integer contraharmonic means
Let and be positive integers. There exist nontrivial cases where their contraharmonic mean
(1) |
is an integer, too. For example, the values have the contraharmonic mean . The only “trivial cases” are those with , when .
The nontrivial integer contraharmonic means form Sloane’s sequence
http://oeis.org/search?q=A146984&language=english&go=SearchA146984.
Proposition 1. For any value of , there are at least two greater values of such that .
Proof. One has the identities
(2) |
(3) |
the right hand sides of which are positive integers and different for . The value is an exception, since it has only with which its contraharmonic mean is an integer.
In (2) and (3), the value of is a multiple of , but it needs not be always so in to be an integer, e.g. we have .
Proposition 2. For all , a necessary condition for is that
Proof. Suppose that we have positive integers such that . Then as well, , since otherwise both and would be divisible by a prime , and thus also one of the factors (http://planetmath.org/Product) and of would be divisible by ; then however would imply that and , whence we would have . Consequently, we must have .
We make the additional supposition that is an integer, i.e. that
is divisible by . Therefore also is divisible by this sum. But because , the factor 2 must be divisible by , which is at least 2. Thus .
The conclusion is, that only the “most trivial case” allows that . This settles the proof.
Proposition 3. If is an odd prime number, then (2) and (3) are the only possibilities enabling integer contraharmonic means.
Proof. Let be a positive odd prime. The values and do always. As for other possible values of , according to the Proposition 2, they must be multiples of the prime number :
Now
and since is prime, either or .
In the former case , one obtains
which is an integer only for and , corresponding (2) and (3).
In the latter case, there must be a prime number dividing both and , whence . The equation
then implies that . So we must have , i.e. necessarily . Moreover, if we had and , then we could write , and thus
which is impossible. We infer, that now , and in any case
Nevertheless, since and , we should have
. The contradiction means that the latter case is not possible, and the Proposition 3 has been proved.
Proposition 4. If is a nontrivial solution of (1) with , then there is always another nontrivial solution with . On the contrary, if is a nontrivial solution of (1) with , there exists no different solution .
For example, there are the solutions and ;
and .
Proof. The Diophantine equation (1) may be written
(4) |
whence
(5) |
and the discriminant of (4) must be nonnogative because of the existence of the real root (http://planetmath.org/Equation) . But if it were zero, i.e. if the equation were true, this would imply for the irrational value . Thus the discriminant must be positive, and then also the smaller root of (4) gotten with “” in front of the square root is positive, since we can rewrite it
and the numerator is positive because . Thus, when the discriminant of the equation (4) is positive, the equation has always two distinct positive roots . When one of the roots () is an integer, the other is an integer, too, because in the numerator of (5) the sum and the difference of two integers are simultaneously even. It follows the existence of , distinct from .
If one solves (1) for , the smaller root
is negative. Thus there cannot be any .
Proposition 5. When the contraharmonic mean of two different positive integers and is an integer, their sum is never squarefree.
Proof. By Proposition 2 we have
Denote
when . Then
whence
(6) |
If is any odd prime factor of , the last equation implies that
and consequently . Thus we see that
This means that the sum is not squarefree. The same result is easily got also in the case that and both are even.
Note 1. Cf. in of
the proof of
this theorem (http://planetmath.org/ContraharmonicMeansAndPythagoreanHypotenuses)
and the Note 4 of http://planetmath.org/node/138this entry.
Proposition 6. For each integer there are only a finite number of solutions of the Diophantine equation (1). The number does not exceed .
Proof. The expression of the contraharmonic mean in (1) may be edited as follows:
In to be an integer, the quotient
must be integer; rewriting this last equation as
(7) |
we infer that has to be a http://planetmath.org/node/923divisor of (apparently for getting values of
greater than ). The amount of such divisors is quite restricted, not more than , and consequently there is only a finite number of suitable values of .
Note 2. The equation (7) explains the result of Proposition 1 (, ). As well, if is an odd prime number, then the only factors of less than are 1 and 2, and for these the equation (7) gives the values and which explains Proposition 3.
References
- 1 J. Pahikkala: “On contraharmonic mean and Pythagorean triples”. – Elemente der Mathematik 65:2 (2010).
Title | integer contraharmonic means |
Canonical name | IntegerContraharmonicMeans |
Date of creation | 2013-12-04 10:25:44 |
Last modified on | 2013-12-04 10:25:44 |
Owner | pahio (2872) |
Last modified by | pahio (2872) |
Numerical id | 45 |
Author | pahio (2872) |
Entry type | Topic |
Classification | msc 11Z05 |
Classification | msc 11D45 |
Classification | msc 11D09 |
Classification | msc 11A05 |
Synonym | integer contraharmonic means of integers |
Related topic | ComparisonOfPythagoreanMeans |
Related topic | DivisibilityInRings |
Related topic | Gcd |
Defines | contraharmonic integer |