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Friday, September 23, 2016

Entry 12

Ramanujan gave a beautiful Diophantine identity. Let ad=bc. Then
64((a+b+c)6+(b+c+d)6(c+d+a)6(d+a+b)6+(ad)6(bc)6)×((a+b+c)10+(b+c+d)10(c+d+a)10(d+a+b)10+(ad)10(bc)10)=45((a+b+c)8+(b+c+d)8(c+d+a)8(d+a+b)8+(ad)8(bc)8)2 known as the 6-10-8 Identity. This is a remarkable relationship by most people’s standards. In the words of Berndt, primary editor of Ramanujan’s Notebooks, “it is one of the most fascinating identities we have ever seen”. One can easily appreciate its simplicity of form and the rather unexpected use of high exponents. However, we can generalize this.

I. Define Fn=xn1+xn2+xn3(yn1+yn2+yn3), where x1+x2+x3=y1+y2+y3=0. If F2=F4=0, then 64F6F10=45F28(Ramanujan) 25F3F7=21F25(Hirschhorn) II. Define Fn=xn1+xn2+xn3+xn4(yn1+yn2+yn3+yn4), where also xi=yi=0. If F1=F3=F5=0, then 7F4F9=12F6F7(yours truly) For example xi=(21,9,13,17) and yi=(23,1,3,21). And so on for similar relations for higher powers.

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