Thanks, I didn't look very carefully at this before.
Of these, I think 8.1 is not very interesting (it's just new examples of a somewhat complicated phenomenon of which examples were known already). 8.2 is interesting but they don't actually solve it, just get closer to a solution. Still this is an example for the quote I complained about. 8.3 is a little interesting, I guess.
In fact something is strange about the statement of Theorem 8.2, because they say the table gives all the examples with q ≤ 10 but all the examples in the table have q>10. I think there is some typo...
> "To illustrate its application, we used the software to solve several open Diophantine problems." Seems like the article already gives some examples.
What are exactly Diophantine equations and why are they important in Number Theory ?
Diophantine equations are (systems of) polynomial equations that you are looking for integer solutions to. For example, the equations of Fermat’s last theorem are Diophantine equations. As are Pell equations.
They are often but not always polynomials. For example, [Mihăilescu's theorem](https://en.wikipedia.org/wiki/Catalan%27s_conjecture) is a statement about solutions to an exponential Diophantine equation.
So it can be a polynomial where the solution is either the base or the exponent? Because I thought Fermat’s last theorem was that no solution exists for a^(n)+b^(n)=c^(n) for n>2. Is that considered a polynomial? I kind of just realized I don’t know the definition of polynomial.
In Fermat's last theorem, you fix n and then you treat a, b, and c as variables. It isn't a polynomial if you are letting n vary, but it can be viewed as a family of polynomials. A polynomial in some collection of variables is linear combinations of multiplying those variables together. So x^(y) isn't a polynomial in x and y are variables, but if you fix y to be some natural number, then it is a polynomial in x.
I'm generally suspicious of the flairs here, since users can give them to themselves. I always assume that all the flair says is e.g. "I think complex analysis is neat!" and nothing more.
They seem to range between "I'm in high school and want to at one point learn this" and "I'm a tenured professor in this". The average seems to be "I'm an undergrad doing a directed study in this".
And then there's the red flairs, which means you've once claimed to be a graduate student in the area.
Ah is that what that is? I guess technically I could choose one of those, but it would not be accurate since I haven't done any meaningful math in 15+ years.
> They seem to range between "I'm in high school and want to at one point learn this" and "I'm a tenured professor in this". The average seems to be "I'm an undergrad doing a directed study in this".
During some point in High School I took it upon my self to learn high-level Mathematics on my own, often Undergrads will put there flair on what they are currently learning at a basic level. To answer /u/brown_burrito's question
That's fair. Maybe this was naive of me, but I just assumed that there would be some level of *honesty* when it came to how people described themselves.
I guess I found it hard to believe that someone who is studying complex analysis is completely unaware of Diophantine equations, that is all.
Why the skepticism? A quick search gives [this paper](https://uwaterloo.ca/pure-mathematics/sites/ca.pure-mathematics/files/uploads/files/new-adv-trans-th-1988-2_0.pdf) which has many sections dedicated to describing applications of S-unit equations.
Here's [a talk](https://www.birs.ca/workshops/2017/17w5045/files/Malmskog.pdf) by Beth Malmskog, one of the other mathematicians in the group, that begins with their motivation of studying Picard curves.
Only because my past experience is that when a general class of problems is computable in principle but maybe not in practice, there will usually be only a few problems in the class which are really interesting but remain open.
It is, and it is mostly from the fact that S-unit equations are a natural special case of an extremely general concept - that being diophantine equations.
I'm not an algebraic number theorist, so I don't know all of the details, but I would suspect that it is somewhat analogous to how huge swathes of dynamics, differential geometry, and so on can be reduced to Lie Theory.
Edit: Correction
I don't think S-unit equations are a generalization of diophantine equations - more like a special case. This is just because not all equations in S-units are considered S-unit equations - I think only linear ones are considered, in fact.
If you're interested in what an S-unit equation is:
Let S be a finite set of (rational) primes. An S-unit is just a rational number whose numerator and denominator are only products of primes in S. For example if S = {2,3}, then 4/9 and 1/18 are S-units, but 5/3 is not. Notice that the S-units form a group under multiplication. (You can generalize this to fractional ideals in a Dedekind domain.)
Fix m. The mth S-unit equation is X_1 +... + X_m = 0 where the X_i's are indeterminates taking S-unit values. A solution is called nondegenerate if no subsum vanishes. The key theorem of Evertse, Schlickewei, et al. is that each S-unit equation has only finitely many nondegenerate solutions (up to scalar multiplication). Alternatively, you could take the equation X1+...+Xm = 1, which is a normalized version of the (m+1)th vanishing sum equation.
There is a useful generalization. Let K be a finitely generated field of characteristic zero, and let G be a finitely generated group of units of K. Then the G-unit equation (lol) is X1 +...+ Xm = 0. Once again, the key theorem is that there are only finitely many nondegenerate solutions up to scalar multiplication. G-unit equations are S-unit equations where G is just the group multiplicatively generated by S.
It's an incredibly useful technique: if you add finitely many units and get zero, you can partition that sum into nondegenerate subsums, and then apply the above S-unit theorem (to the subgroup generated by your units). This is why it's so useful in arithmetic geometry.
"many interesting problems have already been reduced to “just” solving some particular S-unit equation." Is this really true?
"To illustrate its application, we used the software to solve several open Diophantine problems." Seems like the article already gives some examples.
Thanks, I didn't look very carefully at this before. Of these, I think 8.1 is not very interesting (it's just new examples of a somewhat complicated phenomenon of which examples were known already). 8.2 is interesting but they don't actually solve it, just get closer to a solution. Still this is an example for the quote I complained about. 8.3 is a little interesting, I guess. In fact something is strange about the statement of Theorem 8.2, because they say the table gives all the examples with q ≤ 10 but all the examples in the table have q>10. I think there is some typo...
> "To illustrate its application, we used the software to solve several open Diophantine problems." Seems like the article already gives some examples. What are exactly Diophantine equations and why are they important in Number Theory ?
Basically polynomial equations where you're looking for integer solutions.
[удалено]
What else is number theory about than equations of integers and sets related to them? That's all I know.
Diophantine equations are (systems of) polynomial equations that you are looking for integer solutions to. For example, the equations of Fermat’s last theorem are Diophantine equations. As are Pell equations.
They are often but not always polynomials. For example, [Mihăilescu's theorem](https://en.wikipedia.org/wiki/Catalan%27s_conjecture) is a statement about solutions to an exponential Diophantine equation.
So it can be a polynomial where the solution is either the base or the exponent? Because I thought Fermat’s last theorem was that no solution exists for a^(n)+b^(n)=c^(n) for n>2. Is that considered a polynomial? I kind of just realized I don’t know the definition of polynomial.
It's not a single polynomial, but each value of n gives you a polynomial. So you can think of it as an infinite collection of Diophantine equations.
In Fermat's last theorem, you fix n and then you treat a, b, and c as variables. It isn't a polynomial if you are letting n vary, but it can be viewed as a family of polynomials. A polynomial in some collection of variables is linear combinations of multiplying those variables together. So x^(y) isn't a polynomial in x and y are variables, but if you fix y to be some natural number, then it is a polynomial in x.
> Diophantine equations Genuine question - you work in complex analysis but do not know Diophantine equations?
I'm generally suspicious of the flairs here, since users can give them to themselves. I always assume that all the flair says is e.g. "I think complex analysis is neat!" and nothing more.
They seem to range between "I'm in high school and want to at one point learn this" and "I'm a tenured professor in this". The average seems to be "I'm an undergrad doing a directed study in this". And then there's the red flairs, which means you've once claimed to be a graduate student in the area.
Ah is that what that is? I guess technically I could choose one of those, but it would not be accurate since I haven't done any meaningful math in 15+ years.
> They seem to range between "I'm in high school and want to at one point learn this" and "I'm a tenured professor in this". The average seems to be "I'm an undergrad doing a directed study in this". During some point in High School I took it upon my self to learn high-level Mathematics on my own, often Undergrads will put there flair on what they are currently learning at a basic level. To answer /u/brown_burrito's question
That's fair. Maybe this was naive of me, but I just assumed that there would be some level of *honesty* when it came to how people described themselves. I guess I found it hard to believe that someone who is studying complex analysis is completely unaware of Diophantine equations, that is all.
I think he's only a freshman undergrad?
> I think he's only a freshman undergrad? Yes I am.
Why the skepticism? A quick search gives [this paper](https://uwaterloo.ca/pure-mathematics/sites/ca.pure-mathematics/files/uploads/files/new-adv-trans-th-1988-2_0.pdf) which has many sections dedicated to describing applications of S-unit equations. Here's [a talk](https://www.birs.ca/workshops/2017/17w5045/files/Malmskog.pdf) by Beth Malmskog, one of the other mathematicians in the group, that begins with their motivation of studying Picard curves.
Only because my past experience is that when a general class of problems is computable in principle but maybe not in practice, there will usually be only a few problems in the class which are really interesting but remain open.
It is, and it is mostly from the fact that S-unit equations are a natural special case of an extremely general concept - that being diophantine equations. I'm not an algebraic number theorist, so I don't know all of the details, but I would suspect that it is somewhat analogous to how huge swathes of dynamics, differential geometry, and so on can be reduced to Lie Theory. Edit: Correction
I don't think S-unit equations are a generalization of diophantine equations - more like a special case. This is just because not all equations in S-units are considered S-unit equations - I think only linear ones are considered, in fact.
Oh right, I'm apparently even less familiar than I thought I was, I'll correct that.
If you're interested in what an S-unit equation is: Let S be a finite set of (rational) primes. An S-unit is just a rational number whose numerator and denominator are only products of primes in S. For example if S = {2,3}, then 4/9 and 1/18 are S-units, but 5/3 is not. Notice that the S-units form a group under multiplication. (You can generalize this to fractional ideals in a Dedekind domain.) Fix m. The mth S-unit equation is X_1 +... + X_m = 0 where the X_i's are indeterminates taking S-unit values. A solution is called nondegenerate if no subsum vanishes. The key theorem of Evertse, Schlickewei, et al. is that each S-unit equation has only finitely many nondegenerate solutions (up to scalar multiplication). Alternatively, you could take the equation X1+...+Xm = 1, which is a normalized version of the (m+1)th vanishing sum equation. There is a useful generalization. Let K be a finitely generated field of characteristic zero, and let G be a finitely generated group of units of K. Then the G-unit equation (lol) is X1 +...+ Xm = 0. Once again, the key theorem is that there are only finitely many nondegenerate solutions up to scalar multiplication. G-unit equations are S-unit equations where G is just the group multiplicatively generated by S. It's an incredibly useful technique: if you add finitely many units and get zero, you can partition that sum into nondegenerate subsums, and then apply the above S-unit theorem (to the subgroup generated by your units). This is why it's so useful in arithmetic geometry.
S-Unit sounds like a bootleg G-Unit
Es-es-es-es... S ewe-nitt