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
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 27 at 22:08 | vote | accept | Ayan Bhowmik | ||
| Apr 25 at 2:37 | answer | added | Ayan Bhowmik | timeline score: 1 | |
| Apr 25 at 2:32 | comment | added | John Omielan | You're welcome. And, yes, the problem can basically reduce to $x^2 - y^2 = t$, as Will Jagy pointed out earlier. You should consider answering this question yourself with the details. FYI, the solution that I worked out myself is basically the same as that in Post #$7$ of the first AoPS thread I linked to above. | |
| Apr 25 at 2:29 | comment | added | Ayan Bhowmik | Thanks @JohnOmielan didn't know such a search engine existed. I see that the problem reduces to $x^{2}-y^{2}=t$ as pointed out by Will Jagy. | |
| Apr 25 at 2:12 | comment | added | John Omielan | FYI, using an Approach0 search, there's the AoPS threads 2-B-7 polynomials and B.Math - rational solution. Also, Two problems on solving equations in rational and integers has your problem as a second part, but neither of the $2$ answers there deal with it. | |
| Apr 25 at 1:30 | comment | added | Merosity | I don't want to derail from you working with Will here, so this is something to look at later if you're interested. You're essentially given an isotropic vector in a nondegenerate quadratic space, so it is universal. This actually corresponds to an explicit construction here, but you'll have to do a bit of legwork. Here's where I first learned it in case it's helpful to you: Quadratic forms and the local global-principle - swc-math.github.io/aws/2021/index.html#2021Chan | |
| Apr 25 at 1:26 | comment | added | Ayan Bhowmik | It's from ISI TOMATO. Year 2007 question paper. | |
| Apr 25 at 1:22 | comment | added | Will Jagy | Meanwhile, what book is this from? | |
| Apr 25 at 1:19 | comment | added | Will Jagy | Tell you what: how would you show that, given rational $t,$ there is a rational representation $x^2 - y^2 = t \; \; ? \; \; \; $ | |
| Apr 25 at 1:07 | comment | added | Ayan Bhowmik | This might be wrong but can I do something with Bezout's Identity by saying that for $x$ and $y$ be integers and so I can reduce it to for $ap+bq$ using @WillJagy 's idea where $a$ and $b$ have infinite solutions. Would this suffice as a proof because the question is asking for the opposite? | |
| Apr 25 at 0:28 | comment | added | Will Jagy | it does not change the problem to demand $a,b$ integers. After that, there is no damage by demanding $\gcd(a,b) = 1.$ Next, $\frac{-b}{a}$ is a rational square, so..... | |
| Apr 25 at 0:21 | history | asked | Ayan Bhowmik | CC BY-SA 4.0 |