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February 15

Diophantine equation

Is the following Diophantine equation an instance of a well-studied class of equations, and if so, is a solution technique or impossibility proof known?

${\displaystyle -3m^{4}+12mn^{3}-k^{2}=0}$ in positive integers m, n, k with n > m.

Thanks. Loraof (talk) 21:27, 15 February 2018 (UTC)

Not that I'm an expert on Diophantine equations, but I looked at this a little bit and no one else has answered yet. First note that 3|k, so letting k=3l the equation reduces to
${\displaystyle -m^{4}+4mn^{3}=3l^{2}}$
or
${\displaystyle m(4n^{3}-m^{3})=3l^{2}}$
Also note that a solution of the form (m, n, l)=(cm', cn', c2l') implies (m, n, l)=(m', n', l') is also a solution. So wlog we can assume the solution does not have the form (m, n, l)=(cm', cn', c2l') where c>1. This implies, after a bit of work, that m and n are relatively prime. (Basically this exploits the fact that both sides are homogeneous and the variables are separated). From there you can further restrict the form of m by considering its prime divisors. For example, if p is a prime >3, p|m but p2∤m, then p|l, p2 divides the rhs, from which p|n contradicting assumption. This implies m has the form 2a3bu where u is a powerful number relatively prime to 6. There may be something obvious that I'm missing or an applicable theory which I'm not familiar with (algebraic number fields?) but I don't see much else to do with it. --RDBury (talk) 18:18, 17 February 2018 (UTC)
Loraof, I assume you made some brute-force, trial and error attempts at finding solutions. How high did you go? I find no solutions for 1 ≤ m < n ≤ 49796 (at which point I hit up against limits of 64-bit). There might be a probabilistic argument in favor of or against the existence of solutions, but I don't see a way to hook such a proof into this problem. -- ToE 21:36, 19 February 2018 (UTC)

February 19

Algebra

So, Yikes! Haven't done algebra in nearly 20 years.

I was given a sample problem of: 7 + 18f = 20f - 5

The solution says to "Subtract 20f from both sides"

Ok, when I was trying to figure it out, I added 5 to both sides, to clear out the -5.

Why was that not correct? I'm way confused now!

Actually, maybe I am starting to get it...you want to get an F by itself? Ok, if that's right, why not subtract 18f from both sides?

CTF83! 06:14, 19 February 2018 (UTC)

It would be a little easier to subtract 18f from both sides. Then add 5 to both sides, giving 12 = 2f, so f = 6. Bubba73 You talkin' to me? 06:24, 19 February 2018 (UTC)
Oh, ok, I see. Thanks! CTF83! 06:41, 19 February 2018 (UTC)

First Order Logic: Completeness vs. Decidability

Something that has been puzzling me lately: my understanding is that FOL is complete, that all theorems in FOL that are true can be proven to be true. But I know that Turing and Church proved that FOL is not decidable, that there is no algorithm that given any arbitrary theorem can tell whether it is true or false. I know I'm missing something but these two facts seem inconsistent to me. I used to think that the answer was that while all the true theorems can be proven true there are some false theorems that can't be proven false. But someone pointed out to me that if that is the case just take the negation of any theorem. One of them has to be true and hence since FOL is complete you can prove that one which will prove the negation of that theorem to be false. I know I'm making some fundamental error here would appreciate someone helping me understand what it is. --MadScientistX11 (talk) 19:44, 19 February 2018 (UTC)

First-order logic is complete - see Gödel's completeness theorem. It is the higher stuff that is incomplete and indecidable, right? Bubba73 You talkin' to me? 19:52, 19 February 2018 (UTC)
No, unfortunately, the term "complete" is a bit overloaded here. First-order logic is complete in the sense that every statement that is true in all models is also provable. I think this might be called "model-completeness"? Not sure; you usually work it out from context.
That's different from negation-completeness, which says that for every statement, either the statement or its negation is provable. First-order logic is very far from being negation-complete. --Trovatore (talk) 20:02, 19 February 2018 (UTC)
It has been 35 years since I had this stuff. My memory from those days is that first-order logic did not include "there exists" and "for all", and was decidable by building a truth table. Please correct me if I'm, wrong, and it has been a long time... Bubba73 You talkin' to me? 20:50, 19 February 2018 (UTC)
I think you're thinking of propositional logic rather than first-order logic. Propositional logic is indeed negation-complete. But the language is too restrictive to express anything very interesting. --Trovatore (talk) 21:01, 19 February 2018 (UTC) Oops, sorry, that was silly. Propositional logic is not negation-complete. But it is decidable, which first-order logic is not. --Trovatore (talk) 21:05, 19 February 2018 (UTC)
Thanks - it has been a long time. Bubba73 You talkin' to me? 22:28, 19 February 2018 (UTC)
"But someone pointed out to me that if that is the case just take the negation of any theorem. One of them has to be true and hence since FOL is complete you can prove that one which will prove the negation of that theorem to be false."
I believe this is the fundamental error. It is not always the case that either a statement or its negation is provable. For a simple example, suppose we have a single axiom ${\displaystyle \forall x(P(x))}$ and ask whether ${\displaystyle \forall x(Q(x))}$ is provable from that axiom. Obviously it is not, but neither is ${\displaystyle \lnot \forall x(Q(x))}$. The second statement (and its negation) is independent of the axioms. Anon126 (notify me of responses! / talk / contribs) 20:00, 19 February 2018 (UTC)
and @Anon126: Thanks. I think I've got it now. I'm just going to try restating what I think was said to make sure: so I think my error is confusing validity (which I believe is the same as true in all models) with provability. Completeness means that all valid theorems are provable and FOL is complete. But many theorems (e.g., that there is no number between X and it's successor which is true for Natural numbers but not for reals) are not true in all models. And some of those FOL theorems that are not valid are also not provable by the Turing/Church proof. --MadScientistX11 (talk) 22:52, 19 February 2018 (UTC)

Can a light bulb be two functions in two dimensions and one in three dimensions?

Is a pair of functions possible which looks like a light bulb? In two dimensions, to the left of the y-axis is a half circle with radius a, with a curve above the x-axis and a mirror image of it below, ending where the bulb screws in with the value of y approaching b above the x-axis and -b below it. In three dimensions, there is one function where half a sphere is above the xy plane and the radius is again a, with a circle of radius b below the xy plane.— Vchimpanzee • talk • contributions • 21:31, 19 February 2018 (UTC)

I'm not quite sure what you're talking about (English isn't my native language), but you could try something like this. Admittedly that isn't much of a light bulb but playing with coefficients and increasing the polynomial's degree enough you can get as close to the real light bulb as you want, and later it shouldn't be hard to find the function for the rotational shape (sp?). If you want to have a better approximation of a real light bulb, you can pick out enough points and find the polynomial that passes through them using this method, although the polynomial might end up with a really large degree for a passable light bulb.
Although, if you want to have an exact half-circle/hemisphere on the round side of the bulb (x^2+y^2[+z^2]=r^2), you will probably have to define it piecewise (is that what you meant by two functions?) or do something with absolute value function, a polynomial won't do then. 78.1.172.210 (talk) 00:04, 20 February 2018 (UTC)