How do you find integer solutions to x/(y + z) + y/(x + z) + z/(x + y) = 4?
quora.com
quora.com
https://www.amazon.com/Elliptic-Tales-Curves-Counting-Number...
It's not a textbook, which is both good and bad. In my case, it did a good job whetting my appetite for more!
A really well-written and not-extremely-difficult undergrad textbook on elliptic curves:
https://www.amazon.com/Rational-Points-Elliptic-Undergraduat...
(Non-affiliate links, just so you know.)
a) number theory - many questions about number theory boil down to finding points on elliptic curves with rational coordinates
b) algebraic geometry - elliptic curves are very "nice" from this point of view, and they are complicated enough that you can say interesting things, but not so complicated that you can't say anything
c) complex analysis - over the complex plane, an elliptic curve is a torus. You might have seen how a torus can be formed by identifying opposite edges of a square, and an elliptic curve is hence "a quotient of C by a square lattice". Modular forms, which are creatures of complex analysis with deep applications to number theory, are naturally defined in this framework.
and many more things I don't know about. They're in a sort of "sweet spot" and act as a bridge between multiple parts of mathematics.
I tried to crack this problem, and ended up (very naively) resorting to substituting division with modulo operation that is, a%(b+c) + b%(a+c) + c%(a+b) = 4 of which the min solution is a=1, b=2, and c=4. I am glad that the true solution involved EC which, if my understanding is correct, is basically modular operations in high-dimensional space.
Also, you can look at the points on an elliptic curve where you allow the coordinates to be real, complex, rational, or even the integers mod p (any field will do), so the last choice gives you a closer link with modular arithmetic. It's best to treat ECs as their own weird, wonderful beasts!
Quality is valued, and so are expert opinions. I doubt they pay anybody.
http://www.wolframalpha.com/input/?i=(a%2F(b%2Bc))%2B(b%2F(a...
> Solve[x/(y + z) + y/(z + x) + z/(x + y) == 4, {x, y, z}, Integers]
but it gives a very long output in terms of Root[]s and conditional expressions.
In fact, I would expect this to be quite rare as most people would probably use & instead.
What's wrong with that? The same phenomenon occurs on Twitter - people find a way to optimize every character.
Isn't an 80 character limit for titles on a modern web forum really the problem?
＋＋＋＋＋＋＋＋＋＋
++++++++++
x/(y&z) & y/(x&z) & z/(x&y) = 4.
If the / becomes integer (flooring) division, there are many solutions, such as (6, 13, 19).
But if the division results are required to actually be integers, there are no small solutions - at least, none where x, y, and z <= 10000 (checked by brute force with minor optimization). I suspect that unlike the original problem, there are actually no solutions, but it’s just a guess. Anyone want to come up with a proof? :)
Bitwise AND is not a linear function, which is a first obstacle.
x/(y&z) & y/(x&z) & z/(x&y) is even
That means that x/(y&z) is even, or y/(x&z) is even, or z/(x&y) is even.
If x/(y&z) is even, then x is even too, so x&z is even, so y is even (otherwise y/(x&z) wouldn't be integer).
Similarly, we can conclude that all x, y, z are even.
But if we substitute x=x/2, y=y/2, z=z/2, the value of x/(y&z) & y/(x&z) & z/(x&y) doesn't change.
So the (x, y, z) triplet we considered wasn't the smallest. Contradiction.
I thought that equation was degree 4, which aligns with what the author says later on. Am I missing something? It seems odd that he would write out that equation accidentally, maybe just crossed wires though. I'm sure we've all been there.
The degree of the entire equation is the maximum of the degree of the individual terms. However, the degree of a term is the sum of the degree of all of the factors.
For example, the equation "a+b+c=10" is degree 1, but "abc=10" is degree 3.
But with the definition he is using, the degree of a term in the equation is the sum of the degrees of the variables. So (a^2)(b^1)(c^4) has degree 2 + 1 + 4 = 7.
Degree has the property that Deg(P1P2)=Deg(P1)Deg(P2)
How do I escape asterisks?
Solving them or progress? May be I don't care, but these little problem, simple ones does encourage one to start thinking about solving problems, they make math accessible to masses.
I analyzed the problem for a while and theorized ways to collaboratively contribute to the solution. I settled on submitting an integer sequence to OEIS.
What's mind blowing here is that if you ask for base 2, 3, 4 the answers are trivial but for bases 2,3,4,5 somehow you get 82000 and we do not know whether there is another and for 6 it's unknown.
Let s(n) be the sum of all positive divisors of n. E.g., s(2) = 1 + 2, s(4) = 1 + 2 + 4, s(20) = 1 + 2 + 4 + 5 + 10 + 20.
Let H(n) = 1 + 1/2 + 1/3 + 1/4 + ... + 1/n.
Question: is it true that s(n) < H(n) + log(H(n)) exp(H(n)) for all n > 1?
(log is natural logarithm)
It turns out that this simple looking problem is equivalent to the Riemann Hypothesis, which is perhaps the most important unsolved problem in pure mathematics. Many of the best minds in mathematics have attempted it in the approximately 160 years since Riemann posed it.
What equivalent means in this context is that if the Riemann Hypothesis is true, than the inequality above is also true, and if the inequality above is true, then the Riemann hypothesis is true.
For many problems one can easily guess the solution, without necessarily knowing the proof (like Fermat did). The answer is almost certainly negative, and a complete solution requires a very long and tedious proof that cannot be checked without a deep understanding of the field and many years of collective effort.
I am specifically interested in problems which can be understood by a child, have hard solutions which cannot be easily guessed, but are easily verified (ex. factoring the product of two large primes). It seems like these problems would make good candidates for one-way-functions, like the diophantine equations and ECC. But elliptic curve cryptology is too difficult to describe to a child.
You can even show how easy out would be for a simple curve where the left part is an easy equation.
The hard part is what makes an equation easy to solve. (And how some elliptic are broken.) It takes at least some high school math to hint at this.
> 74 = −284650292555885^3 + 66229832190556^3 + 283450105697727^3 [1] [2].
Fermat's last theorem is an example, because it can be shown to be equivalent to, among other things, a very deep statement about elliptic curves (yep, the same thing The Fine Answer is about!) that Andrew Wiles proved.
Does the fractional part of e^n tend to zero?
Take any positive whole number, n.
If it is even, divide it by 2. (n = 2n)
If it odd, multiply it by 3 and add one. (n = 3n + 1)
Repeat. If n = 1, stop.
Does this process always lead to 1? It seems to, but who knows.
Here's one that's genuinely easy:
x^3 + y^3 + z^3 = 29
So how about this one?
x^3 + y^3 + z^3 = 30
Not so easy.
But it does have a solution!
Here it is:
x = -283059965
y = -2218888517
z = 2220422932
This is a great visceral demonstration that Diophantene equations are hard: there can exist no algorithm that solves all of them. Diophantene equations, like the Halting Problem, are undecidable.
y = -390
z = 858
[ http://www.wolframalpha.com/input/?i=(702%2F(-390%2B858))%2B...) ]
Found the above solution using a hillclimbing algorithm.
> This solution is not easy to see by hand, but it’s also not hard to discover with some patience without all the machinery we are reviewing here. It’s the positive solutions that are the lair of dragons.
calc 2.12.4.1
> a=154476802108746166441951315019919837485664325669565431700026634898253202035277999
> b=36875131794129999827197811565225474825492979968971970996283137471637224634055579
> c=4373612677928697257861252602371390152816537558161613618621437993378423467772036
> a/(b+c) + b/(a+c) + c/(a+b) 4
> a/(b+c) ~3.74500615923925922050
> b/(a+c) ~0.23213745990937924275
> c/(a+b) ~0.02285638085136153676
(set-logic QF_UFNIA)
(declare-fun x () Int)
(declare-fun y () Int)
(declare-fun z () Int)
(define-fun left-side () Int
(+ (+
(* (* x (+ x z)) (+ x y))
(* (* y (+ y z)) (+ x y)))
(* (* z (+ y z)) (+ x z)))
)
(define-fun right-side () Int
(* (* (*
4
(+ y z))
(+ x z))
(+ x y))
)
; disallow division by zero
(assert (not (= 0 (+ y z))))
(assert (not (= 0 (+ x z))))
(assert (not (= 0 (+ x y))))
(assert (= left-side right-side))
(check-sat)
(get-model)
(assert (> z 0))
(assert (> x 0))
(assert (> y 0))
I am running this on my machine now. Will report back if it comes up with a solution.
((x + (((2*y*z) + (y*y) + (z*z) + (((z*z*z) - (z*y*y))/(x + y)))/(x + z)))/(y + z))
Edit: of course, this is not a solution. It's now just an example to others to beware of floating point errors.
I realize when you say "isn't quite 4" you mean the solution is not quite equal to 4, but it could also be incorrectly interpreted as a solution that is slightly less than 4.
167820976/41955243 = 4 + 4/41955243
... except that no, its new velocity would be 4m/s. The author, a physicist, ignored sig figs. 4 - 0.00006 = 4, when you're dealing with measurements :)
Still, I'm interested to hear why you think sig figs are high-school only when they're the standard way of writing in scientific notation. Do you genuinely think that the example I gave from the video is reasonable? It sounds like you're a theorist and have no experience in real-world application.
0, 109552575, 29354524 :)