# integer contraharmonic means

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

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

is an integer, too.  For example, the values  $u=3,\;v=15$  have the contraharmonic mean  $c=13$.  The only “trivial cases” are those with  $u=v$,  when  $c=u=v$.

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

The nontrivial integer contraharmonic means form Sloane’s sequence http://oeis.org/search?q=A146984&language=english&go=SearchA146984.

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

 $\displaystyle\frac{u^{2}\!+\!((u\!-\!1)u)^{2}}{u+(u\!-\!1)u}\;=\;u^{2}\!-\!2u% \!+\!2,$ (2)
 $\displaystyle\frac{u^{2}\!+\!((2u\!-\!1)u)^{2}}{u+(2u\!-\!1)u}\;=\;2u^{2}\!-\!% 2u\!+\!1,$ (3)

the right hand sides of which are positive integers and different for  $u\neq 1$.  The value  $u=2$  is an exception, since it has only  $v=6$  with which its contraharmonic mean is an integer.

In (2) and (3), the value of $v$ is a multiple of $u$, but it needs not be always so in 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 (http://planetmath.org/Product) $u$ 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

 $c=\frac{(n^{2}\!+\!1)u}{n\!+\!1}\;=\;\frac{(k^{2}u^{2}\!-\!2ku\!+\!2)u}{ku}\;=% \;ku^{2}\!-\!2\!+\!\frac{2}{k},$

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},  then there is always another nontrivial solution  $(u_{2},\,v,\,c)$  with  $u_{2}.  On the contrary, if  $(u,\,v_{1},\,c)$  is a nontrivial solution of (1) with  $u,  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

 $\displaystyle u^{2}\!-\!cu\!+\!(v^{2}\!-\!cv)\;=\;0,$ (4)

whence

 $\displaystyle u\;=\;\frac{c\!\pm\!\sqrt{c^{2}\!+\!4cv\!-\!4v^{2}}}{2},$ (5)

and the discriminant of (4) must be nonnogative because of the existence of the real root (http://planetmath.org/Equation) $u_{1}$.  But if it were zero, i.e. if the equation  $c^{2}\!+\!4cv\!-\!4v^{2}=0$  were true, this would imply for $v$ the irrational value $\frac{1}{2}(1\!+\!\sqrt{2})c$.  Thus the discriminant must be positive, and then also the smaller root $u$ of (4) gotten with “$-$” in front of the square root is positive, since we can rewrite it

 $\frac{c\!-\!\sqrt{c^{2}\!+\!4cv\!-\!4v^{2}}}{2}\;=\;\frac{c^{2}\!-\!(c^{2}\!+% \!4cv\!-\!4v^{2})}{2(c+\sqrt{c^{2}\!+\!4cv\!-\!4v^{2}})}\;=\;\frac{2(v\!-\!c)v% }{c\!+\!\sqrt{c^{2}\!+\!4cv\!-\!4v^{2}}}$

and the numerator is positive because  $v>c$.  Thus, when the discriminant of the equation (4) is positive, the equation has always two distinct positive roots $u$.  When one of the roots ($u_{1}$) 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 $u_{2}$, distinct from $u_{1}$.

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  $(u,\,v_{2},\,c)$.

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

Proof.  By Proposition 2 we have

 $\gcd(u,\,v)\;=:\;d\;>\;1.$

Denote

 $u\;=\;u^{\prime}d,\quad v\;=\;v^{\prime}d,$

when  $\gcd(u^{\prime},\,v^{\prime})\,=\,1$.  Then

 $c\;=\;\frac{(u^{\prime\,2}\!+\!v^{\prime\,2})d}{u^{\prime}\!+\!v^{\prime}},$

whence

 $\displaystyle(u^{\prime}\!+\!v^{\prime})c\;=\;(u^{\prime\,2}\!+\!v^{\prime\,2}% )d\;\equiv\;[(u^{\prime}\!+\!v^{\prime})^{2}\!-\!2u^{\prime}v^{\prime}]d.$ (6)

If $p$ is any odd prime factor of $u^{\prime}\!+\!v^{\prime}$, the last equation implies that

 $p\nmid u^{\prime},\quad p\nmid v^{\prime},\quad p\nmid[\;\;],$

and consequently  $p\mid d$.  Thus we see that

 $p^{2}\mid(u^{\prime}\!+\!v^{\prime})d\;=\;u\!+\!v.$

This means that the sum $u\!+\!v$ is not squarefree.  The same result is easily got also in the case that $u$ and $v$ both are even.

Note 1.  Cf.  $u\!+\!v=c\!+\!b$  in $2^{\circ}$ 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  $u>0$  there are only a finite number of solutions  $(u,\,v,\,c)$  of the Diophantine equation (1).  The number does not exceed $u\!-\!1$.

Proof.  The expression of the contraharmonic mean in (1) may be edited as follows:

 $c\;=\;\frac{(u\!+\!v)^{2}-2uv}{u\!+\!v}\;=\;u\!+\!v-\frac{2u(u\!+\!v\!-\!u)}{u% \!+\!v}\;=\;v\!-\!u+\frac{2u^{2}}{u\!+\!v}$

In to $c$ be an integer, the quotient

 $w\;:=\;\frac{2u^{2}}{u\!+\!v}$

must be integer; rewriting this last equation as

 $\displaystyle v\;=\;\frac{2u^{2}}{w}\!-\!u$ (7)

we infer that $w$ has to be a http://planetmath.org/node/923divisor of $2u^{2}$ (apparently  $1\leqq w  for getting values of $v$ greater than $u$).  The amount of such divisors is quite restricted, not more than $u\!-\!1$, and consequently there is only a finite number of suitable values of $v$.

Note 2.  The equation (7) explains the result of Proposition 1 ($w=1$,  $w=2$).  As well, if $u$ is an odd prime number, then the only factors of $2u^{2}$ less than $u$ are 1 and 2, and for these the equation (7) gives the values  $v:=(2u\!-\!1)u$  and  $v:=(u\!-\!1)u$  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