Kindly see the question at the end of post. Solutions to the system of three equations,

$$\begin{aligned} a^2+b^2+c^2+d^2\, &= e^2+f^2+g^2+h^2\\ a^4+b^4+c^4+d^4\, &= e^4+f^4+g^4+h^4\\ abcd\, &= efgh \end{aligned}\tag{1}$$

satisfy two nice consequences.

Consequence 1:

Define $n$ as $n(a^2+b^2+c^2+d^2)^2 = a^4+b^4+c^4+d^4$. Then,

$$32(a^6+b^6+c^6+d^6-e^6-f^6-g^6-h^6)\\(a^{10}+b^{10}+c^{10}+d^{10}-e^{10}-f^{10}-g^{10}-h^{10}) = \\15(n+1)(a^8+b^8+c^8+d^8-e^8-f^8-g^8-h^8)^2$$

This generalizes the Ramanujan 6-10-8 Identity which had the case $d=h=0$.

Consequence 2:

Let $(a+b+c+d)(e+f+g+h) ≠ 0$. Then for $k =2,4,6,8$,

$$(-a+b+c+d)^k + (a-b+c+d)^k + (a+b-c+d)^k + (a+b+c-d)^k + \\(2e)^k + (2f)^k + (2g)^k + (2h)^k\\ =\\ (2a)^k + (2b)^k + (2c)^k + (2d)^k + \\(-e+f+g+h)^k + (e-f+g+h)^k + (e+f-g+h)^k + (e+f+g-h)^k$$

This gives a solution to $x_1^8+x_2^8+\dots+x_8^8 = y_1^8+y_2^8+\dots+y_8^8$.

Question: What are other parameterizations of $(1)$ in terms of binary quadratic forms?

(I know of plenty but I used certain assumptions. It would be good to have a fresh approach.)


1 Answer 1


What I write here is not the final answer to this question, but I think it will be helpful.

I notice that the Consequence 1 can be simplified and generalized as below,

Denote $$R_n=(a^n+b^n+c^n+d^n-e^n-f^n-g^n-h^n)/n$$ for any $n<>0$, and $$R_0=2(abcd-efgh)/(abcd+efgh)$$ $$m=(a^2+b^2+c^2+d^2)/(a+b+c+d)^2$$ We have

If $R_0=R_1=R_2=0$, then $$R_3R_5/R_4^2=(m+1)/2$$ $$R_4/R_3=a+b+c+d$$ $$R_3/R_{-1}=-abcd$$ $$\frac{R_{-2}}{R_{-1}}-\frac{R_{-1}}2=\frac1a+\frac1b+\frac1c+\frac1d$$

If $R_1=R_2=R_3=0$, then $$R_4R_6/R_5^2=(m+1)/2$$ $$R_5/R_4=a+b+c+d$$ $$R_4(1+R_0/2)/R_0=-abcd$$ $$\frac{R_{-1}}{R_0}-\frac{R_{-1}}2=\frac1a+\frac1b+\frac1c+\frac1d$$

More similar identities can be found in my site Algebraic Identities.


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