One thing missing from the video is the proof that the tree of solutions generated at the end actually includes all possible solutions. (The proof doesn't actually appear to be too long so it could be a good follow-up video.)
Proof that all solutions can be reached from (1, 1, 1): Start with some solution (x, y, z), and let's assume all three are distinct. Now consider the quadratic f(t) = t^2 - t(3xy) + (x^2 + y^2) Note that z is a root. By Vieta's formulas, the other root z' satisfies z + z' = 3xy and zz' = x^2 + y^2. Thus since z is positive, z' is also positive, and, as in the video, we see that z' = 3xy - z gives another solution. Now, WLOG assume x > y, then take f(x) = 2x^2 + y^2 - 3x^2 y = x^2 ( 2 - 3y ) + y^2. Since y > 0, 2 - 3y 2 + a^2 = a x => 2 = a(x - a). So we see a = 1 or 2. If a = 1 then we get (1, 1, 1) and if a = 2 then we get (1, 1, 2). And from (1, 1, 2) we get to (1, 1, 1) by taking (x, y, 3xy - z).
Thus we see that starting from an arbitrary solution we eventually come to (1, 1, 1), and so these are all the solutions.
nice proof. small mistake: 2 + a^2 = a x should be 2 + a^2 = 3a x. also, I think it would be easier to understand if you said something like "WLOG assume z > x > y" because it then becomes obvious that the maximum decreases as min(z, z') = z' since otherwise z < x.
Thanks for your incredible proof of this.Beforehand I actually regard Vieta's Jumping only as an approach to proving something is impossible(no sol.).I didn't think of that it can also be used to prove an assumption can not hold true forever.
I’m wondering if there is a group of solutions that doesn’t fall into the group starting with (1,1,1). And how many independent groups are there (such that there is no tree of solutions that contains any 2 of the groups)?
Is there a proof that *all* solutions to the Markov Equation are reachable via this tree? Or is it an open question whether there are "extra" solutions? Also, one would have to show that the triples produced by this process are closed to the natural numbers, i.e. that we don't end up with negative (or zero) values at some step.
You can't have zero/negative solutions because the LHS is positive, so if you replace z with 3xy-z, then it must have the same sign as z to maintain the positive sign, which is positive.
@@deinauge7894 But you can easily prove that any solution with negative components comes from flipping two of the signs of a positive solution. So, it would be sufficient to ask if all of the positive solutions come from this method.
@@deinauge7894 Well, it's a diophantine equation, so we are restricting the domain of _acceptable_ solutions to the natural numbers right from the start. Which is why (0,0,0) is not considered a valid solution, although it satisfies the equation. For the same reason, when we say that the equation a^n+b^n=c^n has no solutions for n>2 (Fermat's Last Theorem), we specifically mean integer solutions (a,b,c), and we also require n to be an integer. Note that the trivial solution (0,0,0) is *also* excluded here.
@@andrewbuchanan1506 Precisely! Indeed, it might be easier to chart the Markov Tree, wherever it might lead, and just "prune" any non-acceptable solutions, and only ask whether all positive solutions are a _subset_ of it.
Well, what you can do is prove that if the equation has a solution at all then W is either 1 or 3. We can prove it like this: Suppose, wlog, that xWxy-z), a contradiction, so it must be that z'=(Wxy/2)^2 Taking the extremes of the inequalities and expanding a bit, we obtain: x^2+2y^2>=Wxy^2 (2) But we know that x^2
Michael, could you make a video on a similar equation? A long time ago I saw this one: x^2 + y^2 + z^2 + w^2 = xyzw, which actually has an infinite number of solutions, starting with (2, 2, 2, 2) and then performing a transformation of a generic solution of (x, y, z, w) to (x, y, z, xyz - w) can generate many more (if you change the order inside of the solution as well, due to symmetry).
@@jefftimer1752 Let's assume (x;y;z;w) and (x;y;z;v) are slutions with v != w then x^2+y^2+z^2+v^2 = xyzw-w^2 + v^2 = xyzv --> xyz(w-v)-(w+v)(w-v) = 0 --> xyz-w-v = 0 --> v = xyz-w
At 12:55 when you had switched the second and third terms, you had the resulting value as 9x^2-xz-y. Shouldn't the 3 have been distributed to the xz term as well resulting in 9x^2-3xz-y? Don't think it made a difference to your final calculations as you really didn't use that equation to work anything out but just wondering. This is getting too long so I think this is a good place to stop ✋️.
This was fun! I remember the Markov Equation from a graduate course I took in Diophantine equations. I always loved the tree bit. Actually, I have a different approach that can get you an infinite (but not exhaustive) number of solutions. I'll see if I can put a video together today or tomorrow for that.
Although, to be fair, Michael Penn's opening sentence was " ... to look at the Markov equation and to find positive integer solutions for this equation", which is rather more unequivocal.
@@taprusthe3rd496 That's actually not true, 0 is a Natural number in many contexts regardless of language. In fact if you are using a set theoretic definition of the Naturals then 0 is always a Natural number since the Naturals are defined as the cardinalities of finite sets and 0 is the cardinality of the empty set. Michael typically doesn't include 0 in the Naturals because he's coming from a number theory perspective and for whatever reason in Number Theory 0 is usually excluded (probably because of complications with prime factorization, etc.)
Another, possibly simpler way to see this: the 3xyz equation can be viewed as a 3 input AN, with the x, y, and z as inputs. if odd is high and even is low, then the output of the AND gate is the same as the expected output (parity on right hand side of the equation). However, no logic gate can be constructed for the 2xyz equation that works. If one of the variables is always 1, say z, then you get x*2 - 3xy + y*2 + 1 = 0. Taking y as a free variable, and setting y to every other Fibonacci value, seems to recursively generate x as the next every other Fibonacci value. This might relate to Mathologer video on the secrets of the number walls--needs more review.
At 8:00 we essentially have "And then an absolute miracle happens and we somehow guess the correct answer, which we will now verify." That hardly seems fair. How on Earth would someone guess that?
"Guess and check" is a fairly standard strategy in math. What likely happened historically for this problem is someone (Markov?) just substituted z -> (an arbitrary quadratic in x,y,z) and solved for the coefficients by grinding out algebra.
@@orionspur you can look at the equation as a quadratic in z. If z_1 is a solution, then there exists a second one that satisfies z_1+z_2=3xy, which is exactly the step (x,y,z)->(x,y,3xy-z). In olympiad mathematics this method is called vieta jumping
@@samuelmeyer1884 I wish Michael had included this reasoning for 3xy-z as part of the video, it's quite simple to explain and makes it much less magical.
I like the video, but I feel kind of cheated when it (at least to me) just leaves me with at least two big questions. 1) Will this method find all solutions? 2) Will it even find all solution of this form or will another start (not (1,1,1)) yield other solutions? For example (0,0,0) seems to be overlooked...
x, y, and z are symmetric, meaning you can swap the order of (x, y, z) to get other points like (z, y, x) with six total orderings, which with the identity proven from the video, gives another solution, and so on.
Something similar for fans of non-linear Diophantine equations: Find all positive integer sided cuboids with surface area equal to volume: i.e., find all natural numbers, l, m, n, st 2(l^2) + 2(m^2) + 2(n^2) = lmn
x^2+y^2+z^2=3xyz if (1,1,1) is a solution then equation x^2+1+1=3x*1*1, x^2-3x+2=0 has two solutions one is x=1 from (1,1,1) and second is x=2. (2,1,1) is a solution then equation 2^2+1+z^2=3*2*1*z, z^2-6z+5=0, (1,5) is solution (2,1,5) is a is solution
The equation can be written z^2-3xyz+x^2+y^2. The is a quadratic equation in z. In quadratic equations the sum of the 2 roots is equal to minus of the "coefficients" of the 2nd term, here: z_1+z_2=-(3xy)=3xy So if z_1 is one solution, then 3xy-z_1 is the other. It would have been good if this had been mentioned in the video.
It is important, to show, where w = 3xy-z comes from let's assume: (x;y;z) and (x;y;w) are solutions with w != z x^2+y^2+w^2 = 3xyz-z^2+w^2 = 3xyw -> 3xy(z-w)-(z+w)(z-w) = 0 -> 3xy - z -w =0 -> w = 3xy-z
Can you do a video on how rejecting the axiom of choice lets you construct an uncountable set out of a union of a countable number of countable sets? I think this would really help more people understand the axiom of choice on a deeper level.
I don’t know much about the set theory but I expect the axiom of choice could only ever construct new things, it doesn’t stop you from constructing objects. If you can form such an uncountable set from ZF then you could also do this in ZFC. But in ZFC it is provable that the union is countable, so that’d be a contradiction! (assuming all is consistent and well). It’s more that, in ZF, it’s not possible to prove the countable union of countable sets is countable (I think). Probably one of those: “it is consistent with …”-type unproveables
@@him21016 Not quite so - you can prove that 'in so-and-so model of ZF minus Choice there is no bijection between (this particular union of countable sets) and the natural numbers'. So the set is by definition uncountable. It won't be _every_ model of ZF minus Choice that satisfies this, but there are particular models that do.
That said, the amount of scaffolding necessary to show what models of ZF are, how to construct them, how to build ones without choice, etc. is sufficiently heavy that I'm not sure it would make a great video for this channel. I'd love to be proven wrong, though!
@@bjornfeuerbacher5514 The generalization I proposed does not say anything about natural numbers. This covers all whole number solutions. I haven't check the complex numbers where there might be a solution
@@bjornfeuerbacher5514 There is a solution further down the page. The only cases where there are solutions to this equation are where the right hand coefficient is 1 or 3. This makes the Markov equation much more intseresting
Can also try this. Write the original equation (x^2 + y^2 + z^2 ...) and then write it with z-->a, giving x^2+y^2+a^2=3xya. Subtract the equations. Factor the a^2 - z^2 term on the left hand side. In order for this to be a new solution, (a-z) cannot be zero, so divide both sides by (a-z) to get a+z=3xy. In order for (x,y,a) to be a new solution, we need a=3xy-z.
That's because we know that the right side of the equation is a multiple of 4 and so is 4x1^2 on the left side so it means that y^2 + z^2 must also be factorable by 4, hence it is congruent to 0(mod 4) since it is a multiple of 4.
I found the questions fascinating. But I also found the video rather frustrating because these very non-obvious solutions came out of thin air. Do you just happen to think of these solutions? I would love to see some videos that are not distilled and filtered - i.e. where you work through a problem you don't know already how to solve. I think that kind of video would be very instructive even if it turns out long.
Because: ab (mod n) == (a mod n)(b mod n) mod n. That is, if you multiply two numbers mod 4, that is the same result as taking the individual numbers mod 4 before multiplying, then multiplying them (and then taking mod 4). Since every integer will be congruent to 0, 1, 2, or 3, when taken mod 4, you only need to check up to that far.
i am wondering that can we find a constant replacing 2 and 3 which gives us that there is only one solution yo yhe equation and is it between 2 and 3 and another problem before 2 or after 3 does the pattern keep it like that for example is 4 has infinetly many solution too or before 2 are there still no solution
We know y^2+z^2 is divisible by 4. Clearly it is not possible for just one of y, z to be even and the other odd, since then y^2+z^2 would be an odd number plus an even one, so would also be odd. But it is also not possible for both y and z to be odd: for if they were both odd then y^2+z^2 would be divisible by 2 but not 4. Hence both y and z are even. (This isn't quite how Michael Penn phrased it in the video but the logic of the argument is just the same. I have just formulated it in a way which does not make explicit reference to arithmetic modulo 4)
Interesting, but why just integer solutions. Real solutions are, I think, intersections of 3d spheres and hyperbole. Interesting graphic instructional opportunity?
Guys, can someone help me to find a video? I forgot to save it for a later use. The video I'm looking for is probably not by this channel. It is about the integral from 0 to 1 of arcsin(x) ln(x)/sqrt(x)
this is not the exact same, but you might not be recalling correctly, Michael made a video called "So many Calculus tricks in one integral." The bounds and integral is very similar, if its not, no worries, but I would reccomend checking out none the less :D
@@someone-ol8wc I think i recall the integral perfectly fine, since I did write it down. Thanks for the suggestion anyway, at some point I'll watch it like everything else.
True, since the only solutions, as he says, are numbers who can be divided by all powers of two. 0's absorbant nature fits this criteria. Hence, the only solution is (0, 0, 0).
Taking a cue from Samuel Meyer in the previous thread, you can re-write the equation as (z)² - 3xy(z) + (x² + y²). That is a quadratic in z and you could call the roots z1 and z2. We know that the sum of the roots of a quadratic is -b/a, so we know that z1 + z2 = 3xy. That means that if z1 is a solution (for a given x, y), the only other solution is z2 = 3xy - z1. In other words, if {x, y, z1} is a solution, the only other solution is {x, y, z2} = {x, y, (3xy - z1)}. Of course, that doesn't prove that other solutions not derived from {1, 1, 1} don't exist, but it does prove that all of the solutions derived from {1, 1, 1} have to take that form. I'll have a think about the first bit.
One thing missing from the video is the proof that the tree of solutions generated at the end actually includes all possible solutions. (The proof doesn't actually appear to be too long so it could be a good follow-up video.)
Thank you, I was waiting for him to mention whether this process generates all possible solutions or merely an infinite number of them.
Proof that all solutions can be reached from (1, 1, 1):
Start with some solution (x, y, z), and let's assume all three are distinct.
Now consider the quadratic f(t) = t^2 - t(3xy) + (x^2 + y^2)
Note that z is a root. By Vieta's formulas, the other root z' satisfies z + z' = 3xy and zz' = x^2 + y^2. Thus since z is positive, z' is also positive, and, as in the video, we see that z' = 3xy - z gives another solution.
Now, WLOG assume x > y, then take f(x) = 2x^2 + y^2 - 3x^2 y = x^2 ( 2 - 3y ) + y^2. Since y > 0, 2 - 3y 2 + a^2 = a x => 2 = a(x - a). So we see a = 1 or 2. If a = 1 then we get (1, 1, 1) and if a = 2 then we get (1, 1, 2). And from (1, 1, 2) we get to (1, 1, 1) by taking (x, y, 3xy - z).
Thus we see that starting from an arbitrary solution we eventually come to (1, 1, 1), and so these are all the solutions.
nice proof. small mistake: 2 + a^2 = a x should be 2 + a^2 = 3a x. also, I think it would be easier to understand if you said something like "WLOG assume z > x > y" because it then becomes obvious that the maximum decreases as min(z, z') = z' since otherwise z < x.
Yes so can we solve WITHOIT DAMN MOD AT ALL..Since I don't see most ppl.thinking of mod just factor out the 4
Thanks for your incredible proof of this.Beforehand I actually regard Vieta's Jumping only as an approach to proving something is impossible(no sol.).I didn't think of that it can also be used to prove an assumption can not hold true forever.
Beautiful
No motivation was given for where the mysterious claim came from. Is there a way to intuit the claim?
I’m wondering if there is a group of solutions that doesn’t fall into the group starting with (1,1,1). And how many independent groups are there (such that there is no tree of solutions that contains any 2 of the groups)?
Every triplet below (3000,3000,3000) that matches the equation x² + y² + z² = 3 x y z, are also generated from these generation rules.
according to the mathworld page for markov numbers this process generates all the solutions.
@@dernett To be pedantic, that can't be 100% true because the trivial solution (0, 0, 0) exists but isn't generated by that process.
@@QuantumHistorian The problem is restricted to the natural numbers.
Probably not, but it'not proven in the video. Not even referred.
Is there a proof that *all* solutions to the Markov Equation are reachable via this tree? Or is it an open question whether there are "extra" solutions?
Also, one would have to show that the triples produced by this process are closed to the natural numbers, i.e. that we don't end up with negative (or zero) values at some step.
You can't have zero/negative solutions because the LHS is positive, so if you replace z with 3xy-z, then it must have the same sign as z to maintain the positive sign, which is positive.
and that shows that not all sollutions are reachable through this process. because (1,-1,-1) is a solution ;-)
@@deinauge7894 But you can easily prove that any solution with negative components comes from flipping two of the signs of a positive solution. So, it would be sufficient to ask if all of the positive solutions come from this method.
@@deinauge7894
Well, it's a diophantine equation, so we are restricting the domain of _acceptable_ solutions to the natural numbers right from the start. Which is why (0,0,0) is not considered a valid solution, although it satisfies the equation.
For the same reason, when we say that the equation a^n+b^n=c^n has no solutions for n>2 (Fermat's Last Theorem), we specifically mean integer solutions (a,b,c), and we also require n to be an integer. Note that the trivial solution (0,0,0) is *also* excluded here.
@@andrewbuchanan1506
Precisely! Indeed, it might be easier to chart the Markov Tree, wherever it might lead, and just "prune" any non-acceptable solutions, and only ask whether all positive solutions are a _subset_ of it.
15:37
Maybe a more interesting question would be to find all values of W in Natural numbers such that x^2 + y^2 + z^2 = Wxyz has exactly one solution.
Well, what you can do is prove that if the equation has a solution at all then W is either 1 or 3. We can prove it like this:
Suppose, wlog, that xWxy-z), a contradiction, so it must be that z'=(Wxy/2)^2
Taking the extremes of the inequalities and expanding a bit, we obtain:
x^2+2y^2>=Wxy^2 (2)
But we know that x^2
Nice problem for Lagrangian
Michael, could you make a video on a similar equation? A long time ago I saw this one: x^2 + y^2 + z^2 + w^2 = xyzw, which actually has an infinite number of solutions, starting with (2, 2, 2, 2) and then performing a transformation of a generic solution of (x, y, z, w) to (x, y, z, xyz - w) can generate many more (if you change the order inside of the solution as well, due to symmetry).
But does this process generate ALL possible solutions I wonder?
How does one come up with the generak solution with (xyz-w) as z -component
@@jefftimer1752 Let's assume (x;y;z;w) and (x;y;z;v) are slutions with v != w
then x^2+y^2+z^2+v^2 = xyzw-w^2 + v^2 = xyzv
--> xyz(w-v)-(w+v)(w-v) = 0
--> xyz-w-v = 0
--> v = xyz-w
Markov chain has nothing to do with Markov equation but Markov himself.
This startled me too. Why the title is "The first Markov chain"?
"that was something of a warm up" hahaha you sound just like my favourite math teacher at school :-)
huh why is this particular equation named? must be something important
At 12:55 when you had switched the second and third terms, you had the resulting value as 9x^2-xz-y. Shouldn't the 3 have been distributed to the xz term as well resulting in 9x^2-3xz-y? Don't think it made a difference to your final calculations as you really didn't use that equation to work anything out but just wondering. This is getting too long so I think this is a good place to stop ✋️.
You are correct.
This was fun! I remember the Markov Equation from a graduate course I took in Diophantine equations. I always loved the tree bit. Actually, I have a different approach that can get you an infinite (but not exhaustive) number of solutions. I'll see if I can put a video together today or tomorrow for that.
Are all solutions Fibbonacci numbers? If so then why?
x^3 + y^3 = 3xy is called the folium of Descartes
you forgot about the x,y,z=0 answer
Thank you, great video.
In the anglosphere, it is the convention that 0 isnt a member of N
@@taprusthe3rd496 ahh, ye, this... thank you.
Although, to be fair, Michael Penn's opening sentence was " ... to look at the Markov equation and to find positive integer solutions for this equation", which is rather more unequivocal.
@@taprusthe3rd496 That's actually not true, 0 is a Natural number in many contexts regardless of language. In fact if you are using a set theoretic definition of the Naturals then 0 is always a Natural number since the Naturals are defined as the cardinalities of finite sets and 0 is the cardinality of the empty set.
Michael typically doesn't include 0 in the Naturals because he's coming from a number theory perspective and for whatever reason in Number Theory 0 is usually excluded (probably because of complications with prime factorization, etc.)
The "warm up" problem isn't a warm up at all; the Markov equation is a totally different problem 😅
Another, possibly simpler way to see this: the 3xyz equation can be viewed as a 3 input AN, with the x, y, and z as inputs. if odd is high and even is low, then the output of the AND gate is the same as the expected output (parity on right hand side of the equation). However, no logic gate can be constructed for the 2xyz equation that works. If one of the variables is always 1, say z, then you get x*2 - 3xy + y*2 + 1 = 0. Taking y as a free variable, and setting y to every other Fibonacci value, seems to recursively generate x as the next every other Fibonacci value. This might relate to Mathologer video on the secrets of the number walls--needs more review.
Funny construction ! But, are there other solutions than those given starrting with the triple (1, 1, 1) ?
If not, can we prrove it ?
good question.
At 8:00 we essentially have "And then an absolute miracle happens and we somehow guess the correct answer, which we will now verify." That hardly seems fair. How on Earth would someone guess that?
"Guess and check" is a fairly standard strategy in math. What likely happened historically for this problem is someone (Markov?) just substituted z -> (an arbitrary quadratic in x,y,z) and solved for the coefficients by grinding out algebra.
@@leif_p As a math PhD I find that a little unsettling. We also completely glossed over whether there might be other similar classes of solutions.
@@orionspur you can look at the equation as a quadratic in z. If z_1 is a solution, then there exists a second one that satisfies z_1+z_2=3xy, which is exactly the step (x,y,z)->(x,y,3xy-z). In olympiad mathematics this method is called vieta jumping
@@samuelmeyer1884 oh yeah, like the infamous question 6 😂
@@samuelmeyer1884 I wish Michael had included this reasoning for 3xy-z as part of the video, it's quite simple to explain and makes it much less magical.
I like the video, but I feel kind of cheated when it (at least to me) just leaves me with at least two big questions.
1) Will this method find all solutions?
2) Will it even find all solution of this form or will another start (not (1,1,1)) yield other solutions? For example (0,0,0) seems to be overlooked...
He says at the beginning we are only considering positive integer solutions
@@TheEternalVortex42 Still most of my question is unanswered.
x, y, and z are symmetric, meaning you can swap the order of (x, y, z) to get other points like (z, y, x) with six total orderings, which with the identity proven from the video, gives another solution, and so on.
Not your everyday Markov Chain
Yes, the title ".... Markov Chain..." seems strange
Something similar for fans of non-linear Diophantine equations:
Find all positive integer sided cuboids with surface area equal to volume: i.e.,
find all natural numbers, l, m, n, st 2(l^2) + 2(m^2) + 2(n^2) = lmn
😱(duh!) I meant all natural numbers l, m, n st 2lm + 2ln + 2mn = lmn
@@stephenhamer8192 2lm+2ln+2mn=lmn
2lm+2ln=(l-2)mn
2l(l-2)m+2l(l-2)n=(l-2)²mn
((l-2)m-2l)((l-2)n-2l)=4l²
Say (l-2)m-2l=1,i.e.
(l-2)(m-2)=5,
l=3, m=7, n-6=36, n=42
If m=n
(l-2)m-2l=2l , n=m=lk
(l-2)k=4 , l=3, k=4, m=12=n
l=4, k=2, m=8=n
l=6,k=1, m=6=n
Those are obvious solutions.
x^2+y^2+z^2=3xyz if (1,1,1) is a solution then equation x^2+1+1=3x*1*1, x^2-3x+2=0 has two solutions one is x=1 from (1,1,1) and second is x=2. (2,1,1) is a solution then equation 2^2+1+z^2=3*2*1*z, z^2-6z+5=0, (1,5) is solution (2,1,5) is a is solution
The first part is a nice illustration of proof by infinite descent.
Very nice video once again, thank you very much.
What is the motivation to consider (x,y,3xy-z)?
The equation can be written z^2-3xyz+x^2+y^2. The is a quadratic equation in z.
In quadratic equations the sum of the 2 roots is equal to minus of the "coefficients" of the 2nd term, here: z_1+z_2=-(3xy)=3xy
So if z_1 is one solution, then 3xy-z_1 is the other.
It would have been good if this had been mentioned in the video.
It is important, to show, where w = 3xy-z comes from
let's assume: (x;y;z) and (x;y;w) are solutions with w != z
x^2+y^2+w^2 = 3xyz-z^2+w^2 = 3xyw
-> 3xy(z-w)-(z+w)(z-w) = 0
-> 3xy - z -w =0
-> w = 3xy-z
Shouldn't that be "9x^2y-3xz-y" instead of "9x^2y-xz-y" ? (coeff of 3)
Markov? More like “Magnificent!”
Can you do a video on how rejecting the axiom of choice lets you construct an uncountable set out of a union of a countable number of countable sets?
I think this would really help more people understand the axiom of choice on a deeper level.
I don’t know much about the set theory but I expect the axiom of choice could only ever construct new things, it doesn’t stop you from constructing objects. If you can form such an uncountable set from ZF then you could also do this in ZFC. But in ZFC it is provable that the union is countable, so that’d be a contradiction! (assuming all is consistent and well). It’s more that, in ZF, it’s not possible to prove the countable union of countable sets is countable (I think). Probably one of those: “it is consistent with …”-type unproveables
@@him21016 Not quite so - you can prove that 'in so-and-so model of ZF minus Choice there is no bijection between (this particular union of countable sets) and the natural numbers'. So the set is by definition uncountable. It won't be _every_ model of ZF minus Choice that satisfies this, but there are particular models that do.
That said, the amount of scaffolding necessary to show what models of ZF are, how to construct them, how to build ones without choice, etc. is sufficiently heavy that I'm not sure it would make a great video for this channel. I'd love to be proven wrong, though!
6:48 This _does_ make sense if you allow 0 to be a natural number.
You can get a little more from the.starting equation. If the coefficient on the right hand side is even there are no solutions other than (0,0,0)
The problem says that x,y,z have to be natural numbers. Which usually excludes zero.
@@bjornfeuerbacher5514 The generalization I proposed does not say anything about natural numbers. This covers all whole number solutions. I haven't check the complex numbers where there might be a solution
@@bjornfeuerbacher5514 There is a solution further down the page. The only cases where there are solutions to this equation are where the right hand coefficient is 1 or 3. This makes the Markov equation much more intseresting
@@davidmitchell3881 Yes, I saw that comment, too - thanks.
Its amazing how threatening "I think that's a good place to stop" sounds. Camera operator is going to die if there's one more frame.
Where does 3xy--z come from?
This is the reason I didn't like this video. Instead of solving the problem he proved a "claim" that comes out of nowhere.
From expressing the equation as a quadratic in z and solving for z. If z0 is a root for a fixed x and y, the other root is (3xy - z0).
Can also try this. Write the original equation (x^2 + y^2 + z^2 ...) and then write it with z-->a, giving x^2+y^2+a^2=3xya. Subtract the equations. Factor the a^2 - z^2 term on the left hand side. In order for this to be a new solution, (a-z) cannot be zero, so divide both sides by (a-z) to get a+z=3xy. In order for (x,y,a) to be a new solution, we need a=3xy-z.
As a former physics maniac, I hate equations with unequal dimensions on both sides.
My equations are also full of "oops" 😔
Markov is cool
Could this be generalized into something like x^2 + y^2 + z^2 = (2n+1)xyz for all integers n greater than zero?
I'm not too familar with congruency; At around 3:35, why does it follow that y^2 + z^2 is congruent to 0 mod 4 from the previous equation?
That's because we know that the right side of the equation is a multiple of 4 and so is 4x1^2 on the left side so it means that y^2 + z^2 must also be factorable by 4, hence it is congruent to 0(mod 4) since it is a multiple of 4.
I found the questions fascinating. But I also found the video rather frustrating because these very non-obvious solutions came out of thin air. Do you just happen to think of these solutions? I would love to see some videos that are not distilled and filtered - i.e. where you work through a problem you don't know already how to solve. I think that kind of video would be very instructive even if it turns out long.
What did any of this have to do with Markov chains?
Zero is not a natural number?
I wish I was even half as clever as this guy.
You should make a video on that Fibonacci structure!
for the first problem there still is the solution 0²+0²+0² = 2×0×0×0
I personally see a connection towards primes? ...
4:00 Can someone explain me why he only checked until 3² (mod 4)
Because: ab (mod n) == (a mod n)(b mod n) mod n. That is, if you multiply two numbers mod 4, that is the same result as taking the individual numbers mod 4 before multiplying, then multiplying them (and then taking mod 4). Since every integer will be congruent to 0, 1, 2, or 3, when taken mod 4, you only need to check up to that far.
@@HeavyMetalMouse Now I understand it, thanks for the explanation, it couldn't be better
Wow, thanks.
i am wondering that can we find a constant replacing 2 and 3 which gives us that there is only one solution yo yhe equation and is it between 2 and 3 and another problem before 2 or after 3 does the pattern keep it like that for example is 4 has infinetly many solution too or before 2 are there still no solution
I don't understand why y and z are also even. video time around 4:30
We know y^2+z^2 is divisible by 4. Clearly it is not possible for just one of y, z to be even and the other odd, since then y^2+z^2 would be an odd number plus an even one, so would also be odd. But it is also not possible for both y and z to be odd: for if they were both odd then y^2+z^2 would be divisible by 2 but not 4. Hence both y and z are even.
(This isn't quite how Michael Penn phrased it in the video but the logic of the argument is just the same. I have just formulated it in a way which does not make explicit reference to arithmetic modulo 4)
I can imagine a little recursive lambda to generate them.
8:10 It is fun, because it has zero solutions by the meaning of x=y=z=0 😀
So we were dealing with Markov and then Fibonnaci stepped in...
Hi Dr. Penn!
Okay now do complex solutions
Interesting, but why just integer solutions. Real solutions are, I think, intersections of 3d spheres and hyperbole. Interesting graphic instructional opportunity?
man I gotta shadow switch my life around
3 inputs AND gate.
the thumbnail doesn't match with the video
Guys, can someone help me to find a video? I forgot to save it for a later use. The video I'm looking for is probably not by this channel.
It is about the integral from 0 to 1 of arcsin(x) ln(x)/sqrt(x)
The maths sorcerer channel
@@yourlyrics1682 I recognize the thumbnail style, but I can't find the video. Thanks, I'll look later
this is not the exact same, but you might not be recalling correctly, Michael made a video called "So many Calculus tricks in one integral." The bounds and integral is very similar, if its not, no worries, but I would reccomend checking out none the less :D
Closest I could find: ua-cam.com/video/LEmYPenxIZY/v-deo.html
@@someone-ol8wc I think i recall the integral perfectly fine, since I did write it down. Thanks for the suggestion anyway, at some point I'll watch it like everything else.
for the x^2+y^2+z^2 = 2*x*y*z : you state that there is no solution. While if I plug in (0,0,0), it fits. So there is at least one solution
True, since the only solutions, as he says, are numbers who can be divided by all powers of two. 0's absorbant nature fits this criteria. Hence, the only solution is (0, 0, 0).
{x,y,z € Z: Z>0}
@@baronvonbeandip look at the middle right part of the board. it states x,y,z element of N. As fas as I know 0 is part of N
@@tomwambecq4683 natural numbers in this case are excluding 0, and as such it is not a solution.
@@tomwambecq4683it’s convention if it is in there or not. I think it’s usually not
Man, every recurrence relation's got some Fibonacci thing goin on under the hood, huh?
Gosh I need to learn logic....
have you proved for markov equasion that these are the only solutions?
Taking a cue from Samuel Meyer in the previous thread, you can re-write the equation as (z)² - 3xy(z) + (x² + y²). That is a quadratic in z and you could call the roots z1 and z2. We know that the sum of the roots of a quadratic is -b/a, so we know that z1 + z2 = 3xy. That means that if z1 is a solution (for a given x, y), the only other solution is z2 = 3xy - z1.
In other words, if {x, y, z1} is a solution, the only other solution is {x, y, z2} = {x, y, (3xy - z1)}. Of course, that doesn't prove that other solutions not derived from {1, 1, 1} don't exist, but it does prove that all of the solutions derived from {1, 1, 1} have to take that form. I'll have a think about the first bit.
ayyy who else been to the moroccan mo?
Me
Lmaroc daymne 7adr tbarkllah
nice guys :D
Wrong thumbnail
the moroccan problem has 1 solution, x=y=z=0
The word markov scares me, as I couldn't understand markov chains🥲