contraharmonic means and Pythagorean
hypotenuses∗
pahio†
2013-03-22 2:01:25
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; 92 +402 = 412 .
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◦ . Let the integer c be the contraharmonic mean
c =
u2 +v 2
u+v
of the positive integers u and v with u > v. Then u+v | u2+v 2 = (u+v)2 −2uv,
whence
u+v | 2uv,
and we have the positive integers
a =: u−v =
u2 −v 2
,
u+v
b =:
2uv
u+v
satisfying
a2+b2 =
(u2 −v 2 )2 +(2uv)2
u4 −2u2 v 2 + v 4 +4u2 v 2
u4 +2u2 v 2 +v 4
(u2 +v 2 )2
=
=
=
= c2 .
2
2
2
(u+v)
(u+v)
(u+v)
(u+v)2
2◦ . Suppose that c is the hypotenuse of the Pythagorean triple (a, b, c),
whence c2 = a2 +b2 . Let us consider the rational numbers
u =:
c+b+a
,
2
v =:
c+b−a
.
2
∗ hContraharmonicMeansAndPythagoreanHypotenusesi
(1)
created: h2013-03-2i by: hpahioi
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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±a
are always even and accordingly u and v positive integers. We see also that
u+v = c+b. Now we obtain
c2 +b2 +a2 +2ab+2bc+2ca+c2 +b2 +a2 −2ab+2bc−2ca
4
4c2 +4bc
2c2 +2(a2 +b2 )+4bc
=
= c(c+b)
=
4
4
= c(u+v).
u2 +v 2 =
u2 +v 2
of the different integers u and v.
u+v
(N.B.: When the values of a and b in (1) are changed, another value of v is
obtained. Cf. the Theorem 4 in the parent entry.)
Thus, c is the contraharmonic mean
References
[1] J. Pahikkala: “On contraharmonic mean and Pythagorean triples”. –
Elemente der Mathematik 65:2 (2010).
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