Mallik, A. K.

English

Publications of the IAS Fellows

CLASSROOM
380 RESONANCE  April 2009
Perfect Medians, Euler, and Ramanujan
A K Mallik
Department of Mechanical En-
gineering
Indian Institute of Technology
Kanpur 208 016, India.
Email:akmallik@iitk.ac.in
Keywords
Perfect median, Pythagorean,
triangular number, Pell’s equa-
tion, Euler’s equation, Brahma-
gupta’s equation, Ramanujan,
Mahanalobis, continued fraction.
B ria n H a y e s [1 ] d e ¯ n e s a perfect m ed ia n a s fo llo w s. F o r
th e set o f c o n se c u tiv e n a tu ra l n u m b ers 1 , 2 , 3 , . .. , m ,
. .. , n to h a v e m a s its p e rfe ct m e d ia n w e m u st h a v e
1 + 2 + 3 + ::: + (m ¡ 1 )
= (m + 1 ) + (m + 2 ) + (m + 3 ) + ::: + n ;
o r
m (m ¡ 1 )
2
=
(n ¡ m ) (m + n + 1 )
2
;
o r m 2 =
n (n + 1 )
2
: (1 )
F o llo w in g th e c la ssi¯ ca tio n o f n u m b ers p ra c tic e d b y th e
P y th a g o re a n s, w h o u sed th e sh a p e s p ro d u ce d b y th e
a rra n g e m e n t o f d o ts o r p eb b le s, m 2 a n d n (n + 1 ) = 2 a re
c a lle d sq u a re a n d tria n g u la r n u m b e rs, re sp e c tiv ely [2 ].
T h e d ete rm in a tio n o f th e v a lu e s o f m a n d n sa tisfy -
in g e q u a tio n (1 ) w a s c a rrie d o u t b y H a y e s [1 ] u sin g a
sm a ll co m p u te r p ro g ra m , a n d h e a ttrib u te d th is p u z z le
to D a v id G a le w h o w a s in tere ste d in d e te rm in in g th e
u n d e rly in g p a tte rn in th ese n u m b e rs.
In 1 7 3 3 , L e o n h a rd E u ler a t th e en d o f h is p a p e r en title d
\ O n th e so lu tio n o f p ro b lem s o f D io p h a n tu s a b o u t in te -
g e r n u m b e rs" a sk e d fo r w h a t v a lu es o f n , th e tria n g u la r
n u m b e rs a re a lso sq u a re n u m b e rs. H e a lso p ro v id ed th e
a n sw e r [3 ]: \ T ria n g u la r n u m b e rs w ith n = 1 , 8 , 4 9 , 2 8 8 ,
1 6 8 1 , 9 8 0 0 , e tc., a re sq u a re n u m b ers c o rre sp o n d in g to
m = 1 , 6 , 3 5 , 2 0 4 , 1 1 8 9 , 6 9 3 0 , etc ., resp e ctiv e ly " . T h e se
a n sw e rs ca n b e o b ta in e d a s fo llo w s.
M u ltip ly in g b o th sid es o f (1 ) b y 8 a n d a d d in g 1 , w e g e t
(2 n + 1 )
2 ¡ 8 m 2 = 1 : (2 )
CLASSROOM
381RESONANCE  April 2009
In 1733, Euler
asked about the
triangular
numbers which
are also square
numbers.
B o x 1 .
Following Euler, equations of t he form p 2 ¡ D q 2 = 1 where p and q are integers, and D
is a non-square positive int eger, are historically (but wrongly) known as Pell' s equation.
` Fermat's equation' would have been a more accurate name [3] . But even more accurately,
t his equat ion may be called `Brahmagupta' s equation' .
In about 650 AD Brahmagupta wrote, in his highly in°uential t ext B ra h m a S p h u ta
S id d h a n th a , \A person who can within a year solve the equation x 2 ¡ 92y 2 = 1 is a
mathematician" (see page 252 of the book by Beiler [6] ) . Since the smallest solution is
x = 1 1 51, y = 1 20, one does need to be a great mathematician to solve this equat ion
without the aid of a computer!
If D is a perfect square, then Brahmagupta' s equat ion implies t hat two consecutive
integers are p erfect squares, leading t o the trivial solution p = 1 , q = 0.
If D is a non-square positive integer, then Brahmagupta' s equation has in¯nitely many
solutions. The smallest solutions of p and q vary wildly with the value of D [4] . For
example, the smallest solutions are p = 31, q = 4 wit h D = 60; p = 63, q = 8 with D =
62; and p = 1 766,31 9,049, q = 226, 153,980 with D = 61 .
Di±cult puzzles, ultimately involving solutions to Brahmagupta' s equation, have been
devised with values of D that yield large values for even the smallest solut ion. D = 1 597
and 9781 result in ast ronomically large values for t he smallest solutions of p and q . The
systematic procedure of obtaining the smallest possible solution is given in t extbooks on
number theory.
S u b stitu tin g
p = 2 n + 1 a n d q = 2 m ; (3 )
e q u a tio n (2 ) ca n b e w ritte n in th e fo rm o f th e so -c a lle d
P ell's eq u a tio n (se e B o x 1 ),
p 2 ¡ 2 q 2 = 1 ; o r
³
p +
p
2 q
´ ³
p ¡
p
2 q
´
= 1 : (4 )
T h is h a s in ¯ n itely m a n y so lu tio n s [4 ]. It is ea sy to see
(b y in sp ec tio n ) th a t th e sm a lle st p o ssib le so lu tio n to
(4 ) is p 1 = 3 ; q 1 = 2 . H e n c e, fro m (3 ) w e g e t n 1 =
1 ; m 1 = 1 : T o g e t th e n e x t so lu tio n , ¯ rst w e w rite e q u a -
tio n (4 ) w ith th e sm a lle st v a lu e s o f p a n d q a s³
3 + 2
p
2
´ ³
3 ¡ 2
p
2
´
= 1 : (5 )
CLASSROOM
382 RESONANCE  April 2009
In 650 AD,
Brahmagupta wrote,
“A person who can
within a year solve the
equation x2– 92y2 = 1
is a mathematician”.
The smallest solution
is x = 1151, y = 120.
S q u a rin g b o th sid es o f (5 ), w e g et³
1 7 + 1 2
p
2
´ ³
1 7 ¡ 1 2
p
2
´
= 1 :
T h u s, w e o b ta in th e sec o n d so lu tio n a s p 2 = 1 7 ; q 2 = 1 2 :
U sin g (3 ), o n e g e ts n 2 = 8 ; m 2 = 6 :
T a k in g th e c u b e o f b o th sid e s o f (5 ), w e g e t³
9 9 + 7 0
p
2
´ ³
9 9 ¡ 7 0
p
2
´
= 1 ;
y ie ld in g p 3 = 9 9 ; q 3 = 7 0 . T h u s, fro m e q u a tio n (3 ) o n e
o b ta in s n 3 = 4 9 ; m 3 = 3 5 : A ll th e in ¯ n ite ly m a n y so lu -
tio n s c a n b e o b ta in ed in th is w a y b y ta k in g su cc e ssiv e ly
h ig h e r p o w e rs o f b o th sid e s o f (5 ).
K a n ig e l [5 ] te lls u s th e fo llo w in g sto ry a b o u t th e m a th -
e m a tic a l g en iu s S R a m a n u ja n . In D e ce m b e r 1 9 1 4 , R a -
m a n u ja n w a s a sk e d b y h is frie n d P C M a h a la n o b is (fo u n -
d e r o f th e In d ia n S ta tistica l In stitu te) to so lv e a p u z zle,
\ P u zz le s a t a V illa g e In n " , th a t a p p e a re d in th e p o p -
u la r E n g lish m a g a zin e S tra n d . T h e p u zz le sta te d th a t
n h o u se s o n o n e sid e o f a stre et a re n u m b ere d se q u e n -
tia lly sta rtin g fro m 1 . T h e su m o f th e h o u se-n u m b e rs
o n th e left o f a p a rtic u la r h o u se h a v in g th e n u m b er m ,
e q u a ls th a t o f th e h o u se s ly in g o n th e rig h t o f th is p a r-
tic u la r h o u se . It is g iv e n th a t n lie s b e tw e en 5 0 a n d
5 0 0 a n d o n e h a s to d e term in e th e v a lu e s o f m a n d n .
T h u s, th e p u z z le b a sic a lly b o ile d d o w n to ¯ n d in g th e
p e rfe ct m e d ia n m fo r n ly in g b e tw e e n 5 0 a n d 5 0 0 , w h ich
is 2 0 4 w ith n = 2 8 8 . R a m a n u ja n , w h ile stirrin g v eg eta -
b les o v e r th e g a s ¯ re, ra ttle d o u t a co n tin u ed fra c tio n
g e n e ra tin g a ll th e p e rfe c t m e d ia n s. W h e n h is a sto u n d e d
frie n d a sk ed h o w h e h a d so lv e d th e p ro b lem , R a m a n u -
ja n a n sw e red , \ Im m e d ia te ly I h e a rd th e p ro b le m it w a s
c le a r th a t th e so lu tio n sh o u ld o b v io u sly b e a c o n tin u e d
fra c tio n ; I th e n th o u g h t, w h ich c o n tin u e d fra ctio n ? A n d
th e a n sw e r c a m e to m y m in d " .
R a m a n u ja n g o t th e ¯ rst tw o v a lu es o f m (n a m e ly , 1 a n d
CLASSROOM
383RESONANCE  April 2009
Suggested Reading
[1] B Hayes, American Scientist, Vol.96, p. 362, 2008.
[2] M Kline, Mathematics and the Physical World, Dover Publications, Inc.,
New York, p.27–28, 1981.
[3] C E Sandifer, The Early Mathematics of Leonhard Euler, The MAA
Tercentenary Euler Celebration, The Mathematical Association of America,
Washington, D.C., p.105, 2007.
[4] I Stewart, Math Hysteria, Oxford University Press, Oxford, p.50, 2004.
[5] R Kanigel, The Man Who Knew Infinity, A Life of the Genius Ramanujan,
Simon and Schuster Inc., New York, p. 214–215, 1991.
[6] Albert H Beiler, Recreations in the theory of numbers, 2nd ed. Dover
Publications Inc., New York, 1964.
“... [When] I heard the
problem, it was clear
that the solution
should be a continued
fraction. I then
thought, which
continued fraction?
And the answer came
to my mind.”
S Ramanujan
6 ) m e n ta lly , a n d g a v e th e fo llo w in g c o n tin u e d fra c tio n :
1
6 ¡ 1
6 ¡ 1
6 ¡ :::
T h e n u m e ra to r a n d th e d en o m in a to r o f ev ery ra tio n a l
c o n v e rg e n t o f th is c o n tin u ed fra ctio n g iv e tw o c o n se c -
u tiv e v a lu e s o f m . T h e ¯ rst ra tio n a l a p p ro x im a tio n is
1 / 6 , 1 a n d 6 b ein g th e ¯ rst tw o p e rfe ct m e d ia n s. T h e
se co n d ra tio n a l a p p ro x im a tio n 1
6 ¡ 1
6
= 6 / 3 5 a g a in g iv e s
tw o c o n sec u tiv e v a lu e s o f m , v iz ., 6 a n d 3 5 . T h e n ex t
o n e w ill g iv e 3 5 a n d 2 0 4 , a n d so o n .
It m a y b e m en tio n e d th a t th e c o n tin u e d fra c tio n o f R a -
m a n u ja n ,
x =
1
6 ¡ 1
6 ¡ 1
6 ¡ :::
=
1
6 ¡ x ;
w h e n so lv e d y ie ld s x = 3 § 2
p
2 . S in c e x < 1 ; x =
3 ¡ 2
p
2 is th e c o rre ct so lu tio n ; th e o th e r ro o t is m o re
th a n u n ity . T h e rec ip ro c a l o f th is ro o t is 1x = 3 + 2
p
2 :
S o , w e g et b a ck e q u a tio n (5 ), w h ich g en e ra te s a ll th e
so lu tio n s.


Perfect medians, Euler, and Ramanujan

http://repository.ias.ac.in/93454/1/perfect.pdf

Perfect medians, Euler, and Ramanujan

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