Entry 12

Ramanujan gave a beautiful Diophantine identity. Let \(ad=bc\). Then
$$\small64\big((a+b+c)^6+(b+c+d)^6-(c+d+a)^6-(d+a+b)^6+(a-d)^6-(b-c)^6\big)\times\\\small\big((a+b+c)^{10}+(b+c+d)^{10}-(c+d+a)^{10}-(d+a+b)^{10}+(a-d)^{10}-(b-c)^{10}\big)=\\\small45\big((a+b+c)^8+(b+c+d)^8-(c+d+a)^8-(d+a+b)^8+(a-d)^8-(b-c)^8\big)^2$$ 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 \(F_n = x_1^n+x_2^n+x_3^n-(y_1^n+y_2^n+y_3^n),\,\) where \(\,\small x_1+x_2+x_3=y_1+y_2+y_3 = 0.\,\) If \(F_2 = F_4 = 0\), then $$64F_6F_{10} = 45F_8^2\quad \text{(Ramanujan)}$$ $$25F_3F_{7} = 21F_5^2\quad \text{(Hirschhorn)}$$ II. Define \(F_n = x_1^n+x_2^n+x_3^n+x_4^n-(y_1^n+y_2^n+y_3^n+y_4^n),\,\) where also \(\,\small \sum x_i =\sum y_i= 0.\,\) If \(F_1 = F_3 = F_5 = 0\), then $$7F_4F_9 = 12F_6F_7\quad \text{(yours truly)}$$ For example \(x_i = (21,\,9,\,-13,\,-17)\) and \(y_i = (23,\,1,\,-3,\,-21)\). And so on for similar relations for higher powers.