The First IMO Problem | International Mathematical Olympiad 1959 Problem 1

#Math #MathOlympiad #Algebra

In this video we are going to solve the first IMO problem, problem 1 in IMO 1959.

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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.

[theevilmathematician]

I solved using a different method. By Bezout's Lemma, 3*(14n+3) -2 * (21n+4) =1, so GCD of numerator and denominator is 1 and (21n+4)/14n+3 is irreducible for all natural numbers, n.

[marcod]

What about this super simple method:

if an integer p divides two integers, then p divides their difference too.

So if 21n+4 and 14n+3 share a common factor p, this p is also a factor of their difference 7n+1.

However, if p divides 14n+3 and 7n+1, it divides their difference too, which is 7n+2.

Going on, p has to divide the difference between 7n+2 and 7n+1, too, which is 1.

Only 1 can do that, so no other common factors are possible.

[WasTaken]

Me: I can solve an IMO problem
The problem:

[petersievert]

Totally different approach, not very elegant, I guess, but with simple tools:

Assume 14n+3 = xy with x, y being positive integers and x > 1.
Then 21n + 4.5 = 3xy/2
21n + 4 = 3xy/2 - 0.5
(21n + 4)/x = 3y/2 - 0.5/x
As 14n+3 is odd, so must y be, thus the fractional part of 3y/2 is 0.5
Because 0.5/x < 0.5, 3y/2 - 0.5/x clearly cannot be an integer, because it's fractional part is 0.5 - 0.5/x
Thus (21n+4)/x is not an integer, thus x does not divide 21n+4
Thus 14n+3 cannot have a common divisor x > 1 with 21n+4

[richardfredlund]

gcd(2*(21n+4), 3*(14n+3)) = gcd(42n+8, 42n+9) =1 (because the numbers differ by 1). It follows that gcd(21n+4, 14n+3)=1 because if they shared a common factor when you multiply one of them by 2 and the other by 3 they would still share a common factor.

[pdpgrgn]

Is this solution valid ? :
ax+by = c has integer solution x, y if and only if gcd(a, b) divides c.
(21n+4).(-2) + (14n+3).(3) = 1
gcd (21n+4, 14n+3) divides 1
gcd (21n+4, 14n+3) is 1

[ayushrudra]

Bruh if only i qualified for imo in 1959 id get gold lmao

[punkrockeris]

I relied on the same fact about the GCD but there is a minor issue with the delivery here that can help understand the reason why you choose to go down the path of analysing the GCD.
Since the problem asks to provide a PROOF, you may want to actually build a proper proof by contradiction. You would assume that the fraction is not irreducible, therefore the GCD between numerator and denominator shouldn't be 1 for any value of n first, AND THEN you start doing what you showed here. Finally, since the GCD turned to be 1, you found your contradiction stems from assuming the fraction is reducible so the opposite should be true.

[vincentlaw]

Mathematical induction should work as well?