54. Two diophantine equations studied by Ramanujan. Introduction In his second notebook [3, p.225], Ramanujan wrote. (6n2 + (3n3. â n)). 3. + (6n2. â (3n. 3.
54. Two diophantine equations studied by Ramanujan
Introduction
In his second notebook [3, p.225], Ramanujan wrote
3 3 2 6n2 + (3n3 − n) + 6n2 − (3n3 − n) = 6n2 (3n2 + 1)
(1)
and 3 m7 − 3m4 (1 + p) + m(3(1 + p)2 − 1) + + =
3 2m6 − 3m3 (1 + 2p) + (1 + 3p + 3p2 ) 3 m6 − (1 + 3p + 3p2 ) 3 m7 − 3m4 p + m(3p2 − 1) . (2)
Clearly, these represent solutions of the diophantine equations A3 + B 3 = C 2
(3)
A3 + B 3 + C 3 = D 3 .
(4)
and It appears ([2 pp.578-9]) that (3) was first studied by Euler, and the general solution found by R. Hoppe, and ([2 pp.550-4]) that (4) was studied by Vieta and Fermat, and the general solution found by Euler. Both equations, particularly (4), have aroused considerable interest over the centuries. But Ramanujan knew none of this when he composed [3]. It is, of course, easy to verify Ramanujan’s solutions. The question that arises is, how did Ramanujan obtain them? My object in this note is to give a plausible answer to this intriguing question.
The diophantine equation A3 + B 3 = C 2
Suppose
A3 + B 3 = C 2 . Write A = x + y, B = x − y. Then A3 + B 3 = (x + y)3 + (x − y)3 = 2x3 + 6xy 2 = 2x(x2 + 3y 2 ). Now write y = a − b, x2 = 12ab, for then x2 + 3y 2 = 3(a + b)2 and A3 + B 3 = 6x(a + b)2 . Now let x = 6m2 . Then A3 + B 3 = 36m2 (a + b)2 = (6(a + b)m)2 = C 2 ,
2 with C = 6(a + b)m. Also 12ab = x2 = (6m2 )2 = 36m4 so ab = 3m4 . Thus we obtain the following result. Theorem 1. If m is an integer and if ab = 3m4 then 3 3 6m2 + (a − b) + 6m2 − (a − b) = (6(a + b)m)2 . Indeed, a straightforward calculation yields the identity 3 3 6m2 + (a − b) + 6m2 − (a − b) = (6(a + b)m)2 + 144m2 (3m4 − ab)
(5)
of which Theorem 1 is an immediate corollary. If we set a = 3m2 n, b = m2 /n, multiply by n6 and delete the factor m6 , we obtain Ramanujan’s result (1). If instead we set a = 3m4 /n2 , b = n2 and multiply by n6 we obtain Euler’s solution of (3), (3m4 + 6m2 n2 − n4 )3 + (−3m4 + 6m2 n2 + n4 )3 = (18m5 n + 6mn5 )2 .
(6)
Yet again, if we set a = 3n, b = m4 /n and multiply by n6 we obtain (3n3 + 6m2 n2 − m4 n)3 + (−3n3 + 6m2 n2 + m4 n)3 = (18mn4 + 6m5 n2 )2 . Now set n = m2 − c and we obtain (8m6 − 20cm4 + 15c2 m2 − 3c3 )3 + (4m6 − 4cm4 − 3c2 m2 + 3c3 )3 = (24m9 − 84cm7 + 114c2 m5 − 72c3 m3 + 18c4 m)2 . Now put c = 2p and divide throughout by 43 = 82 to find (m6 − 2pm4 − 3p2 m2 + 6p3 )3 + (2m6 − 10pm4 + 15p2 m2 − 6p3 )3 = (3m9 −21pm7 +57p2 m5 −72p3 m3 +36p4 m)2 . (7)
The diophantine equation A3 + B 3 + C 3 = D 3 A3 + B 3 + C 3 = D 3 .
Suppose
3 Write A = y − x, B = u + v, C = u − v and D = y + x. Then u(u2 + 3v 2 ) = x(x2 + 3y 2 ) or, u x2 + 3y 2 = 2 . x u + 3v 2 If we set both sides equal to M we find u = Mx
and
x2 + 3y 2 = M (u2 + 3v 2 ) = M (M x)2 + 3M v 2 = M 3 x2 + 3M v 2
or, 3(y 2 − M v 2 ) = (M 3 − 1)x2 . Now let M = m2 and we obtain 3(y + mv)(y − mv) = (m6 − 1)x2 . Up to this point, I have been heavily indebted to Bruce Berndt [1, p.198]. However, at this point our paths diverge. Set x = 3m and then (y + mv)(y − mv) = 3m2 (m6 − 1). Now write y + mv = am, y − mv = bm, where ab = 3(m6 − 1). Then y=
1 1 (am + bm), v = (a − b), while x = 3m, u = 3m3 . 2 2
So we have A=
1 1 1 1 (a+b−6)m, B = 6m3 + (a − b) , C = 6m3 − (a − b) , D = (a+b+6)m. 2 2 2 2
Thus we obtain the following result. Theorem 2. If m is an integer and if ab = 3(m6 − 1) then 3 3 ((a + b − 6)m)3 + 6m3 + (a − b) + 6m3 − (a − b) = ((a + b + 6)m)3 . Once again, a straightforward calculation yields the identity 3 3 ((a + b − 6)m)3 + 6m3 + (a − b) + 6m3 − (a − b) = ((a + b + 6)m)3 + 144m3 3(m6 − 1) − ab ,
(8)
4 of which Theorem 2 is an immediate corollary. If we write a = 3n, b = (m6 − 1)/n and multiply through by n3 we obtain 3mn2 − 6mn + m7 − m
3
+ =
3 3 3n2 + 6m3 n − m6 + 1 + −3n2 + 6m3 n + m6 − 1 3 3mn2 + 6mn + m7 − m .
Next set n = m3 − c and we obtain 4m7 − (6c + 6)m4 + (3c2 + 6c − 1)m
3
+ =
3 3 8m6 − 12cm3 + (3c2 + 1) + 4m6 − (3c2 + 1) 3 4m7 − (6c − 6)m4 + (3c2 − 6c − 1)m .
Now put c = 2p + 1 and divide throughout by 43 to find m7 − (3p + 3)m4 + (3p2 + 6p + 2)m
3
+ +
3 2m6 − (6p + 3)m3 + (3p2 + 3p + 1) 3 3 m6 − (3p2 + 3p + 1) = m7 − 3pm4 + (3p2 − 1)m ,
which is Ramanujan’s result (2).
References [1] Bruce C. Berndt, Ramanujan’s Notebooks, Part III, Springer-Verlag, New York, 1991. [2] L. E. Dickson, History of the Theory of Numbers, Vol. 2, Chelsea, New York. [3] S. Ramanujan, Notebooks, Tata Institute of Fundamental Research, Bombay, 1957, Vol.2.
