3
$\begingroup$

While reading this PDF, they introduced a theorem by Ramanujan:


Theorem

If $$\alpha^2+\alpha\beta+\beta^2=3\lambda\gamma^2$$

Then$$(\alpha+\lambda^2\gamma)^3+(\lambda\beta+\gamma)^3=(\lambda\alpha+\gamma)^3+(\beta+\lambda^2\gamma)^3$$


I was wondering if there is a simple way to prove this? You could just expand both sides and show that they are equivalent, but that doesn't give me insight to how Ramanujan came up with this formula in the first place.

Note: $\alpha,\beta,\gamma,\lambda\in\mathbb{N}$

$\endgroup$
  • 2
    $\begingroup$ You must add that the symbols represent natural numbers. $\endgroup$ – Masacroso Oct 3 '16 at 16:27
  • 2
    $\begingroup$ Even Ramanujan could often not explain how he came up with his results. The memoir by G.H.Hardy about him is fascinating. $\endgroup$ – DanielWainfleet Oct 3 '16 at 17:15
4
$\begingroup$

A proof without expanding both sides.

Using that $$A^3-B^3=(A-B)(A^2+AB+B^2)$$ we get $$(\alpha+\lambda^2\gamma)^3-(\beta+\lambda^2\gamma)^3+(\lambda\beta+\gamma)^3-(\lambda\alpha+\gamma)^3$$


$$=(\alpha-\beta)((\alpha+\lambda^2\gamma)^2+(\alpha+\lambda^2\gamma)(\beta+\lambda^2\gamma)+(\beta+\lambda^2\gamma)^2)$$$$\qquad+(\lambda\beta-\lambda\alpha)((\lambda\beta+\gamma)^2+(\lambda\beta+\gamma)(\lambda\alpha+\gamma)+(\lambda\alpha+\gamma)^2)$$


$$=(\alpha-\beta)(\color{red}{\alpha^2+\alpha\beta+\beta^2}+3\alpha\lambda^2\gamma+3\beta\lambda^2\gamma+3\lambda^4\gamma^2)$$$$\qquad+\lambda(\beta-\alpha)(\lambda^2(\color{red}{\alpha^2+\alpha\beta+\beta^2})+3\lambda\beta\gamma+3\lambda\alpha\gamma+3\gamma^2)$$


$$=(\alpha-\beta)(\color{red}{3\lambda\gamma^2}+3\alpha\lambda^2\gamma+3\beta\lambda^2\gamma+3\lambda^4\gamma^2)$$$$\qquad+\lambda(\beta-\alpha)(\lambda^2\cdot \color{red}{3\lambda\gamma^2}+3\lambda\beta\gamma+3\lambda\alpha\gamma+3\gamma^2)$$


$$=(\alpha-\beta)(3\lambda\gamma^2+3\alpha\lambda^2\gamma+3\beta\lambda^2\gamma+3\lambda^4\gamma^2)$$$$\qquad-(\alpha-\beta)(3\lambda\gamma^2+3\alpha\lambda^2\gamma+3\beta\lambda^2\gamma+3\lambda^4\gamma^2)$$

$$\boxed{=0}$$

$\endgroup$
  • $\begingroup$ Okay, then why won't this work: If$$\beta^3+3\beta\lambda\gamma+\gamma^3=\lambda^3\tag1$$Then$$(\beta\epsilon-\gamma^2)^3+(\beta^2+\gamma\lambda)^3=(\beta\lambda+\gamma^2)^3+(\beta\epsilon-\gamma^2)^3$$ $\endgroup$ – Frank Jan 3 '17 at 22:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy