Integer Contraharmonic Means

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

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 are in Sloane's sequence A146984.

Proposition 1. For any value of , there are at least two greater values of such that .

Proof. One has the identities

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 order 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 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

whence

Also the smaller root 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 .