Integer Solutions
of
a1h
+ a2h + a3h
= b1h + b2h
+ b3h
( h = 1, 3, 4)
- Ajai Choudhry
obtained solutions to this system in 1991. [59]
- [ -3254, 5583, 5658 ] = [ -1329, 2578, 6738 ]
- Chen Shuwen found the following 5-7-2-6 identity in 2017.
- 9216 (a5 + b5 + c5 - d5 - e5
- f5 )( a7 + b7 + c7 - d7
- e7 - f7 ) = 35 ((a2 + b2 + c2
- d2 - e2 - f2 )3 -16 ( a6
+ b6 + c6 - d6 - e6 - f6
))2
- This ( Chen 5-7-2-6 identity) can also be expressed by defining
- P2 = (a2 + b2 + c2
- d2 - e2 - f2 ) / 2
- P5 = (a5 + b5 + c5
- d5 - e5 - f5 ) / 5
- P6 = (a6 + b6 + c6 - d6 - e6 - f6 )
/ 6
- P7 = (a7 + b7 + c7 - d7 - e7 - f7 )
/ 7
- Then ( P23 /
12 - P6 )2 = P5 P7
, where ah + bh + ch = dh +
eh + fh ( h = 1, 3, 4 )
- Similarly, Ramanujan 6-10-8 Identity and Hirschhorn 3-7-5 Identity can be
expressed as
- R3 = (a3 + b3 + c3
- d3 - e3 - f3 ) / 3
- R5 = (a5 + b5 + c5
- d5 - e5 - f5 ) / 5
- R6 = (a6 + b6 + c6 - d6 - e6 - f6 )
/ 6
- R7 = (a7 + b7 + c7 - d7 - e7 - f7 )
/ 7
- R8 = (a8 + b8 + c8 - d8 - e8 - f8 )
/ 8
- R10 = (a10 + b10 + c10 - d10 - e10 - f10 )
/ 10
- Then R8 2
= 4/3 R6 R10 and R5 2
= R3 R7
, where ah + bh + ch = dh +
eh + fh ( h = 1, 2, 4 )
- Noted by Chen Shuwen, in 2017.
Last revised Nov,5, 2017.
Copyright 1997-2017, Chen Shuwen