Two from Mongolia.

[@danielflynn3067]

When he gets the purple equality at 6:58, he can just insert that into the absolute equality below and get two cases: - |f(0)| = 0 => f(0) = 0 - |f(0)| = |-2f(0)| => f(0) = 2f(0) or -2f(0) => f(0) = 0 either way

[@captainsnake8515]

How other KZbinrs title their video: "OMG INSANE MATH PROBLEM [I CAN'T BELIEVE I SOLVED IT] [NOT CLICKBAIT] [99.99% OF MONGOLIAN HIGH SCHOOLERS CAN'T DO THIS] How Michael Penn titles his videos: "Two from Mongolia"

[@MichaelPennMath]

What can I say... I appreciate understated things.

[@prithujsarkar2010]

@@MichaelPennMath sir michael please try problems from the indian national maths olympiad , especially number theory ones , its an humble request

[@prithujsarkar2010]

thats the speciality of prof michael , he believes in quality rather than quantity and fame , and we all love it !

[@tonmuns161]

Indeed there is a very simple solution to the second problem. Just fill in the a,b,c,d for x,y and z and you get that the equation fulfills for all a,b,c,d s.t. ad=bc But I like Michael's solution better..

[@roboto12345]

This is soooo true. Literally I love your videos. I wanna study pure math and abstract algebra due to your videos

[@samassamongkol888]

Here's a solution with basic algebra. In the original equation xy = z^2+1, WLOG assume x < y (they can't be equal). As long as z > 1, we have x < z (otherwise xy >= z(z+1) > z^2+1) and the original equation can be rewritten as x(x+y-2z) = (z-x)^2+1. By setting X and Y (using upper case to denote different variables) equal to the smaller/bigger of x and x+y-2z on the LHS and setting Z = z-x, we have the original pattern XY = Z^2+1 with X, Y, Z being natural numbers. By repeating this process, the resulting Z in each step will be smaller until it is 1. We'll then have the final equation (1)(2) = 1^2+1 which works with (A, B, C, D) = (1, 0, 1, 1). Now we reverse the process. If in each prior step, x is less than x+y+2z, we'll have (x, x+y-2z, z-x) = (X, Y, Z) = (A^2+B^2, C^2+D^2, AC+BD). By solving for it, we have x = X = A^2+B^2; z = Z+x = A(A+C)+B(B+D); y = Y+2z-x = (A+C)^2+(B+D)^2. By setting (a, b, c, d) = (A, B, A+C, B+D), we have (x, y, z) = (a^2+b^2, c^2+d^2, ac+bd) as required. Similarly if in each prior step x is greater than x+y-2z, we'll have (x, y, z) = (C^2+D^2, (A+C)^2+(B+D)^2, C(A+C)+D(B+D)) which works by setting (a, b, c, d) = (C, D, A+C, B+D). Repeat until we return to the original equation.

[@jatloe]

Nicee! I really like how this can be solved by saying “it will eventually boil down to something” :D

[@jatloe]

A question though, why can’t Z go overboard, like if Z=2-3 or something?

[@shohamsen8986]

let A=(a,b) B=(c,d). Then using Lagrange's identity, we know (A.B)^2+(AxB)^2=|A|^2|B|^2. Set (AXB)^2=1, |A|^2=x, |B|^2=y and A.B=z and viola. The existence of integer solutions to (AXB)^2=1 is guaranteed by Bezout' identity (or Euclidean algorithm or Linear Diophantine equations/method)

[@VaradMahashabde]

Nice, but it shows that x,y,z exist for some a,b,c,d, which could be shown with algebra, but we need to show that for any x,y,z, some a,b,c,d exist. This method may leave some x,y,z as forbidden and unobtainable for all a,b,c,d

[@shohamsen8986]

@@VaradMahashabde good point. Let me think about it

[@bambarasamsara7116]

hi greetings from Mongolia! great to see our flag on my favorite channel!

[@MijiratHere]

For number 2: Just plug in x,y and z in terms of a,b,c,d into original equation and get the following: (a^2+b^2)(c^2+d^2)=(ac+db)^2+1 (ac)^2 + (ad)^2 + (bc)^2 + (bd)^2 = (ac)^2 + 2(ac)(db) + (db)^2 + 1 Move to the left everything except 1 (some of terms cancels out): (ad)^2 - 2(acdb)+(bc)^2=1 We can see it’s a complete square in the left: (ad - bc)^2 = 1 --> ad - bc = 1 or -1 And that appears to be a condition on a, c, d, b, that gives you infinite number of solutions - just take any 2 conseqetive numbers, factor it into 2 factors and name it a,c,b,d. For example: 25 - 24 =1 --> 5*5 - 4*6 = 1 --> let a=d=5, b=4, c=6, then: x=5*5+4*4= 41 y = 6*6 + 5*5 = 61 z = 5*6 + 4*5 = 50 So pasting in original equation (x*y = z^2+1), we get: 41*61 = 50*50+1 2501 = 2501 BINGO. And all a,b,c,d, that are set under condition ad - bc = 1 or -1, would work

[@Cor97]

Easy even for students :)

[@indisputablefacts8507]

I came up with this answer too. When I was in grad school (in computer science, many years ago), I was studying this arcane correctness theory in one of my compsci classes; in the same semester I was taking an undergrad-level math class. In the math class, we got some homework that asked for a proof; I decided to attack it with the fancy compsci stuff. My proof took up a whole page of dense formulae. I was really happy with the proof. I turned it in. I was rewarded with a large, red circle across the whole page. I was pretty grumpy since he clearly hadn't even read it. The prof went up to the board and showed a one sentence proof that you barely had to be taking the class to understand. That convinced me that my grade was correct regardless of whether my proof was correct or not. Hook 'em.

[@aqeel6842]

That does not show that all solutions for x, y, and z can be expressed in that way, it only shows that if they could be expressed in that way, there will be an infinite number of solutions. There could still be some solutions that cannot be expressed in that way, according to your 'proof'

[@MijiratHere]

@@aqeel6842 yes, but the goal was to show that there exists a solution, not to find every one of them

[@aqeel6842]

@@MijiratHere Oh sorry I didn't see that. I've been too used to seeing Michael solve problems where every solution is required

[@goodplacetostop2973]

17:34 Энэ бол зогсооход тохиромжтой газар юм Homework... Let fn(x) = cos x cos 2x ... cos nx. For which n in the range 1, 2, ... , 10 is the integral from 0 to 2π of fn(x) dx non-zero?

[@jamiewinter2134]

For (any) non-negative odd n, one can see by making the substitution u=x-pi that the integral in question becomes an integral over a symmetric interval around zero of an odd function, hence zero. A guess would be that this does not happen for n positive and even, but I dont know :)

[@goodplacetostop2973]

short answer= 3, 4, 7 and 8.

[@VerSalieri]

@@goodplacetostop2973 I think I saw this on a JEE paper....which btw was going to be my suggestion for the good dr. to start featuring. But.. I think I am having more fun with him choosing the topics.

[@R0M4ur0]

I tried to turn the product(a=1;n;cos(ax)) into a sum of cosines of various multiples of x. It is easy to proof that cos(u)*cos(v) = 1/2*[cos(u+v)+cos(u-v)] By adding a factor you get cos(u)*cos(v)*cos(z)=1/4*[cos(u+v+z)+cos(u+v-z)+cos(u-v+z)+cos(u-v-z)] In general fn(x) as defined in the problem can be expressed as follows 1/2^(n-1)*cos[x*(n+-(n-1)+-(n-2)...+-2+-1] where the horrible +-'s mean the combination of all the 2^(n-1) possible way to add/subtract the first n-1 natural numbers to n. As a result, either you get cosines of an integer multiple of x or you get cosines of a 0 multiple of x (cos0x=1). These latter are the only ones producing a non 0 result when integrated in the 0 - 2pi interval. For n=1 there's no way to get cos(0*x) For n=2 there's no way as well as the sum of the first 2 numbers is 3 which is odd For n=3 one of the cosines is cos[x*(3-2-1)] which is 1. The result of the integral should be pi/2 For n=4 one of the cosines is cos[x*(4-3-2+1)] which is 1. The result of the integral should be pi/4 For n=5 there is no way to get cos(0*x) as the sum of the first 5 natural numbers is odd For n=6 there is no way to get cos(0*x) for the same reason For n=7 there are many of such cosines: cos[x*(7+6-5-4-3-2+1)] cos[x*(7-6+5-4-3+2-1)] cos[x*(7-6-5+4-3+2+1)] cos[x*(7-6-5+4+3-2-1)] If I nailed them all, the result of the integral is 8*pi/2^6=pi/8 For n=8 there are some of the 0-cosines: cos[x*(8+7-6-5-4+3-2-1)] cos[x*(8+7-6-5-4-3+2+1)] cos[x*(8-7+6-5+4-3-2-1)] cos[x*(8-7+6-5-4+3-2+1)] cos[x*(8-7-6+5+4-3-2+1)] cos[x*(8-7-6+5-4+3+2-1)] cos[x*(8-7-6-5+4+3+2+1)] If I got them all the result of the integral should be 14*pi/2^7 = 7*pi/64 For n=9 and n=10 there is no way to get cos(0*x) as both sums are odd. Hope I manages to be clear... being not a native speaker doesn't help...

[@ashwinvishwakarma2531]

complex exponential could be useful, but the solution is probably alot more clever

[@siddharthabhattacharya3787]

Hello Michael Penn. Love your videos They're amazing 👍😍 They help me fir my Indian Olympiads Thank you❤🌹😊 Keep it up

[@siddharthabhattacharya3787]

@giraffemathcoder yeah you can do it if you are aiming for OCSC

[@prithujsarkar2010]

@@siddharthabhattacharya3787 arre arre panditji aap bhi idhar , charan sprash !

[@siddharthabhattacharya3787]

@giraffemathcoder probablity shouldn't be that important Ig But you can maybe do Inequalities Combi Number theory That may suffice

[@siddharthabhattacharya3787]

@@prithujsarkar2010 Aree aap har jagah phaile hue hai 🙏 Aapko mera saadar praanam

[@quirtt]

@@siddharthabhattacharya3787 inequalities 😂

[@duduong]

At 12:13, by requiring z+i=(a+bi)*(c+di), you made an implicit assumption that such factorization is unique, but it is not. In general, A^2+B^2 can have many choices in factorizing into (a+bi)(a-bi). The original question can be reduced to "Prove that z+i = (a_1+-i b_1)(a_2+-i b_2)...for some choice of plus-minus signs, where (a_n+i b_n)*(a_n-i b_n) is a component in the gaussian prime decomposition of z^2+1."

[@roboto12345]

I wanna recomend a problem IMO SL 2016 A3 it is a really interesting question

[@JM-us3fr]

I don't think the second problem is very obvious without some intense number theory. Gaussian integers was probably the best route. For those looking for a simpler solution, note you only need to show x=a^2+b^2 for coprime a and b. Then you can just choose c and d to be solutions to ad+b(-c)=1 (using Bezout's identity), and the rest follows.

[@willnewman9783]

You and I have very different understandings of what "intense number theory is."

[@JM-us3fr]

@@willnewman9783 Intense BASIC number theory, for sure, but still a long series of steps. Certainly no L functions, sieve theory, or number fields.

[@ChefSalad]

I think you′re all making this harder than it needs to be. All you need to do is find a solution. That shows it′s possible to find solutions. I posted elsewhere how to do that, but I′ll repeat it here so you don′t have to go looking. If you stare at it for a while, you realize that it might make sense to get rid of some of the variables. So let's "get rid of" x by setting a=0, b=1. Then x=0²+1²=1 and z=0*c+1*d=d, which simplifies our equation to 1*y=z²+1, but if we substitute, we get c²+d²=d²+1, which means c²=1 thus c=1 and d∈ℤ. Of course d≠0 because if it did, then z=0∉ℕ, but it can be any other integer. So now we have an infinite family of solutions: a=0, b=c=1, d∈ℤ≠0. We only need one, and so we′re done.

[@willnewman9783]

@@ChefSalad What you are saying does not solve the problem. What is needed to show that for every triple of numbers (x,y,z) with xy=z^2+1 there exist a,b,c,d with x=a^2+b^2, y=c^2+d^2 and z=ac+bd. What you show is that for infinitly many solutions (x,y,z) there exists a,b,c,d, namely (1,d^2+1,d). But not every solution is of this form. For example (2,13,5) is a solution for which you have not shown there exist a,b,c,d.

[@ChefSalad]

@@willnewman9783 I disagree. It doesn't say we need to show it for all x,y,z∈ℕ, just for some x,y,z∈ℕ, so we only need one solution. Or at least, that′s the way I read it. If he wanted it for all x,y,z, he should have written ∀x,y,z∈ℕ.

[@rmschad5234]

For question 2, I used the cross terms in z and the magnitude terms in x and y as motivation to write s_x=a+ib,s_y=d+ic,z=IM(s_x*s_y) (I'm not sure if I need a -i out front, whatever.) So the original equation becomes |s_x*s_y|^2=|IM(s_x*s_7)|^2 + 1 and we are know looking for a complex number with one part equal to 1. This leads to a pretty immediate solution of a,b,c,d=1,1,1,2 => x=2,y=5,z=3. Finally, 2*5=10 is indeed equal to 3^2+1=10.

[@s4623]

For the 2nd problem, can't we just put xy = a²c²+b²c²+a²d²+b²d² and z²+1=a²d²+2abcd+b²c²+1 and if xy=z+1 then b²c²+ a²d² = 2abcd+1 (subtracted a²d²+b²c² from both sides) move 2abcd to the left and factor (bc-ad)²=1 And then you can choose a=3, d=5, b=8, c=2 (the question did only ask for at least one solution). To arrive at (a,b,c,d)= (3,8,2,5) just look for consecutive composite natural numbers (I believe the smallest is 15 and 16 if you need all 4 distinct factors and none of them equal to 1). To verify, substitute back: x = 3² + 8² = 73 y = 2² + 5² = 29 z = 3×2 + 5×8 = 46 xy = 73×29 = 2117 z² = 46² = 2116

[@dudewaldo4]

Jsuk some of the different colored chalk that you use to differentiate stuff look exactly the same in the video

[@bhanusri3732]

I want to do post graduation in math after being part of this channel

[@prithujsarkar2010]

Sir michael .. could you please try number theory problems from the indian national maths olympiad please ? thanks in advance

[@quirtt]

Yo

[@shohamsen8986]

Small correction. American's don't use the term 'sir' to refer to their teacher, you just call them by their first names. They appreciate it much more if u just call them by their 1st names.

[@danielmilyutin9914]

Fir second task we use equality: (a^2+b^2)(c^2+d^2) = (ac)^2 + (bc)^2 + (ad)^2 + (bd)^2 = (ac)^2 + 2abcd + (bd)^2 + (bc)^2 - 2abcd + (ad)^2 = (ac+bd)^2 + (bc-ad)^2 from which follows that (bc-ad)^2 = 1. so matrix with integer numbers [a b; c d] must have determinant of +-1. Actually we must show existence, so only one set of values is enough. Ex. a=d=1, c=d=0. But there are actually infinitely many solutions. for this we may build matrices from finite series of multiplications of basis matrices: [1 N; 0 1], [1 0; M 1], [-1 0; 0 1], [1 0; 0 -1] for integers N,M which may or may not differ for each multiplication. BTW, the most famous example is Fibonacci numbers matrix [F(n+1) F(n); F(n) F(n-1)].

[@sergiogiudici6976]

I wrote as x/(z-i) = (z+i)/y this means that both side are equal to m/n where m = a+ib and n= c+id are some gaussian integers. Now, derive z-i and z+i and impose that the complex coniugated of z-i Is equal to Z+i. This implies (aa+bb)x = (cc+dd)y suggesting the thesis to be proved. Is that correct?

[@elunedssong8909]

Before vid: Q1 states that a-b>_f(a)-f(b) The only equations i can think of that satisfy this condition are f(x)=(- or +)x+-z where z is any number you like. Defintely, any polynomial with more than one exponent will not satify this inequality for some real value. It's either decimals for fractions(1/x) and roots, x^(z , 0 symbol for? Logs don't work, .5^x seems to work for positive values, but it fails for negative values. .5^-1 = 2, -2=4.4-2>2-1 Man this problem is hard. No clue how to make sense of this > symbol. I guess i state without justifcation that this polynomial is the only type of real equation to satify this inequality, since i've eliminated all other types i can think of, i think? Now if f(f(f(0))=0 for the polynomials is true, then it must be the case that z=0. so we have y=+-x, x=0, y=0. But i'm severly lacking clearly. perahps 1/n! is another valid solution, but it cheats around the idea of real->real since it only accepts integer inputs, i think its not acceptable. What is that > for??? i must be missing a type of equation. After video: LOL you weren't supposed to think about its maping at all. Its mostly just a disguised logic puzzle. Nice.

[@samuelbam3748]

A solution to the teacher's question that is equivalent to this one, but doesn't require complex numbers: If we can show that x and y can be written as the sum of two squares, we are done. Suppose there is a prime p that divides x or y. Since xy=z^2+1 we have that z^2=-1 mod p so -1 is a quadratic residue mod p. This means that p /= 3 mod 4 (by eulers criterion we have for odd p 1 = (-1/p) = (-1)^(p-1/2) so p=1 mod 4). But it is well known that all numbers were primes of the form p=4k+3 have an even exponent in the prime factorization can be expressed as the sum of two squares. Because here the exponent is 0 for all these primes, x and y can be expressed as the sum of two squares

[@noahtaul]

You can find a,b,c,d with x=a^2+b^2 and y=c^2+d^2, but why do the a, b, c, d always line up perfectly to give you z=ac+bd?

[@VaradMahashabde]

This is not simpler in my opinion, but it sure is interesting

[@MichaelPennMath]

Using quadratic residues is definitely a nice approach!

[@riadsouissi]

this is a nice proof.

[@georget2063]

This does not really work. If you dont take my word, here is a counter example: 325 = 5²•13 = 1 + 18² = (5²+0²)(2² + 3²) Now since a = 0 or b = 0, then z = +_ 10, +_15, which definitely isnt +_ 18.

[@forever10rch99]

Are there any equal channels about for example geometry or other types of olympiad maths?

[@philkeyouz2157]

This remember me , if xy+1=z^2 show that it exist 2 primes p and q such that pq=x+2z+y. And x or y are not primes

[@willnewman9783]

This is not true, 4*2+1=3^2, but x+2z+y=4+6+2=12 which is not pq for any primes p,q

[@philkeyouz2157]

@@willnewman9783 you are right, I forgot to specify that x or y must not be prime

[@willnewman9783]

@@philkeyouz2157 It is still not true. 4*6+1=5^2. And if you want to say that z cannot be prime as well, there is still 13*15+1=14^2, none of which are prime.

[@coc235]

@@willnewman9783 But 13 is a prime

[@quirtt]

Why don't you ever do combinatorics?

[@prithujsarkar2010]

yo

[@prithujsarkar2010]

bhai thodi respect deke somment kar xD like "plz try combi" or so xdxdxd , guruji hai hamare xdxdxd

[@siddharthabhattacharya3787]

@@prithujsarkar2010 true 😆😆😆😆

[@quirtt]

@@prithujsarkar2010 xdxdxd k next time

[@KyleDB150]

I think I used a simpler (but equivalent) path for the end of the first question: Once you have: |f(0)| = |f(f(0)) - f(0)| = |f(f(0))| Notice that the right equality is only possible if either: Case 1: f(0)=0 So that's done Or case 2: f(0)=2f(f(0)) Taking the absolute value of both sides, and taking oh the factor of 2: |f(0)| = 2*|f(f(0))| But from the original equality, we have |f(0)| = 1*|f(f(0))| This is only possible if |f(f(0))| = 0 And since that's equal to |f(0)|, it gives our f(0)=0 Maybe that's more confusing to some, but it avoids carrying two different +/- around

[@cmielo9529]

can you give some examples of other coutries having olympiads for teachers?

[@MORISENSEIISGOD]

Here is how I did the second equation: We start with the subbing x=a^2+b^2, y=c^2+d^2 and z=ac+bd into xy=z^2+1 We get a^2*c^2+a^2*d^2+b^2*c^2+b^2*d^2=a^2*c^2+b^2*d^2+2abcd+1 Cancelling terms on both sides: a^2*d^2+b^2*c^2=2abcd+1 Therefore a^2*d^2+b^2*c^2-2abcd=1 factorising: (ad-bc)^2=1 So abs(ad-bc)=1 So any integers a,b,c,d that satisfy abs(ad-bc)=1 will satisfy x=a^2+b^2, y=c^2+d^2 and z=ac+bd and xy=z^2+1.

[@fullfungo]

But if you start with the substitution, it means you are assuming it is true, which we have to prove

[@hans-juergenbrasch3683]

Nearly true, a bit more correct would be: for any a,b,c,d with abs(ad-bc)=1 and z=ac+bd you can fix x and y as required. But you still would have to show that such a,b,c,d exist.

[@ismailsaglam2876]

in the second problem, there is a problem. you can not wrte a number x as a pruduct of two gaussian integers which are conjugate. 3 is such an example.

[@shaevor5680]

numbers like 3 are what he calls "regular primes": numbers that are both prime in Z and in Z[i]. Starting at 11:50, he explains why they cannot occur in the product.

[@llunaecy]

Edit: Seems like I misunderstood the question... it's asking to prove that for any set of x,y,z that satisfies the equation, there exists at least one set of a,b,c,d... Somehow I thought it was as easy as proving there existed a set of a,b,c,d that would give you x,y,z that works... I'll just leave the original comment here so you can laugh at my mistakes :) 11:03 I don't know if I made some assumptions here that I shouldn't have, but what I did was... just insert the given x, y and z into the original equation to get some values of a,b,c,d that work: xy=z^2+1 (a^2+b^2)(c^2+d^2)=(ac+bd)^2+1 (ac)^2+(ad)^2+(bc)^2+(bd)^2=(ac)^2+2abcd+(bd)^2+1 (ac)^2 and (bd)^2 cancel each other out (ad)^2+(bc)^2=2abcd+1 (ad)^2-2abcd+(bc)^2=1 (ad-bc)^2=1 ad-bc=(+/-)1 and from there it's quite clear that there are integer solutions of a,b,c,d

[@tomaszkinowski4090]

The issue is that we have to show that the natural number solutions to the equation xy=z^2+1 can be expressed as a sum of squares, which is not always true for all natural numbers (for example 6 cannot be expressed as a sum of two squares).

[@tonyhaddad1394]

I solved the same way you solved its more easier

[@tonyhaddad1394]

@@tomaszkinowski4090 but our goal is only show there are a natural number satisfy this equations Or im wrong ??

[@tomaszkinowski4090]

@@tonyhaddad1394 The idea is that we have to show that natural numbers that satisfy the equation, can be expressed in that form. The issue is, not every natural number can be expressed as the sum of two squares. Hence, we have to show that all the natural numbers which satisfy the equation, can be written as the sum of two squares.

[@tonyhaddad1394]

@@tomaszkinowski4090 ahhhh oke bro now i get it thank u man

[@OchirBFolo]

Монгол 😁🎊🇲🇳

[@prabhatsharma5751]

Love from Nepal ❤

[@particleonazock2246]

Sain baina yy!

[@Tekashiixine-yb5kh]

the simple solution is just factoring and combining like terms, end up with (ad-bc)^2=1, so as long as ad and bc are consecutive natural numbers, they will satisfy xy=z^2+1

[@MichaelPennMath]

This was my problem though, how do we know we can find a,b,c,d satisfying ad-bc=1 along with all of the other conditions?

[@ahzong3544]

I feel like any tuple (a, b, c, d) that satisfies ab - cd = +-1 will satisfy all of the other conditions. That being said, there should always exists 2 numbers whose difference is 1. For example, take ab = 9 and cd = 8. One possibility is the tuple (3, 3, 4, 2). We can verify that this is true by calculating x, y and z and plugging them in the original equation.

[@R0M4ur0]

@@MichaelPennMath @Michael Penn I used the same approach as Tekashi and got to the same equation. To answer your question, do you think it would suffice to take into consideration the factorization over two numbers of any two consecutive integers? I mean: let's take two consecutive numbers u and v=u+1. I can always express both numbers as u=u1*u2 and v=v1*v2 where u1 and u2 are any two divisors of u (and are equal to1 and u if u is prime) and v=v1*v2 (same as before). Now by setting a=u1, b=v1, c=v2 and d=u2, you'll find that xy=z^2+1. May you help me to identify what's wrong whit this proof?

[@MichaelPennMath]

You have to be careful here though. You are given x,y, and z then have to produce a,b,c,d. Maybe I am reading this wrong, but it feels like you are finding x,y,z for special a,b,c,d...

[@Tekashiixine-yb5kh]

@@MichaelPennMath the ()2 means to the second power, also the proof kind of shows that in order for xy=z^2+1, (ad-bc)^2=1. It's not finding special abcd but rather showing that there is a specific set of abcd for each unique combination of x,y,z. In contrast, if given x,y,z can not be written in terms of a,b,c,d where ad and bc are consecutive then xy doesn't not equal z^2+1. Sorry if that sounds confusing I'm not too familiar with the vocabularies in proofs.

[@MAREKROESEL]

2,1,1,1 ?

[@АртурГалиуллин-ш2э]

a=1, b=0, c=z, d=1?

[@brunocaf8656]

This way you would be saying that x must always be 1, and y must always be z²+1, which is just a particular case

[@beautifulworld6163]

From 🇲🇳 ❤

[@keriank5883]

For the first one, as f(f(f(0)))=0 can we say that f(f(...f(0))...)=0 (infinite inclusion) so f(0)=0 ?

[@shohamsen8986]

This is not correct. let f[n](0) be short hand for f convolved with itself n times and then applied to 0. So f[1](0)=f(0), f[2](0)=f(f(0)), f[3](0)=f( f( f(0) ) ), etc. The infinite convolution that you claim is lim n-> infty f[3n]= 0. It could so happen that f[2n](0)=\= 0 for all n (provided n is not a multiple of 3). A similar argument could then be made about f[n](0). In short your result is sequence dependent.

[@keriank5883]

My bad I forgot the fact that it was f[3n](0) which was equal to 0

[@KyleDB150]

I mean I solved the second one by simply subbing in the x,y,z equations into xy=z^2 + 1 and simplifying, which shows that we need bc-ad=1 So pick any (a,b,c,d) which satisfies that, like (2,1,1,1), and you've shown by example that "for some x,y,z in N, there exist some a,b,c,d in Z which satisfy that system of equations" *the question didn't say "for any x,y,z", so one example is enough* You can also express x=a^2 + b^2 = (a+bi)(a-bi), the same for y, and xy = z^2 + 1 = (z+i)(z-i), and result in the same requirement that bc-ad=1 But I guess if you're showing that it works for any x,y,z, then you need to use those Gaussian primes to show that it's always possible to find a,b,c,d that satisfy bc-ad=1

[@thomshrike]

I did the same.

[@taransingh5026]

hey micheal PLSSSSSSS can U just solve 2020 JEE ADVANCED maths paper................I REALLY LIKE YOUR TEACHING PLSSSSSSSSS..

[@blazedinfernape886]

Stop asking for jee advanced solution every single time. There are already a lot of videos of people doing that.

[@prithujsarkar2010]

stop destroying the decorum of olympiad maths please , jee adv maths is way more different than the beautiful maths of olympiads

[@samuelemorreale7510]

Here in Italy we barely have maths books

[@massipiero2974]

Well math isn't exactly the most popular subject in italy, but that's a bit exaggerated 😂😂 in which city do you live? (i'm still writing in english so other people can read this)

[@samuelemorreale7510]

@@massipiero2974 yeah, obviously it's an exaggeration, but they could do so much better

[@TheoH54]

I bet there must still be some ol' books of Fibonacci around at the antique markets. Just kidding :-). Anyway, I am not sure what math level you are looking for. For starters: there is a great series of books by Piergiorgio Odifreddi containing lots of (basic) mathematics . But there are many books by other authors. For more advanced level books you probably need to know some english. Any decent common library or bookstore here in Italy usually has a a series of books on mathematics. For still higher level: any university library or math department library has more than enough books. Good luck!

[@samuelemorreale7510]

@@TheoH54 mine was a joke about the somewhat poor math teaching in Italy

[@TheoH54]

​@@samuelemorreale7510 Si può lamentarsi di come viene insegnato la matematica in Italia. Però, con tutto rispetto, le posso assicurare che i libri scolastici (parlo del ciclo scuole medie/superiori) sono di ottimo livello. Per quanto riguarda la matematica insegnata alle università, la 'Scuola Italiana' non è inferiore a nessun altro paese. Semmai è un problema di riconoscenza nazionale perché per gran parte della gente (politici inclusi..) la matematica, e la Scienza in generale, è qualcosa di alieno. Questo vale ovviamente per molti paesi, non solo per l'Italia. Chiudo dicendo che i libri ci sono, anche qui in Italia; basta la volontà di andar a scovarli! Buona serata :-)

[@AlephThree]

My “solution” for the second one was trivial by comparison. Have I messed something up? I started by expanding z^2=(ac+bd)^2=a^2c^2 +2abcd +b^2d^2, and xy= (a^2+b^2)(c^2+d^2)=a^2c^2 a+a^2^d^2+b^2c^2+b^2d^2 Then noted a couple of terms cancel out on each side when setting xy=z^2+1, giving a^2d^2+b^2c^2=2abcd+1 This looked like the first expansion and factored as (ac-bd)^2=1, which solves as ac-bd=+/-1. But then clearly this solves eg a=c=4 and b=3, d=5.

[@hans-juergenbrasch3683]

The second one is an ill-posed problem. Setting a=z, b=1,c=1,d=0 always yields the identity and for some z (such that z*z+1 is prime, e.g. z=2 or 4) this is the only type of solution.

[@hans-juergenbrasch3683]

@@angelmendez-rivera351 you are absolutely right

[@srijanbhowmick9570]

Please do one from India . A humble request from a long time viewer .

[@AVM81]

For question 2, First I reinterpreted the question as follows Given xy = z^2+1, if x=a^2+b^2, y=c^2+d^2, z=ac+bd, then there exists a solution such a,b,c,d are positive intigers Thus I put later equations into the first and get (a^2+b^2)(c^2+d^2) = (ac+bd)^2 + 1 expanding out, I get a^2c^2+a^2d^2+b^2c^2+b^2d^2 = a^2c^2 + 2abcd + b^2d^2 + 1 Eliminating like terms and rearranging slightly gives (ac)^2 + (bd)^2 = 2 (ac) (bd) + 1 As right side is two times and integer plus one, it is odd. For two numbers added together to give and odd number, one must be odd and one must be even, Thus I will assume (ac)^2 is even and (bd)^2 is odd. Since even squares give even numbers and odd squares give odd numbers, (ac) is even and (bd) is odd. I am going to make the following substitutions. (ac) = 2m, (bd) = 2n+1, and thus replacing, expanding, and simplifying (2m)^2 + (2n+1)^2 = 2 (2m) (2n+1) + 1 4m^2 + 4n^2 + 4n + 1 = 8mn + 4m + 1 4m^2 + 4n^2 + 4n = 8mn + 4m m^2 + n^2 + n = 2mn + m Trying simple solution of m=1, n=1 gives 1 + 1 + 1 = 2(1) + 1 which is equal Returning above, I assumed (ac) = 2m, (bd) = 2n+1, and replacing in, (ac) = 2m = 2(1) = 2, (bd) = 2n+1 = 2(1) +1 = 3 thus ac = 2, bd = 3 Using this and trying cases, I get a=1, b=3, c=2, d=1 which gives x = a^2 + b^2 = 1 + 9 = 10, y = c^2 + d^2 = 4 + 1 = 5 and xy = 50 z = ac + bd = 1 + 6 = 7, z^2 +1 = 49+1 = 50 Thus as I have shown there exists a solution that satisfies everything, a solution exists.

[@ChefSalad]

On the second problem, if you stare at it for a while, you realize that it might make sense to get rid of some of the variables. So let's "get rid of" x by setting a=0, b=1. Then x=0²+1²=1 and z=0*c+1*d=d, which simplifies our equation to 1*y=z²+1, but if we substitute, we get c²+d²=d²+1, which means c²=1 thus c=1 and d∈ℤ. Of course d≠0 because if it did, then z=0∉ℕ, but it can be any other integer. So now we have an infinite family of solutions: a=0, b=c=1, d∈ℤ≠0. We only need one, and so we′re done.