integer contraharmonic means
Let $u$ and $v$ be positive integers. There exist nontrivial cases where their contraharmonic mean
$c:={\displaystyle \frac{{u}^{2}+{v}^{2}}{u+v}}$ | (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 $c=u=v$.
$u$ | $2$ | $3$ | $3$ | $4$ | $4$ | $5$ | $5$ | $6$ | $6$ | $6$ | $6$ | $7$ | $7$ | $8$ | $8$ | $8$ | $9$ | $9$ | $\mathrm{\dots}$ |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
$v$ | $6$ | $6$ | $15$ | $12$ | $28$ | $20$ | $45$ | $12$ | $18$ | $30$ | $66$ | $42$ | $91$ | $24$ | $56$ | $120$ | $18$ | $45$ | $\mathrm{\dots}$ |
$c$ | $5$ | $5$ | $13$ | $10$ | $25$ | $17$ | $41$ | $10$ | $15$ | $26$ | $61$ | $37$ | $85$ | $20$ | $50$ | $113$ | $15$ | $39$ | $\mathrm{\dots}$ |
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 $u>2$, there are at least two greater values of $v$ such that $c\in \mathbb{Z}$.
Proof. One has the identities
$\frac{{u}^{2}+{((u-1)u)}^{2}}{u+(u-1)u}}={u}^{2}-2u+2,$ | (2) |
$\frac{{u}^{2}+{((2u-1)u)}^{2}}{u+(2u-1)u}}=\mathrm{\hspace{0.33em}2}{u}^{2}-2u+1,$ | (3) |
the right hand sides of which are positive integers and different for $u\ne 1$. 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
$$\mathrm{gcd}(u,v)>1.$$ |
Proof. Suppose that we have positive integers $u,v$ such that $\mathrm{gcd}(u,v)=1$. Then as well, $\mathrm{gcd}(u+v,uv)=1$, 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 $\mathrm{gcd}(u,v)\geqq p$. Consequently, we must have $\mathrm{gcd}(u+v,uv)=1$.
We make the additional supposition that $\frac{{u}^{2}+{v}^{2}}{u+v}$ 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 $\mathrm{gcd}(u+v,uv)=1$, 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 $\mathrm{gcd}(u,v)=1$. 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^{} $u$:
$$v=nu,n\in \mathbb{Z}$$ |
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
$$c=\frac{({n}^{2}+1)u}{n+1}=\frac{({k}^{2}{u}^{2}-2ku+2)u}{ku}=k{u}^{2}-2+\frac{2}{k},$$ |
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
$${n}^{2}+1={(4m-1)}^{2}+1=\mathrm{\hspace{0.33em}16}{m}^{2}-8m+2\not\equiv 0\phantom{\rule{veryverythickmathspace}{0ex}}(mod4),$$ |
which is impossible. We infer, that now $\mathrm{gcd}(n+1,{n}^{2}+1)=2$, and in any case
$$\mathrm{gcd}(n+1,{n}^{2}+1)\leqq \mathrm{\hspace{0.33em}2}.$$ |
Nevertheless, since $n+1\geqq 3$ and $n+1\mid {n}^{2}+1$, we should have
$\mathrm{gcd}(n+1,{n}^{2}+1)\geqq 3$. 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 $$, then there is always another nontrivial solution $({u}_{2},v,c)$ with $$. On the contrary, if $(u,{v}_{1},c)$ is a nontrivial solution of (1) with $$, there exists no different solution $(u,{v}_{2},c)$.
For example, there are the solutions $(2,\mathrm{\hspace{0.17em}6},\mathrm{\hspace{0.17em}5})$ and $(3,\mathrm{\hspace{0.17em}6},\mathrm{\hspace{0.17em}5})$;
$(5,\mathrm{\hspace{0.17em}20},\mathrm{\hspace{0.17em}17})$ and $(12,\mathrm{\hspace{0.17em}20},\mathrm{\hspace{0.17em}17})$.
Proof. The Diophantine equation^{} (1) may be written
${u}^{2}-cu+({v}^{2}-cv)=\mathrm{\hspace{0.33em}0},$ | (4) |
whence
$u={\displaystyle \frac{c\pm \sqrt{{c}^{2}+4cv-4{v}^{2}}}{2}},$ | (5) |
and the discriminant^{} of (4) must be nonnogative because of the existence of the real root (http://planetmath.org/Equation) ${u}_{1}$. But if it were zero, i.e. if the equation ${c}^{2}+4cv-4{v}^{2}=0$ were true, this would imply for $v$ the irrational value $\frac{1}{2}(1+\sqrt{2})c$. Thus the discriminant must be positive, and then also the smaller root $u$ of (4) gotten with “$-$” in front of the square root is positive, since we can rewrite it
$$\frac{c-\sqrt{{c}^{2}+4cv-4{v}^{2}}}{2}=\frac{{c}^{2}-({c}^{2}+4cv-4{v}^{2})}{2(c+\sqrt{{c}^{2}+4cv-4{v}^{2}})}=\frac{2(v-c)v}{c+\sqrt{{c}^{2}+4cv-4{v}^{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-4{u}^{2}}}{2}=\frac{2(u-c)u}{c+\sqrt{{c}^{2}+4cu-4{u}^{2}}}$$ |
is negative. Thus there cannot be any $(u,{v}_{2},c)$.
Proposition 5. When the contraharmonic mean of two different positive integers $u$ and $v$ is an integer, their sum is never squarefree^{}.
Proof. By Proposition 2 we have
$$\mathrm{gcd}(u,v)=:d>\mathrm{\hspace{0.33em}1}.$$ |
Denote
$$u={u}^{\prime}d,v={v}^{\prime}d,$$ |
when $\mathrm{gcd}({u}^{\prime},{v}^{\prime})=\mathrm{\hspace{0.17em}1}$. Then
$$c=\frac{({u}^{\prime \mathrm{\hspace{0.17em}2}}+{v}^{\prime \mathrm{\hspace{0.17em}2}})d}{{u}^{\prime}+{v}^{\prime}},$$ |
whence
$({u}^{\prime}+{v}^{\prime})c=({u}^{\prime \mathrm{\hspace{0.17em}2}}+{v}^{\prime \mathrm{\hspace{0.17em}2}})d\equiv [{({u}^{\prime}+{v}^{\prime})}^{2}-2{u}^{\prime}{v}^{\prime}]d.$ | (6) |
If $p$ is any odd prime factor of ${u}^{\prime}+{v}^{\prime}$, the last equation implies that
$$p\nmid {u}^{\prime},p\nmid {v}^{\prime},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{2{u}^{2}}{u+v}$$ |
In to $c$ be an integer, the quotient
$$w:=\frac{2{u}^{2}}{u+v}$$ |
must be integer; rewriting this last equation as
$v={\displaystyle \frac{2{u}^{2}}{w}}-u$ | (7) |
we infer that $w$ has to be a http://planetmath.org/node/923divisor of $2{u}^{2}$ (apparently $$ 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 $2{u}^{2}$ 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^{} |