Timeline for answer to Parameterization for diophantine $a^4+28a^3b+70a^2b^2+28ab^3+b^4=c^4+28c^3d+70c^2d^2+28cd^3+d^4$ by David
Current License: CC BY-SA 4.0
Post Revisions
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S 6 hours ago | history | edited | kjetil b halvorsen | CC BY-SA 4.0 |
Improved MathJax formatting.
|
| S 6 hours ago | history | suggested | M. A. | CC BY-SA 4.0 |
Improved MathJax formatting.
|
| 9 hours ago | review | Suggested edits | |||
| S 6 hours ago | |||||
| Mar 8 at 12:11 | comment | added | Wolfgang | @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. | |
| Mar 6 at 3:57 | comment | added | David | @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). | |
| Mar 6 at 2:59 | comment | added | Guruprasad | @David , your parametrizations are all Trivial because $f(a,b)-f(c,d)\neq 0 $ , except already specified points in your post. | |
| Mar 5 at 20:07 | comment | added | Aleksandr | $A^4 + 28A^3 B + 70A^2 B^2 + 28A B^3 + B^4 = C^4 + 28C^3 D + 70C^2 D^2 + 28 C D^3 + D^4$ 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*} | |
| Mar 5 at 20:01 | vote | accept | Aleksandr | ||
| Mar 5 at 19:05 | history | answered | David | CC BY-SA 4.0 |