Function math please help on how to find a+b from the equation?

Question

For any rational number x, y, function f(x) is a real number and f(x+y) = f(x)f(y) - f(xy) + 1. Again f(2017) ≠ f(2018). f(2017/2018) = a/b, where a , b are co-prime. What is a+b?

Answer 1

In function problems it is often useful to start thinking about some specific examples involving the additive identity (0) and the multiplicative identity (1)

When x=y=0, you have that f(0)=f^2(0)-f(0)+1 which has solution f(0)=1

When x = any number and y=0, you get f(x)=f(x) which isn't so helpful.

That means we have no obvious good way to get to f(1) so lets assume f(1)=r and see if we can build a sequence from that

Note that when y=1, we have

f(x+1)=f(x)r-f(x)+1 .................................................................................................(*)

Then

f(2)=r^2-r+1

f(3)=r^3-2r^2+2r

f(4)=r^4-3r^3+4r^2-2r+1

etc.

But wait! We can get at f(4) not only as f(3+1) but also as f(2+2). Lets see what happens when we set x=y=2:

f(2+2)=f(2)f(2)-f(2*2)+1 ie

2f(4)=f^2(2)+1

Now use the expressions above for f(2) and f(4) to get

2(r^4-3r^3+4r^2-2r+1)=(r^2-r+1)^2+1

Open brackets and move everything to one side to get

r^4-4r^3+5r^2-2r=0

Factor:

r(r-1)^2(r-2)=0

This means that r=0, 1 or 2! Lets go through all the cases:

Case 1: r=0 Then note that f(even)=1, f(odd)=0, ie, for integer n we can write

f(b)=0.5(*1+(-1)^n)

Also (*) becomes

f(x+1)=1-f(x)

Now we need to look into fractions:

Let x=1/2 and y=0 then in the original equation

f(1/2)=f(0+1/2)=f(0)f(1/2)-f(0)+1

f(0)=f(even)=1 f(1)=f(odd)=0 so the above becomes

f(1/2)=f(1/2)+1 which means we have no solution in this case

Case 2: r=1

Then (*) becomes f(x)=1 which means f(2017)=f(2018) which is not allowed so there is no solution

Case 3: r=2

Then (*) becomes

f(x+1)=f(x)+1

For integers n then f(n)=n+1 (you can show that with mathematical induction for example). We will show that f(x)=x+1 for all rational x not just integers. To do so, we will consider the function g(x)=f(x)-x-1.

Note that g(x+1)=f(x+1)-(x+1)-1=f(x+1)-x-2=f(x)+1-x-2=f(x)-x+1=g(x) So g(x) has a period of 1.

Furthermore g(1)=f(1)-1-1=2-1-1=0

This means that for any integer n g(n)=0,

You can show that

g(x+y)=g(x)g(y)+(y+1)g(x)+(x+1)g(y)-g(xy) ....................................................(**)

Now let p and q be co-prime integers with g>0. Let x=p/q and y=q

g(x+y)=g(p/q+q)=g(p/q) because of periodicity of 1

g(x*y)=g(p/q*q)=g(p)=0

Substitute in (**) to get

g(p/q)=g(p/q)g(q)+(q+1)g(p/q)+(p/q+1)g(q)-g(p/q*q)

recall that g(q)=0 since q is an integer thus

g(p/q)=(q+1)g(p/q)

Since the above is true for all integer q>0, that means g(p/q)=0 ie

f(x)=x+1 for all rational x

Then f(2017/2018)=2017/2018+1 =(2017+2018)/2018 =4035/2018 and from there you can get any combination of a and b such as a+b