International Mathematical Olympiad 2017 Problem 2

[@DaveyJonesLocka]

I love that you not only show clever moves, but you also try to show the insight and motivation behind them. When I teach, I never just do something because it’s the next move. I try to show students how to analyze a current situation for motivation behind the next move.

[@jwy4264]

ok dave

[@devd_rx]

Ok

[@BucifalulR]

A fix for the hole in the proof at @10:50 where r and s are said to exist and also be REAL. This fix was taken from the comments below (@vindex7) and a friend's suggestion: we can assume a < b (we're considering aribtrary a, b such that a is different than b). Then we have rs=a-1 and r+s=b, so by Vieta's formulas, r and s are the roots of the equation z^2-bz+a-1=0, which has discriminant b^2-4(a-1) > b^2-4(b-1)=(b-2)^2>=0, so r,s do exist and are indeed real numbers.

[@davidblauyoutube]

On the assumption that f(x) is differentiable, there's a shortcut as follows: 1. Let y=0 to give f(0) = f(f(0)f(x)) + f(x). 2. Differentiate to give 0 = f'(f(0)f(x))*f(0)f'(x)+f'(x). 3. Case I: f'(x)=0 identically, so f(x)=c. Recover f(x)=0 as in the video. 4. Case II: f'(x)!=0, so -1 = f'(f(0)f(x))*f(0). Since f'(x) is not identically zero, f(x) is multivalued and therefore f'(y) is constant (on setting y=f(0)f(x)). Write f(x) = ax + b; then -1 = f'(b(ax+b))*b = a*b so a = -1/b and f(x) = b - x/b. Back sub into the original formula yields b^2 = 1 and we recover the two solutions f(x) = 1 - x and f(x) = -1 + x. //

[@angelmendez-rivera351]

This is true, but you are not allowed to actually assume f is differentiable.

[@-sab5226]

no you can't do that

[@bourhinorc1421]

nope no differentiable

[@pepefrogic3034]

You also assumed that f is onto if you have f is onto then you get setting f(x) to be a, in the case f(1) is 0 and f(0) is 1 that f(a) = 1-a

[@prod_EYES]

Calculus not allowed

[@hodesdjole1771]

most of these functional equations have constant or linear solutions, in rare cases they have more exotic ones (solutions i mean), i guess that kinda means that imposing crazy functional conditions only results in trivial answers but then again there have been functional problems that have had very exotic solution, for example "the functional square root" and such, from which one can imagine a fruitfull amount of similair investigations of functions

[@klarkunskap1090]

thanks for the functional equations, love them! Thanks for listening to your followers, keep up the great content.

[@letsthinkcritically]

Thank you very much!!

[@Stelios2711]

Not so intuitively motivated, but I guess that it's becuase of the difficulty of the problem.

[@nikhileshkrishna889]

Genius trick

[@nikhileshkrishna889]

Testing

[@nikhileshkrishna889]

Testing

[@swipingeverymoment]

@@nikhileshkrishna889 ?

[@liuyxpp]

Let y=0 to give f(0) = f(f(0)f(x)) + f(x), let a = f(0), then f(af(x)) = a - f(x). Let t = f(x), then f(at) = a - t. Let t = s/a, then f(s) = a - s/a. s is in the domain, so we just have f(x) = a - x/a. From the video, we have a = +/-1, so f(x) = 1 - x or f(x) = x - 1.

[@mrityunjaykumar4202]

when you took f(x)=t and then came to conclude that f(at)=a-t.. this is wrong because here you're considering t can be any real number from the domain but its actually f(x) hence it is the image of x and not all real numbers from domain can be represented by f(x).. it can only be possible when every element in the domain is the image of some elements in the domain itself.

[@mokouf3]

Another trick: Let f(x) = Sum(a_k*x^k, k=1 to n) put y=x, do polynomial order check, you will find that only possible value for n is n=1. Now let f(x) = ax+b, solve for a first, then b. I think this method is faster, but I don't know if this is a method that should be used.

[@DungNguyen-ti4hg]

Your solution not good. Because we have to find all of f. Not only polynomials

[@mokouf3]

​@@DungNguyen-ti4hg My concept: All functions can be expressed with Taylor Series. Sounds good now?

[@Deathranger999]

@@mokouf3 That's not true though.

[@nicholaskhawli4330]

@@mokouf3 all analytic functions over C can be represented as Taylor series for all x in R which isn't ALl solutions

[@Mega11041104]

fが単射であることを示すためにf(a)=f(b)、a=rs+1、b=r+sと置いてますが、このようなr,sが存在するかどうか示していません。

[@ImaginaryMdA]

I could prove the three solutions over the rational numbers, but I had no idea how to increase the scope of my solution to include all real numbers. The step I didn't see was the "a=r+s, b=rs+1" trick, and because of that I couldn't prove injectivity.

[@videolome]

You didn’t see it because the proof is wrong. Take for example a=5 and b=2. Then (x-r)(x-s)=x^2-2x+4 There are no real numbers that are roots of this polynomial. Will he recognize his mistake? I don’t think so...

[@noctnightcr9918]

can't u just swap a and b by symmetry? So rs+1=2, r+s=5, which obviously has solutions as setting rs=1 gives r+1/r=5 where LHS is continuous for r>0

[@Deathranger999]

@@videolome The commenter who mentioned we can swap by symmetry is right, but another argument is this. I think because of the linearity (f(x + 1) = f(x) + 1), you can just let b' = b + k for some large integer k, and a' = a + k. Then clearly f(b') = f(a') still. The discriminant is b^2 - 4a + 4, so since b^2 grows much faster than -4a, we will be able to find a k such that the discriminant b'^2 - 4a' + 4 is positive (this argument can be made more rigorous). At that point we can use the same argument to prove that b' = a', from which it follows that b = a.

[@drewmichael3986]

how do you prove that there exists r,s such that a = rs+1 and b = r+s? How about when a = 2, b = 1?

[@letsthinkcritically]

As f(x+1) - f(x) = 1, we can translate a,b to higher or lower values together, and we can always translate to certain values such that such r,s exist.

[@drewmichael3986]

oh nice thanks :)

[@vindex7]

An easier way: swap a and b, making a=1, b=2. If ab^2-4(b-1)=(b-2)^2 is positive and r,s exist.

[@dm9696]

@@letsthinkcritically But how can we guarantee that just because we can translate a and b as we like?

[@videolome]

The proof is wrong. This crucial step was not done correctly. He doesn’t understand what he is doing. There are no real r, s that satisfy the equations for a=5 and b=2. It is very basic math, but he doesn’t see it...

[@pratikmaity4315]

If f(f(0)^2))=f(1)=0 then why is it necessary that f(0)^2=1?

[@aniruddhvasishta8334]

He assumed that f(1) is the only way to get 0

[@franolich3]

@@aniruddhvasishta8334 But why is this assumption valid? Originally he showed that there exists an "a" such that f(a)=0. He did not prove there is only one such zero of f.

[@maxmuster6390]

He says we have at least one a that satisfies f(a)=0 because of f(f(0)^2)=0. Then he creates this construct, which includes the term a/(a-1). If the construct is valid f(x)=0 follows as shown. If the construct is invalid because a/(a-1) is invalid f(x)=0 does not follow. So he has to recalculate the whole thing just for a=1 for all other cases the construct holds and f(x)=0.

[@harisshaqiri4785]

fun fact my teacher made this problem and he posted it to the imo

[@tonyha8888]

Thanks for a very nice solution. please can you solve "British Mathematical Olympiad Round 2" 2012 problem 2. Thanks in advance!

[@hodesdjole1771]

this is some real fucking multidimensional chess shit right here dude smh

[@SONUKUMAR-mb2sp]

Which book , should i prefer for such functional equation problems

[@brendanchamberlain9388]

Although not specifically for functional equation problems, the book from AOPS on Intermediate Algebra has a good chapter on functional equations.

[@letsthinkcritically]

I learn functional equations by reading problems and solutions on AoPS.

[@timetraveller2818]

At 6:33 you assumed f is injective f(f0)²)=0=f(1) Does NOT imply f(0)²=1

[@Kettwiesel25]

He did not assume injectivity. He knew that 1 was the only root of f because otherwise there is a other than 1 s.t. f(a)=0 in which case he had proved before that f(0)=0 and based on that f=0

[@fipillo4658]

Why if a = 1 then f(1)=0 ? It isn't clear to me

[@mcwulf25]

Lost me on the injective bit. Why f(a) - 1?

[@tomykill5232]

Nice

[@saatvik7643]

THANKS FOR THE VIDEOS SIR ....REALLY YOUR OP !!! YOUR CHANNEL REALLY HELPS ME TO THINK CRITICALLY. CAN U SUGGEST ME A BOOK FOR NUMBER THEORY I AM A BEGINNER IN NUMBER THEORY .....

[@lovrodrofenik1455]

This is the book Modern Olympiad Number Theory by Aditya Khurmi

[@gastoncastillo9946]

How do you know that 1 is the only number which its image under f is 0?

[@angelmendez-rivera351]

Because it was proven in the video that if f(a) = 0, then a + a/(a - 1) = a^2/(a - 1) or a = 1. a + a/(a - 1) = a^2/(a - 1) implies f(x) = 0 for all x, so any other solutions necessarily satisfy that if f(a) = 0, then a = 1.

[@justins1146]

Thanks

[@bayezidx]

How did you replace f(b) by f(a) at 12:42??

[@thomy2562]

We are proving injectivity so we chose such a and b that f(a)=fb)

[@bidish2224]

Great👍👍👍

[@jofx4051]

I think we should watch Dr Peyam lecturers for this 😂

[@dominiquebercot9539]

Si f(0)=0 on a f(0*f(x))+f(x)=0, donc f(x)=0 pour tout x Si f(0)=a, on obtient f(f(x)*a)+f(x)=a, puis on pose f(x)=t/a, on a alors f(t)+t/a=a, donc f(t)=a-t/a puis’ f(x*y)=a-xy/a et f(f(x)*f(y))+ f(x+y)=a-xy/a^3 (apres4lignes de calculs) Il y a égalité si a^3=a, donc à =1 ou -1 Mes calculs me semblent moins compliqués que les votres. Sont.ils corrects? Amitiés de France

[@chris-ph6so]

I am currently in high school and I am asking myself if those kind of problems are also hard for people who actually studied mathematics ?

[@bayezidx]

practice makes it easier

[@marcusrees5364]

Hi chris, I just finished a bachelor's degree in maths. These sorts of problems are still hard; they require totally different tools and ways of thinking than the sort of content usually covered at university.

[@chris-ph6so]

@@marcusrees5364 okay thanks for the answer becuase im in highschool and i realy like math but thoose problems are really hard

[@vtk5581]

could you solve me this one: find all integers x;y;z such that: 1+2^x=3^y+2^2z+1

[@rendyadinata5534]

The only solution is (x,y,z)=(1,0,0), (2,1,0) since for x,y,z > 1the LHS will be odd but the RHS will be even. (For x,y,z positive) ( what's left Is just to prove that there are no solutions x,y for x,y>1 2^x=3^y+1).

[@vtk5581]

@@rendyadinata5534 well I already solved it 3 weeks ago. Anyways, thank you!

[@tonystarklive9018]

If a=rs + 1 then f(rs)=f(a-1) instead of f(a)-1 i think there something wrong

[@elfek009]

yy i think so

[@cosimodamianotavoletti3513]

We know that for all x f(x+1)-1=f(x), so f(a)-1=f(a-1)

[@tonystarklive9018]

@@cosimodamianotavoletti3513 oh yes,I forget this part

[@sarvendrashukla8051]

How can we prove the function is surjective?

[@bertrandviollet8293]

I don't understand the beginning,let f(x)=c then c+c =c

[@bertrandviollet8293]

Oops i understand now

[@Nothingeverything192]

Tell me I don't get it

[@ИбадатЖұмабек]

Hard problems

[@adchayansivakumar1667]

f(x)

[@chiranjibimahapatra708]

What is going on😕😕😕

[@shafikbarah9273]

Easy

[@timehasstoppedandthefunbeg4467]

God, i really hate these kind of question