an infinitely long solution.

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[@FreeAsInFreeBeer]

At 9:45 I think you reversed the indexes. It should be x_n = 2 x_(n+1) otherwise the x_(n+1) < x_n inequality at 10:22 doesn't hold.

[@giovanicampos4120]

Exactly. If there was x_{n+1} = 2x_n then the solution would be geometric sequences for x, y and z with ratio 2.

[@martintoilet5887]

I had the same question.

[@thesecondderivative8967]

I keep seeing this in the comments and I'm confused. When he showed a, b and c were even, he then represented them individually as a= 2x_1, b=2y_1 etc. Plugging this back in the equation, he discovered that x_1 and the rest were also even. Therefore x_1 = 2x_2. And like that, he formed an infinite sequence.

[@tonysplodge44]

I didn't realise until halfway through that this was going to be proof by infinite descent....which I first learned at school in the proof that Fermat doesn't work for n=4.... but this video was much better, and Prof Penn's had a haircut as well....

[@adrien7933]

I guessed it was gonna be by infinite descent just with the title of the video😅

[@axelperezmachado3500]

wait what! Learned that at school?!

[@vanandreyambalong628]

woah good place to stop is nowhere

[@musik350]

It'd be hard to stop if you have an infinite number of solutions!

[@pavlos712]

Nicely observed.

[@wyattstevens8574]

Literally!

[@demenion3521]

the cases at the start could have been compressed a bit by noticing that a²b² can only be either 0 or 1 (mod 4). therefore we have two cases: if a²b² is congruent 0, then a,b,c all need to be even since no other combinations or three numbers in {0,1} could add up to 0 (or other multiples of 4). in the second case a²b² is 1 (mod 4) and therefore exactly one of the three terms on the LHS needs to be odd, but we need to have both a and b odd to get a²b² being odd in the first place which is a contradiction. the rest of the argument is the same as in the video

[@afignisfirer4675]

in fact, if a²b² = 1 mod(4), both "b" and "a" need to be odd, if one of them is even this product will be divisible by 4, which implies that a²b² = 0 mod(4)

[@pratikmaity4315]

As Michael missed the case c=0 but it is easy to see that when c=0 then there will be no solution. Because then a²+b²=a²b².WLOG let a²≥b² if b²>2 then a²b²>2a²≥a²+b² which is a contradiction. So b²=1 which implies a²+1=a² which is again a contradiction. So no solution exists. One more method to prove the same thing goes as followes : since a²+b²=a²b² this shows that a and b are both even by mod4 argument. So taking a=2a_1 and b=2b_1 we get 4a_1²+4b_1²=16a_1²b_1². Or a_1²+b_1²=4a_1²b_1². Now we can again see that a_1 and b_1 are even and we can see that this is infinite descent again which shows that a=b=0 but Michael pointed this triple already. So c≠0. One more method to prove the same thing goes as follows : since c=0 so a²+b²=a²b². So (a²-1)(b²-1)=1 so from here also get that a²+1 is 1 or -1 and b²+1 is also 1 or -1. So we get again that a=b=0 and we are done!!

[@gaufqwi]

At the end your sequence indices are reversed. Should be x_n = 2x_{n+1}, etc.

[@blackkk07]

Logically you did not tells about those cases about SOME of {a, b, c} is zero. For example, how about a, b > 0 and c = 0? Any way this is just a small mistake, because these missing cases can easily merge with case 2. Nice video!

[@Bodyknock]

For case 2 just assume a>0 and even. This leads to the same infinite descent for a as in the video, as thus a>0 is impossible. Likewise you can show that b>0 and b even leads to an infinite descent for b. Which means a=b=0, and thus c^2 = 0 so c=0 as well.

[@DitDede]

The logic works for all nonnegative a,b,c. You just need one of them to be positive to guarantee a sequence of strictly decreasing natural numbers. Basically, you can phrase it as a proof by contradiction by extending the claim to: The only solution to a^2+b^2+c^2=4^k*a^2*b^2 is a=b=c=0. If there's a nontrivial solution, take a smallest such solution. Smallest in the sense that a,b,c0. Then you get another nontrivial solution with a,b,c

[@megauser8512]

If c = 0, then c is congruent to 0 mod 4, which is taken care of in the video by the infinitely decreasing sequence of even numbers.

[@billh17]

It is clear that a = 0 or b = 0 implies a = b = c = 0. Consider the case where c = 0 (and a > 0 and b > 0). Then, a^2 + b^2 = a^2 * b^2. Thus, a divides b and b divides a. Thus, a = b. Thus, 2 * a^2 = a^4. Hence, 2 = a^2. Contradiction.

[@oida10000]

If a, b>0 and c=0 then you have a^2+b^2=a^2b^2 which implies b^2=a^2/(a^2-1)=a^2/((a+1)(a-1) ==> b=±a/sqrt(a^2-1). But now you need a^2-1 to be a perfect square and its root a factor of a itself. It is obvious that there aren't that many integers like this. Aside from 1 no perfect square is just one more than a smaller perfect square.

[@noumanegaou3227]

We can preuve by introduction for all n positive integer 2^n | a, b, c This Contradiction We must check case when c = 0

[@jimskea224]

This has a neat geometrical interpretation. It's saying that the length of a nonzero line segment OP from the origin to a point P is never numerically equal to the area of the rectangle with diagonal OP projected onto the plane of two coordinates.

[@kruksog]

Love it. Thank you for pointing that out.

[@moonlightcocktail]

Yes, if the lengths of the sides are all integers. If not we could have a = 3 b = 3 c = 3rt7 for instance

[@jimskea224]

@@moonlightcocktail Yes, of course.

[@hassanalihusseini1717]

Thank you Prof. Penn! I always have problems when it comes to number theory used in proofs, but this one you explained so clearly that I could follow without difficulties!

[@goodplacetostop2973]

10:51 Well, that’s a loss for me today 😂 Anyway, I’ve uploded some homeworks on my channel. One homework will be uploaded every odd day for the next couple of weeks.

[@user-sh1ce3yw9f]

T brawl is better

[@someperson188]

Suppose (*) a² + b² + c² = (2^n)a²b², where a, b, c, n (n 1) are non-negative integers. Calculating Mod 4, we show that a, b, and c must be even, by considering the cases n = 0 and n >= 2. Therefore , (a/2)² + (b/2)² + (c/2)² = (2^(n+2))(a/2)²(b/2)². Thus, if (*) had a solution for which a > 0, it would have a solution for which a was smaller and positive. The method of descent then proves that (*) has no solution with a > 0. Hence a = 0, which easily implies b = c = 0.

[@wojteksocha2002]

Another slightly different way is to write a, b, c as a=2^k*x, b=2^l*y, c=2^m*z, where x, y, z are odd numbers. Since klm > 0, we can divide both sides by 2^min{k, l, m} and if 2 of the number k, l, m are not equal to min, then notice that one side of the equation is odd and one is even. But if 2 of the numbers k, l, m are equal to min, then after diving both sides by 2^min{k,l,m} we see that right side is divisible by 4 and the left one is not.

[@saiputcha1730]

But both sides are still even, right?

[@a_llama]

if 2 of k,l,m were equal to the min, how would that work out?

[@wojteksocha2002]

@@a_llama ok, then is not working. But in all other cases it's ok. I will quickly edit my comment and then everything should be fine.

[@manucitomx]

I loved this! Thank you, professor!

[@vh73sy]

In the complex domain C, an easy solution by inspection is a= 1, c= i or a=1, c= -i for any value for b, where i=sqrt(-1)

[@trueriver1950]

The easy soln is the only solution; but proving that could take you forever... In computing terms you have to spot that you have entered an infinite loop and step back from that loop to apply that all important second fact

[@ZXLegend1]

I have a very fast end to case 2 after one figures out those earlier conditions The equation is homogeneous so if a solution exists then we can also find one with gcd(a, b, c) = 1 But those conditions before the construction show that 2|gcd(a,b,c), contradiction

[@cbbuntz]

This reminds me of the types of things that pop out of Vieta's formulas. I still haven't really figured out how to work with them how how they really make solving polynomials any easier because it just seems like they add to the complexity rather than reducing it. But anyway, Vieta's formulas are kind of interesting. I was long aware of the fact that the second coefficient of a monic polynomial is the sum of the negative roots, and the last coefficient was the product of the negative roots. Vieta's formulas will tell you how the other coefficients are related, and it makes a light go off in your head saying "ohhh so that's why binomial coefficients are also n choose k".

[@littlefermat]

My (wrong) Solution:😂 We want T=a^2b^2-a^2-b^2 to be a perfect square. Clearly if a,b are even solutions then a/2,b/2 are solutions as well. So let us keep dividing both a,b by twos until we can't. Then WLOG a is even and b is odd So T=0*1-0-1=-1 (mod4) which can't be a perfect square. Or a,b are odd, so T=1*1-1-1=-1 which again can't be a perfect square Hence the solution is (0,0) and so we are done!

[@xc0xupx]

"Clearly if a,b are even solutions then a/2,b/2 are solutions as well." Good way of thinking, but it's not true, because T is not a homogeneous polynomial.

[@littlefermat]

@@xc0xupx Oops! My mistake 😅 So This is a one way for how not to solve the problem😂

[@gasun1274]

why would it be clear that 2 | a, b implies that them being halved also makes T a perfect square?

[@brussbass]

Alternatively, at the point it goes like X1 = 2 * X2, may be we can use mathematical induction, which will lead to this that X1 does not have 1 in its factorisation. Which is non-sense ;)

[@SlidellRobotics]

Actually, that a>0, b>0, c>0 was never used to eliminate case 1, nor to determine that all three must be even for case 2. After that, all that really matters are the absolute values anyway, apart from determining any symmetric cases.

[@dimy931]

I prefer writing this as let's assume a = 2^i x , b = 2^j y, c = 2^k z, where x,y,z are odd. And get a contradiction that way to the fact x,y and z are odd. I find the infinite descent proof arguments are always much theoretically murky and they can always be recast into normal proof by contradiction

[@ARKGAMING]

10 minutes and 53 seconds is infinitely long? I mean I guess if you use Zeno's paradox but...

[@jaimeduncan6167]

IT's infinite because of the solution, I believe

[@ARKGAMING]

@@jaimeduncan6167 yeah....it was a joke....

[@jaimeduncan6167]

@@ARKGAMING 🤣 English is not my first language, and did not see it.

[@Pablo360able]

In the final step, you said x_n+1=2x_n, which is the opposite of what you want. That would be an infinite sequence of *increasing* natural numbers, which can totally exist. Case in point: The natural numbers in the standard order. (Also, some subsequence of every infinite sequence of natural numbers.)

[@pavlos712]

We love you Michael. You increase my interest in mathematics!

[@jamesfortune243]

Nicely done and easy to understand.

[@leif1075]

Wjat about case 3 when both a and b are even but c is odd .he didnt rule it out explicitly..though i can see because on the LSH you would have an odd number and on the right an even..

[@yuseifudo6075]

He explicitly said it in the video. If a is even then b²+c²=0(mod 4) which is only possible when b and c are both even.

[@noumanegaou3227]

We X(n+1) = 2X(n)?

[@leetf8522]

I believe he meant to write X(n) = 2X(n+1), thus the sequence is always decreasing (which was shown to be impossible).

[@noumanegaou3227]

@@leetf8522 ok thanks you

[@Happy_Abe]

What about when C=0?

[@Happy_Abe]

I know same logic as case 2 applies, just saying we should include that case there too

[@Bodyknock]

Really he could have simply assumed a>0 for case 2 and done the exact same steps and you still get the same infinite descent with respect to a. Likewise the proof that b>0 leads to an infinite descent for b the same way. Therefore a=b=0, which means c^2 =0 and thus c=0 .

[@pratikmaity4315]

It's easy to solve when c=0. When c=0 then our equation becomes a²+b²=a²b². WLOG let a²≥b². If b²>2 then a²b²>2a²≥a²+b² which is a contradiction. So b²=1 and we get a²+1=a² which is a contradiction so c cannot be 0.

[@aaronsmith6632]

Cool proof!

[@lisandro73]

I am sorry but if Xn+1 = 2Xn that means that the sequence is increasing not decreasing, or am I wrong?

[@abusoumaya8469]

yes it should increase but in that sequence Xn+1 is less than Xn so there is contradiction

[@orenfivel6247]

what about a,b>0 & c=0

[@crehenge2386]

can you have an infinite long solution without an explicit formula or pattern?

[@tnekkc]

What does "mod" mean?

[@milanstevic8424]

modular arithmetic en.wikipedia.org/wiki/Modular_arithmetic

[@blueblackpenn6368]

That is good place to view

[@RaffoPhantom]

a=0, b=1, c=i

[@TwilightBrawl59]

10:51

[@user-sh1ce3yw9f]

How

[@jazzjohn2]

The statement in the graphic is not the problem solved.

[@MichaelPennMath]

oops, my thumbnail making time is just after i wake up and I am often still half asleep! I'll fix it now

[@jazzjohn2]

@@MichaelPennMath that explains it. Thanks!

[@rialtho_the_magnificent]

that is a bit of an anti-climax

[@dodokgp]

a sad ending :(

[@ayoubabid8783]

It's easy Can you do some hard problem ::)

[@milanstevic8424]

four-eyes! xD

[@aliderkaoui9663]

Hello Michel i need You pleas

[@GreenMeansGOF]

This video is infinitely long too. So many MasterCard ads. Please stop!

[@advaykumar9726]

1

[@user-sh1ce3yw9f]

Second