Pythagorean hypotenuses as contraharmonic means

$c^2=a^2+b^2$	$c={\displaystyle \frac{u^2+v^2}{u+v}}$	N.B.
$5^2=3^2+4^2$	$5=(2^2+6^2)/(2+6)$	3
${10}^2=6^2+8^2$	$10=(6^2+{12}^2)/(6+12)$
${13}^2=5^2+{12}^2$	$13=(3^2+{15}^2)/(3+15)$	5
${15}^2=9^2+{12}^2$	$15=(9^2+{18}^2)/(9+18)$
${17}^2=8^2+{15}^2$	$17=(5^2+{20}^2)/(5+20)$	5
${20}^2={12}^2+{16}^2$	$20=(8^2+{24}^2)/(8+24)$
${25}^2=7^2+{24}^2$	$25=(4^2+{28}^2)/(4+28)$	7
${26}^2={10}^2+{24}^2$	$26=(6^2+{30}^2)/(6+30)$
${29}^2={20}^2+{21}^2$	$29=({14}^2+{35}^2)/(14+35)$	7
${30}^2={18}^2+{24}^2$	$30=({12}^2+{36}^2)/(12+36)$
${34}^2={16}^2+{30}^2$	$34=({10}^2+{40}^2)/(10+40)$	5
${35}^2={21}^2+{28}^2$	$35=({14}^2+{42}^2)/(14+42)$
${37}^2={12}^2+{35}^2$	$37=({30}^2+{42}^2)/(30+42)$	7
${39}^2={15}^2+{36}^2$	$39=(9^2+{45}^2)/(9+45)$
${40}^2={24}^2+{32}^2$	$40=({16}^2+{48}^2)/(16+48)$
${41}^2=9^2+{40}^2$	$41=({36}^2+{45}^2)/(36+45)$	3
${50}^2={14}^2+{48}^2$	$50=({42}^2+{56}^2)/(42+56)$	7
${51}^2={24}^2+{45}^2$	$51=({15}^2+{60}^2)/(15+60)$
${52}^2={20}^2+{48}^2$	$52=({12}^2+{60}^2)/(12+60)$	3
${53}^2={28}^2+{45}^2$	$53=({18}^2+{63}^2)/(18+63)$	7
${55}^2={33}^2+{44}^2$	$55=({22}^2+{66}^2)/(22+66)$
${58}^2={40}^2+{42}^2$	$58=({30}^2+{70}^2)/(30+70)$	7
${60}^2={36}^2+{48}^2$	$60=({24}^2+{72}^2)/(24+72)$
${61}^2={11}^2+{60}^2$	$61=(6^2+{66}^2)/(6+66)$	11
${61}^2={11}^2+{60}^2$	$61=({55}^2+{66}^2)/(55+66)$	11
${65}^2={39}^2+{52}^2$	$65=({26}^2+{78}^2)/(26+78)$
${68}^2={32}^2+{60}^2$	$68=({20}^2+{80}^2)/(20+80)$
${70}^2={42}^2+{56}^2$	$70=({28}^2+{84}^2)/(28+84)$
${73}^2={48}^2+{55}^2$	$73=({40}^2+{88}^2)/(40+88)$	11
${74}^2={24}^2+{70}^2$	$74=({14}^2+{84}^2)/(14+84)$	7
${75}^2={21}^2+{72}^2$	$75=({12}^2+{84}^2)/(12+84)$	7
${78}^2={30}^2+{72}^2$	$78=({18}^2+{90}^2)/(18+90)$
${80}^2={48}^2+{64}^2$	$80=({32}^2+{96}^2)/(32+96)$
${82}^2={18}^2+{80}^2$	$82=({10}^2+{90}^2)/(10+90)$	3
${85}^2={40}^2+{75}^2$	$85=({25}^2+{100}^2)/(25+100)$
${87}^2={60}^2+{63}^2$	$87=({42}^2+{105}^2)/(42+105)$	7
${89}^2={39}^2+{80}^2$	$89=({24}^2+{104}^2)/(24+104)$	13
${90}^2={54}^2+{72}^2$	$90=({36}^2+{108}^2)/(36+108)$
${91}^2={35}^2+{84}^2$	$91=({21}^2+{105}^2)/(21+105)$
${95}^2={57}^2+{76}^2$	$95=({38}^2+{114}^2)/(38+114)$	19
${97}^2={72}^2+{65}^2$	$97=({45}^2+{117}^2)/(45+117)$	13
${100}^2={60}^2+{80}^2$	$100=({40}^2+{120}^2)/(40+120)$

Title	Pythagorean hypotenuses as contraharmonic means
Canonical name	PythagoreanHypotenusesAsContraharmonicMeans
Date of creation	2013-11-20 20:23:01
Last modified on	2013-11-20 20:23:01
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	10
Author	pahio (2872)
Entry type	Data Structure
Classification	msc 11Z05
Classification	msc 11D45
Classification	msc 11D09
Classification	msc 11A05
Related topic	PythagoreanTriple
Related topic	ContraharmonicMean
Related topic	FirstPrimitivePythagoreanTriplets
Related topic	ContraharmonicProportion