The 60th International Mathematical Olympiad (IMO) held its closing ceremony on June 21, 2019 in Bath, United Kingdom. The Chinese team (6 gold, 227 total points) and the US team (6 gold, 227 total points) tied for first place in the group. The Canadian team (1 gold/1 silver/4 bronze, with a total score of 144) won the team’s 24th place (the best result in the Canadian team’s history was the team’s 5th place in 2012).

The International Mathematics Competition (IMO) is composed of 6 questions, each with 7 points and a full score of 42 points. The competition is held in two days. Every day, participants need to solve 3 math problems within 4.5 hours. Usually the first question (that is, questions 1 and 4) per day is the simplest, the second question (that is, questions 2 and 5 is of medium difficulty, and the third question (that is, questions 3 and 6 is the most difficult. In general, IMO questions It is difficult and flexible. It requires participants to have a correct way of thinking, good mathematical literacy, solid basic skills, perseverance and certain creativity.

The 2019 IMO test questions consist of the following: The first question is a function equation question (the easier question in this year); the second question is a plane geometry question (medium difficulty); the third question is a graph theory question (the most difficult two questions) (1); Question 4 is a number thesis (relatively easy question); Question 5 is combinatorial mathematics (medium difficulty); Question 6 is a plane geometry question (one of the two most difficult questions).

Let’s take the first question as an example, through the solution of Teacher ICICAL Li for everyone to appreciate the charm of IMO.

2019 IMO Question 1: Use ℤ to denote the set of whole integers. Find all functions f: ℤ → ℤ , satisfying any integer a and b, have f(2a)+2f(b)=f(f(a+b))

solution:

We write the following function equation as (*)

f(2a)+2f(b)=f(f(a+b)) (*)

In (*), we take a=0, b=x, then (*) becomes

f(0)+2f(x)=f(f(x)) (1)

In (*), we take b=0, a=x, then (*) becomes

f(2x)+2f(0)=f(f(x)) (2)

Where x is any integer.

Comparing the right side of equations (1) and (2), both are f(f(x)). So their left sides must also be equal, that is,

f(2x)+2f(0)= f(0)+2f(x) (3)

Subtract 2f(0) on both sides of equation (3), and remember c=f(0), equation (3) is simplified as

f(2x)=2f(x)-c (4)

In equation (1), taking x=a+b, then (1) becomes

c+2f(a+b)=f(f(a+b)) (5)

Comparing (*) and (5), the right side is the same, so the left side is also equal, ie

f(2a)+2f(b) = c+2f(a+b) (6)

From equation (4), we know

f(2a) = 2f(a)-c

Let F(x)=f(x)-c, so (6) is simplified as

F(a)+F(b) = F(a+b) (7)

For any integer a, b holds. So F(x) is additive, then F(x) must be homogeneous linear in ℤ , that is to say

F(x)=kx,

Where k=F(1)

and so

f(x)=kx+c (8)

Taking the expression (8) of f(x) into (*), we simplify to get

2k(a+b)+3c=k^2 (a+b)+(k+1)c

For any integer a, b holds.

and so

2k=k^2, 3c=(k+1)c (9)

So k=0 or k=2

If k=0, then c=0. At this time, for any x, f(x)=0.

If k=2, c can be any value. At this time, for any x, f(x)=2x+c.

It has been verified that both f(x)=0 and f(x)=2x+c satisfy the function equation (*). Therefore, all f: ℤ → ℤ functions that satisfy (*) are f(x)=0 and f( x)=2x+c.

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