# contraharmonic means and Pythagorean hypotenuses

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

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 (http://planetmath.org/PythagoreanTriple), 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 (http://planetmath.org/IntegerContraharmonicMeans).)

## References

• 1 J. Pahikkala: “On contraharmonic mean and Pythagorean triples”.  – Elemente der Mathematik 65:2 (2010).
 Title contraharmonic means and Pythagorean hypotenuses Canonical name ContraharmonicMeansAndPythagoreanHypotenuses Date of creation 2013-11-03 21:13:57 Last modified on 2013-11-03 21:13:57 Owner pahio (2872) Last modified by pahio (2872) Numerical id 24 Author pahio (2872) Entry type Theorem Classification msc 11D09 Classification msc 11D45 Classification msc 11Z05 Classification msc 11A05 Synonym contraharmonic integers Synonym Pythagorean hypotenuses are contraharmonic means Related topic FirstPrimitivePythagoreanTriplets Related topic ProofOfPythagoreanTriplet2 Related topic SquareOfSum Related topic PythagoreanTriple Related topic DerivationOfPythagoreanTriples Related topic LinearFormulasForPythagoreanTriples