Timeline for Prove that for any rational number $t$ , there is a solution of the equation $ax^2+by^2=t$.
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 27 at 22:08 | vote | accept | Ayan Bhowmik | ||
| Apr 25 at 3:25 | comment | added | Will Jagy | alright, you have reduced it to $x^2 - y^2 = t.$ Can you give a convincing proof that this can be done in rationals? A beginning: $x^2 - y^2 = (x+y)(x-y) $ Tell you what: you might as well demand $x-y = 1$ and $x+y = t$ | |
| Apr 25 at 2:43 | history | edited | Ayan Bhowmik | CC BY-SA 4.0 |
Added link.
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| S Apr 25 at 2:37 | review | First answers | |||
| Apr 25 at 3:00 | |||||
| S Apr 25 at 2:37 | history | answered | Ayan Bhowmik | CC BY-SA 4.0 |