Solving an IMO Problem in 10 Minutes! | International Mathematical Olympiad 2006 P4

#NumberTheory #MathOlympiad #IMO

The International Mathematical Olympiad is the biggest Mathematics Contest for high school students, with the most number of participating countries. The contest is held over 2 days, when contestants work on 6 problems within a total of 9 hours. Now we are going to smash one of them in 10 minutes!

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I share maths problems and maths topics from well-known contests, exams and also from viewers around the world. Apart from sharing solutions to these problems, I also share my intuitions and first thoughts when I tried to solve these problems.

[sevenflowers]

hi so i solved before watching the video, and how i did it was to use y^2-1 on LHS and 2^x ( 1+2^x+1) on RHS. note that RHS is alreaedy resolved into two factors with even part and odd part seperated. and y^2-1 has factors (y-1)(y+1). now call y-1 as 2^s.h with h being odd, and compare the even and odd parts. solving a bit gives you that the only possible case is s=1. in this case i assume h+1 as 2^t.k and write the whole eqn in h and k, and then solve as diophantine. basically you get h = (k+8)/(k^2-8) which gives k=3 as the only value for which h is positive integer.

[pratikmaity]

Keep going bro these kind of math problems motivate me a lot!!

[ahongkongmathlover]

Nice job! The video is awesome! Keep going!

[toppersclashing]

I m sure ur channel will grow much larger just because the Content is awesome :)

[CHector]

You should have done the case x<0, however it is clear that you don't get any solution, because clearly 1+2^(-1)+2^(2(-1)+1) =1+1/2+1/2=2=y^2 is impossible and if x<-2 then 1<1+2^x+2^(2x+1)<2.
Good video !

[karolkurek]

I didn't grab a finer point: how you went from 2^(x-1) congruent to 0 mod 8 to y+1+2^(x-1)=7*2^(2x-3). Can someone explain why k must be 2x-3, not 2x-1 or 2x+7 and so on?

[leonivanov]

4:14 i think you can immediately say that system 1 is not going to work because for big enough x in original equation LHS is odd thus y is odd, while in upper equation of system 1 RHS is odd, so y should be even, therefore contradiction.

[mikesyd]

Systems 1 and 2 can be got much more simpler. F.e. just move 1 from left to the right. Everything else is superior, keep it up 👍

[pedrojose]

Good morning, from Brasil! I Allways give my like before watch the video. Most of time, I try to solve first.
2^x + 2^(2x+1)=y^2-1
We heave at least the trivial answer for x=0 and y=2 or y=-2.
No negative answers for x as sqr(2^x+2^(2x+1)+1) will be not a integer. As 2^x+2^(2x+1)+1=2 for x=-1 and less than 2 for x<-1
Let's find for x>=1
2^x(2^x+1)=(y+1)(y-1)
d | (y+1) and d | (y-1) ==> d|2; So 2^(x+1)+1=a.b and g.c.d(a, b)=1 and as y=1 or y=3 (mod4) we have two possibilities:
(I) y+1=2a and y-1=b*2^(x-1) or (II) y+1=b*2^x-1 and y-1=2a
First (I)
(i) y+1=2a
(ii) y-1=b*2^(x-1)
y^2 =2^x + 2^(2x+1) +1<2^(2x+2); as 2^(2x+1) > 2^x+1
So y < 2^(x+1) ==> y+1 <= 2^(x+1) and (i) we have that b<=4. As both a, b are odd b=1 or b=3.
Trying for b=1
y+1=2^(x-1)+2=2a; As ab=2^(x+1) and b=1 then a= 2^(x+1)
2^(x-1) + 2 = 2*(2^(x+1) +2) contradiction as 2*(2^(x+1) +2)>2(x-1) + 2
Trying b=3
y+1=3*2^(x-1)+2=2a, multiplying by b=3 9*2^(x-1)+6=2*(2^(x+1) +1)
2^(x-1)(9-2^3)= -4 contradiction.
Now (II)
(i) y+1=b*2^(x-1)
(ii)y-1=2a
as we have already seen before y < 2^(x+1) ==> y-1 < 2^(x+1) and b<2^2; b=1 or b=3 again
Trying b=1
y-1= 2^(x-1)-2=2a. b=1==> a= 2^(x-1); 2^(x-1)-2=2*(2^(x+1)+1) contradiction.
Trying b=3
y-1=3*2^(x+1) -2=2a. Multiplying by b=3: 9*2^(x-1)-6=2^(x+2)+2
2^(x-1)*(9-2^3)=8 ==>2^(x-1)=8; x=4
b=3 ==> x=4 and y=-23 or y=23
We have four solutoions (0, -2); (0, 2); (4, -23); (4, 23)
But I have burned a lot of time solving!!!
Keep on rocking, go straighy ahead! I really like so much this site! Congratulations!

[vvosbmu]

applying that a^2+b^2=c^2 has a solution, I can only get the answer of x=0.

[mcwulf]

Blimey I thought I solved it by taking the trivial X=0 case and moving on to solve the difference of squares to get the X=4 solution. But my work wasn't done!!!

[alainrogez]

2:30 I don't understand where this 8 come from.

[jinhuiliao]

1) x=0 leads to (0, +/-2)

2) x=-1 no integer solution for y

3) x<-1, 1<LHS<2, no integer solution for y

4) x>0
Because 1+2^x+2^(2x+1)=1=y^2 mod 2. y=2m+1. Substitute and simplify.
2^x(1+2^(x+1))=4m(m+1)
m and m+1 have different parity. Right side is divisible by 8. x>=3.
Divide both sides by 4
2^(x-2)(1+2^(x+1))=m(m+1)
Here are 2 subcases: 2^(x-2) divides m or 2^(x-2) divides m+1
Let t=2^(x-2) and t>=2.

Subcase 1: 2^(x-2)=t divides m
Let m=kt, k are odd integers. (Since if y is a solution, -y is also a solution. For the purpose of simplicity, we only consider y>0 or k>0).
Substitute:
t(1+8t)=tk(kt+1)
t=(k-1)/(8-k^2)>=2. No positive odd k satisfies.

Subcase 1: 2^(x-2)=t divides m+1
m+1=kt. k are odd integers. Similarly consider k>0 only.
Substitute
t(1+8t)=(kt-1)kt
t=(k+1)/(k^2-8)>=2
k=3 is the only positive odd integer that satisfies.
t=(3+1)/(3^2-8)=4=2^(x-2)
Solve x=4 any y=+/-23

Taken together
(0, 2), (0, -2), (4, 23), (4, -23) are the only integer sets that satisfy the original equations.

[emilsriram]

Hi, I might sound stupid to some of you with this question - why can't x or k be negtive (3:58)?
But overall, I was able to easily keep up with everything else and I'm only a seventh grader. Cool solution.

[alfeusdelroy]

I came to see solutions, because i cant even understand the problem




*I regret my decisions*

[juuso]

I tried solving

2^k[2^(x-k)+2^(2x+1-k)] = (y+1)(y-1)

for some intermediate k between 0...x

In other words equating 2^k and 2^(something else) with y+1 and y-1. However this doesnt give solutions other than the trivial x=0.

It turns out the coefficient '7' is key solving this way because 7 cannot be distibuted into factors so it has to be in either term y-1 or y+1 together with 2^k, and only this way one can get the other solution (4, 23).

So the lesson is that one should try to modify the equations to get a prime factor, which will give unique solution for the (y+1) and (y-1) equations.

[benYaakov]

Assume 2^x as 'm' for some integer 'm'
Now, equations become
2m² + m + 1 = y²
Use quadratic formula .
For 'm' to be an integer, the determinant of formula must be a square number .
So,
[1 – 4(8) ] must be a square number .
But it is imaginary .
XD 😂😂😂😂