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How to find new parametric forms for the OP diophantine equation $$A^4+28A^3B+70A^2B^2+28AB^3+B^4=C^4+28C^3D+70C^2D^2+28CD^3+D^4$$ The smallest parametric solution can be found in this collection by Tito Piezas III.

$$A=n^3-2n^2+1$$ $$B=n^3+2n^2-1$$ $$C=n^3-n-1$$ $$D=n^3-n+1$$

By searching, I discovered a polynomial of the third degree defining at least two more parametric families. $A=10n^3+15n^2+7n+1$ I would like to mention Tomita, who found an interesting solution to the OP equation for the 2nd degree $$A=-3n^2-96n+576$$ $$B=-7n^2+224n-1216$$ $$C=-19n^2+224n-448$$ $$D=9n^2-96n-192$$

The purpose of the question is to obtain new parametric families for a given diophantine equation.

I don't speak English, please correct any mistakes if there are any.

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    $\begingroup$ How Elliptic curves tag related to your post ? $\endgroup$ Commented Mar 5 at 15:27
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    $\begingroup$ @Guruprasad it's a K3 surface and a common technique is to find an elliptic fibration and use the group law on an elliptic curve to find new rational curves ( = partial parametrizations) from known ones. $$ $$ [The surface is a quotient of the 8-th degree Fermat surface: start from $(A+B)^8 + (A-B)^8 = (C+D)^8 + (C-D)^8$ and write $(a,b,c,d) = (A^2,B^2,C^2,D^2)$.] $\endgroup$ Commented Mar 5 at 15:45
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    $\begingroup$ @NoamD.Elkies , Thank you. Your comment explains a lot to me. Can you write it as an answer by Computing new rational curves ? $\endgroup$ Commented Mar 6 at 2:50
  • $\begingroup$ As mentioned by Noam D Elkies, the "OP" polynomial, is represented by an 8th degree (8-2-2). Hence the numerical solution of "OP' (a,b,c,d)=(7,5,15,1) is of the form [(p+q)^8+(p-q)^8=(r+s)^8+(r-s)^8], where (p,q,r,s) are square root of the integer set, (a,b,c,d). And the equation of the 8th degree is in radicals. $\endgroup$ Commented Mar 6 at 4:08
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    $\begingroup$ @Aleksandr, There is the parametric solution of degree $2.$ \begin{align*} A &=-3(n^2+32n-192),\\ B &=-(7n^2-224n+1216),\\ C &=-(19n^2-224n+448),\\ D &=3(3n^2-32n-64). \end{align*} $\endgroup$ Commented Mar 7 at 4:13

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Equation given by "OP":

$$\Bigr(a^4 + 28 a^3 b + 70 a^2 b^2 + 28 a b^3 + b^4\Bigr) = \Bigl(c^4 + 28 c^3 d + 70 c^2 d^2 + 28 c d^3 + d^4\Bigr). \tag{1} \label{eq:1} $$

In equation \eqref{eq:1}, we take:

$$ \begin{align} a &= n^3 - 3 n^2 + 1, \\ b &= n^3 + 3 n^2 - 1, \\ c &= n^3 - 2 n + 1, \\ d &= n^3 + 2 n - 1, \end{align} $$

and $n=2/3$ is solution.

Next we take:

$$ \begin{align} a &= n^3 - 3 n^2 + 1, \\ b &= n^3 + 3 n^2 - 1, \\ c &= n^3 - 4 n + 1, \\ d &= n^3 + 4 n - 1, \end{align} $$

and $n=4/3$ is solution

Next we take:

$$ \begin{align} a &= n^3 - 4 n^2 + 1, \\ b &= n^3 + 4 n^2 - 1, \\ c &= n^3 - 5 n + 1, \\ d &= n^3 + 5 n - 1, \end{align} $$

and $n=5/4$ is a solution.

Hence, equation \eqref{eq:1} above is satisfied for

$$n=\left(\frac23,\frac43,\frac54 \right), $$ when represented by polynomials shown above.

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    $\begingroup$ $A^4 + 28A^3 B + 70A^2 B^2 + 28A B^3 + B^4 = C^4 + 28C^3 D + 70C^2 D^2 + 28 C D^3 + D^4$ It might also be interesting, this other group of solutions that alternates: \begin{align*} A &= 10n^3 + 15n^2 + 7n + 1 \\ B &= 44n^3 + 93n^2 + 65n + 15 \\ C &= 23n^3 + 43n^2 + 26n + 5 \\ D &= 25n^3 + 49n^2 + 32n + 7 \end{align*} \begin{align*} A &= 10n^3 + 15n^2 + 7n + 1 \\ B &= 44n^3 + 39n^3 + 11n + 1 \\ C &= 23n^3 + 26n^2 + 9n + 1 \\ D &= 25n^3 + 26n^2 + 9n + 1 \end{align*} $\endgroup$ Commented Mar 5 at 20:07
  • $\begingroup$ @David , your parametrizations are all Trivial because $f(a,b)-f(c,d)\neq 0 $ , except already specified points in your post. $\endgroup$ Commented Mar 6 at 2:59
  • $\begingroup$ @Guruprasad. My point was to conjecture, that the parametrization is going in the general direction of representing variables, (a,b,c,d), as a polynomial function, of the form, (n^3+pn^2+q). $\endgroup$ Commented Mar 6 at 3:57
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    $\begingroup$ @Aleksandr Seeing the binomial coefficients $\binom 8{2k}$ here, I wouldn't be surprised that the patterns generalize quite easily. It might be interesting to examine the "smallest non trivial polynomial", as here $A = 10n^3 + 15n^2 + 7n + 1$, supposing it is well-defined. $\endgroup$ Commented Mar 8 at 12:11

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