharmonic mean of which is the hypotenuse of the triangle (see Pythagorean triangleMathworldPlanetmath (http://planetmath.org/PythagoreanTriangle)).

The following table allows to compare the values of u, v, c, h when  1<u<v.

u 2 3 3 4 4 5 5 6 6 6 6 7 7 8 8 8 9 9
v 6 6 15 12 28 20 45 12 18 30 66 42 91 24 56 120 18 45
c 5 5 13 10 25 17 41 10 15 26 61 37 85 20 50 113 15 39
h 3 4 5 6 7 8 9 8 9 10 11 12 13 12 14 15 12 15

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

v1:=(u-1)u,v2:=(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  is that

gcd(u,v)> 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 squarefreeMathworldPlanetmath.

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

Proposition 6 follows also from the inequalityMathworldPlanetmath

1h=12(1u+1v)>12u

which yields the estimation

0<h< 2u (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 2u2u+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.

Title integer harmonic means
Canonical name IntegerHarmonicMeans
Date of creation 2013-11-06 17:18:49
Last modified on 2013-11-06 17:18:49
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 20
Author pahio (2872)
Entry type Topic
Classification msc 11Z05
Classification msc 11D45
Classification msc 11D09
Classification msc 11A05
Related topic HarmonicMean
Related topic HarmonicMeanInTrapezoid