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# contraharmonic means and Pythagorean hypotenuses

One can see that all values of $c$ in the table of the parent entry are hypotenuses in a right triangle^{} with integer sides. E.g., 41 is the contraharmonic mean of 5 and 45; $9^{2}\!+\!40^{2}\;=\;41^{2}$.

Theorem. Any integer contraharmonic mean of two different positive integers is the hypotenuse of a Pythagorean triple. Conversely, any hypotenuse of a Pythagorean triple is contraharmonic mean of two different positive integers.

Proof. $1^{\circ}.$ Let the integer $c$ be the contraharmonic mean

$c\;=\;\frac{u^{2}\!+\!v^{2}}{u\!+\!v}$ |

of the positive integers $u$ and $v$ with $u>v$. Then $u\!+\!v\,\mid\,u^{2}\!+\!v^{2}\,=\,(u\!+\!v)^{2}-2uv$, whence

$u\!+\!v\,\mid\,2uv,$ |

and we have the positive integers

$a\;=:\;u\!-\!v\;=\;\frac{u^{2}\!-\!v^{2}}{u\!+\!v},\quad b\;=:\;\frac{2uv}{u\!% +\!v}$ |

satisfying

$a^{2}\!+\!b^{2}\;=\;\frac{(u^{2}\!-\!v^{2})^{2}\!+\!(2uv)^{2}}{(u\!+\!v)^{2}}=% \frac{u^{4}\!-\!2u^{2}v^{2}+v^{4}\!+\!4u^{2}v^{2}}{(u\!+\!v)^{2}}=\frac{u^{4}% \!+\!2u^{2}v^{2}\!+\!v^{4}}{(u\!+\!v)^{2}}=\frac{(u^{2}\!+\!v^{2})^{2}}{(u\!+% \!v)^{2}}\;=\;c^{2}.\\$ |

$2^{\circ}.$ Suppose that $c$ is the hypotenuse of the Pythagorean triple $(a,\,b,\,c)$, whence
$c^{2}=a^{2}\!+\!b^{2}$. Let us consider the rational numbers^{}

$\displaystyle u=:\frac{c\!+\!b\!+\!a}{2},\quad v=:\frac{c\!+\!b\!-\!a}{2}.$ | (1) |

If the triple is primitive, then two of the integers $a,\,b,\,c$ are odd and one of them is even; if not, then similarly or all of $a,\,b,\,c$ are even. Therefore, $c\!+\!b\!\pm\!a$ are always even and accordingly $u$ and $v$ positive integers. We see also that $u\!+\!v=c\!+\!b$. Now we obtain

$\displaystyle u^{2}\!+\!v^{2}$ | $\displaystyle=\;\frac{c^{2}\!+\!b^{2}\!+\!a^{2}\!+\!2ab\!+\!2bc\!+\!2ca\!+\!c^% {2}\!+\!b^{2}\!+\!a^{2}\!-\!2ab\!+\!2bc\!-\!2ca}{4}$ | ||

$\displaystyle=\;\frac{2c^{2}\!+\!2(a^{2}\!+\!b^{2})\!+\!4bc}{4}=\frac{4c^{2}\!% +\!4bc}{4}=c(c\!+\!b)$ | |||

$\displaystyle=\;c(u\!+\!v).$ |

Thus, $c$ is the contraharmonic mean $\displaystyle\frac{u^{2}\!+\!v^{2}}{u\!+\!v}$ of the different integers $u$ and $v$. (N.B.: When the values of $a$ and $b$ in (1) are changed, another value of $v$ is obtained. Cf. the Proposition^{} 4 in the
parent entry.)

# References

- 1
J. Pahikkala: “On contraharmonic mean and Pythagorean triples”. –
*Elemente der Mathematik*65:2 (2010).

## Mathematics Subject Classification

11D09*no label found*11D45

*no label found*11Z05

*no label found*11A05

*no label found*

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## Comments

## strange @@2@@

What means @@2@@? Sometimes, \PMlinkname{}{} produces such =o(

## Re: strange @@2@@

Maybe it will work if you use the canonical name PythagoreanTriple (instead of the non-canonical PrimitivePythagoreanTriplet).

## Re: strange @@2@@

Yark, I have already tried both names =o(

## Re: strange @@2@@

I don't know what the problem is, but I did some experiments and it seems that if you use \[ ... \] rather than $$ ... $$ in the line above then it works OK!

## Re: strange @@2@@

Yes, it is true, Yark, now it's o.k.! Thank you very much for your experiments! You are super -- I would never had discovered it! BTW, I have neve before used \[...\], only dollars =o)

Regards,

Jussi