Find all functions f : R-->R such that f (f (x) + y^2) = x −1 + (y + 1)f (y) holds for all real numbers x, y
Problem
Source: IMOC 2021 A4
Tags: function, algebra, functional equation
11.08.2021 14:37
iflugkiwmu wrote: Find all functions f:R→R such that f(f(x)+y2)=x−1+(y+1)f(y) holds for all real numbers x,y. Let P(x,y) denote the given assertion. P(x,−1) implies f(f(x)+1)=x−1, hence f is bijective. P(x,0) yields f(f(x))=x−1+f(0). Since f is surjective, we get f(z+1)=f(z)−f(0) for all z. In particular, plugging in z=0 gives f(1)=0. Next, P(1,x) implies f(x2)=(x+1)f(x). Hence, we get (1−x)f(−x)=(1+x)f(x) for all x. Now pick a∈R with f(a)=1. P(a,x) yields f(x2)−f(0)=f(x2+1)=f(x2+f(a))=a−1+(x+1)f(x)=a−1+f(x2)⟹a=1−f(0).Hence, we obtain that f(1−f(0))=1 and f(−f(0))=1+f(0). Now 1−f(0)2=(1−f(0))f(−f(0))=(1+f(0))f(f(0))=f(0)2−1⟹f(0)2=1⟹f(0)∈{−1,1}.We first consider the case f(0)=1. Then f(f(x))=x for all x. Hence, P(f(x),y) yields f(x+y2)=f(y2+f(f(x)))=f(x)−1+(y+1)f(y)=f(x)+f(y2)−1.In particular, f(y2)+f(−y2)=2 for all y. This implies f(x)+f(−x)=2 for all x. Therefore, we get (1+x)f(x)=(1−x)f(−x)=(1−x)(2−f(x))⟹f(x)=1−x,which is indeed a solution. On the other hand, if f(0)=−1 then f(x+1)=f(x)+1 and f(f(x))=x−2 for all x. Hence, P(f(x),y) yields f(x+y2−2)=f(x)+f(y2)−1⟹f(x+y2)=f(x)+f(y2)+1.As before, we obtain f(x)+f(−x)=−2 and, together with (1−x)f(−x)=(1+x)f(x), the solution f(x)=x−1. Thus there are two solutions: f(x)=x−1 and f(x)=1−x.
11.08.2021 20:45
iflugkiwmu wrote: Find all functions f:R→R such that f(f(x)+y2)=x−1+(y+1)f(y) holds for all real numbers x,y. Let P(x,y) be the assertion f(f(x)+y2)=x−1+(y+1)f(y). P(x,0)⇒f(f(x))=x−1+f(0), so f is bijective and f(f(0))=f(0)−1. P(x,−1)⇒f(f(x)+1)=x−1 Then f(f(f(x))+1)=f(x)−1, so f(x+f(0))=f(x)−1. Moreover, we have f(x−1)=f(f(f(x)+1))=f(x)+f(0), so f(x+1)=f(x)−f(0). Setting x=0 in the last one, we obtain f(1)=0. Also, f(1−f(0))=1 and f(−f(0))=1+f(0) from the first equation. P(0,x)⇒f(x2)=(x+1)f(x)⇒(−x+1)f(−x)=(x+1)f(x) Setting x=f(0), we have: (1−f(0))f(−f(0))=(f(0)+1)f(f(0))⇒(1−f(0))(1+f(0))=(1+f(0))(f(0)−1)⇒f(0)∈{−1,1}. Case 1: f(0)=1 What we previously obtained becomes f(f(x))=x and f(x+1)=f(x)−1. P(f−1(x−y2),y+1)−P(f−1(x−y2),y)⇒Q(x,y):f(x+2y)=f(x)+f(y)−y−1 Q(0,x2)⇒f(x)+x2=f(x2) Q(x,y2)⇒f(x+y)=f(x)+f(y)−1 Let g(x)=f(x)−1, then g(x+y)=g(x)+g(y) and g(x+1)=g(x)−1. Also, g(0)=g(1)−1=0. P(x,y) becomes g(g(x)+y2)=x+y+(y+1)g(y) Using additivity, this becomes g(g(x))+g(y2)=x+y+(y+1)g(y). Setting x=0 and y↦x+y, we have g(x2)=x+(x+1)g(x) and g(x2+2xy+y2)=x+y+(x+y+1)g(x+y). Then g(2xy)=xg(y)+yg(x), and setting y=12 we test to get g(x)=−x. Then f(x)=−x+1, which works. Case 2: f(0)=−1 Then f(f(x))=x+2 and f(x+1)=f(x)+1. P(f−1(x−y2),y+1)−P(f−1(x−y2),y)⇒R(x,y):f(x+2y)=f(x)+f(y)+y+1 R(0,x2)⇒f(x)−x2=f(x2) R(x,y2)⇒f(x+y)=f(x)+f(y)+1 Now h(x)=f(x)+1 is additive. Now we have: h(h(x)+y2)=x−y+(y+1)h(y),since h(x+1)=h(x)+1. Setting y=0, we have h(h(x))=x, then additivity gives h(x2)=−x+(x+1)h(x). Then h(x2+2xy+y2)=−x−y+(x+y+1)(h(x)+h(y)), so: h(2xy)=xh(y)+yh(x),and as before we test to find h(x)=x. Then f(x)=x−1, which also works.
11.08.2021 22:45
iflugkiwmu wrote: Find all functions f : R-->R such that f (f (x) + y^2) = x −1 + (y + 1)f (y) holds for all real numbers x, y Sorry for bad formatting, this is my first long post Let g(x)=f(x+1). The given equation is equivalent to g(g(x−1)+y2−1)=x−1+(y+1)g(y−1)Which is equivalent to g(g(x)+y2+2y)=x+(y+2)g(y)Let P(x,y) denote this equation. P(x,−2 gives that g is bijective. Therefore there must exist a real number α such that g(α)=0. P(α,−1 gives g(−1)=α+g(−1). Therefore α=0 and g(0)=0. Using this, P(x,0) gives g(g(x))=x. Also, P(0,y) gives g(y2+2y)=(y+2)g(y)Now, we can get P(g(x),y) is equivalent to g(g(g(x))+y2+2y)=g(x)+(y+2)g(y)g(x+y2+2y)=g(x)+g(y2+2y)Therefore g(a+b)=g(a)+g(b) holds for all values of a∈R and b∈R+. This implies g(−x)+g(x)=g(0)=0, and therefore g(−x)=−g(x) for all real values of x and g is additive over the real numbers. Finally, combining this with g(y2+2y)=(y+2)g(y) we get g(y2+2y)=(y+2)g(y)g(y2)+2g(y)=yg(y)+2g(y)g(y2)=yg(y)And setting y:=g(y) is this equation gives g(g(y)2)=g(y)yWe combine this with g(y2)=yg(y) to get that g(g(y)2)=g(y2)g(y)2=y2g(y)=±ySuppose there exist 0≠x≠y≠0 such that g(x)=x and g(y)=−y. This would imply ±1(x+y2+2y)=x−y2−2y±2(y+x2+2x)=y−x2−2xWhich doesn't hold for any nonzero value of x,y and any election of signs for the ± (the unique nonzero solution for the first equation is (x,y)=(x,−2) and for the second is (x,y)=(−2,y). Therefore x=y=−2, but x and y have to be different and therefore there's no solution) Therefore g(x)=±x with the same sign for all x. And this gives the solutions f(x)=x−1 and f(x)=1−x, which clearly work.
11.08.2021 22:53
@above g(y)2=y2 does not imply g(y)=y for all y or g(y)=−y for all y. There can be a solution such that g(1)=1 and g(2)=−2.
11.08.2021 23:06
SerdarBozdag wrote: @above g(y)2=y2 does not imply g(y)=y for all y or g(y)=−y for all y. There can be a solution such that g(1)=1 and g(2)=−2. Oh, you are right. Thanks for pointing out, I think it's now fixed
11.08.2021 23:19
±(x+y2+2y)=x−y2−2yFor +, y=-2
11.08.2021 23:20
For -, x=0.
11.08.2021 23:37
SerdarBozdag wrote: ±(x+y2+2y)=x−y2−2yFor +, y=-2 Oh you are right again, i mistakely assumed y≠0 implied y2+2y≠0. I think its corrected now, thanks again for pointing out (I think your solution for - doesnt work since I assumed x,y≠0)
12.08.2021 14:36
Let P(x.y) be f(f(x)+y2))=x−1+(y+1)f(y) P(x,0)⇒f(f(x))=x−1+f(0)Then f is a bijection Take u such that f(u)=0 P(u,0)⇒f(0)=u−1+f(0)⇒u=1Then f(1)=0 P(1,y)⇒f(y2)=(y+1)f(y)...(1)(1,−y)⇒f(y2)=(−y+1)f(−y)...(2)P(1−(y+1)f(y),y)⇒f(1−(y+1)f(y))+y2=1take y=0 f(1−f(0))=1let v=1−f(0) P(0,v)⇒f(f(0)+v2)=vNow applies f to both sides: f(0)+v2−1+f(0)=1⇒v2−2v=0⇒v=2,0⇒f(0)=−1,1Case 1, f(0)=−1: We have f(f(x))=x−2 P(f(x+1),−1)⇒f(x)=f(x+1)−1⇒f(x+1)=f(x)+1Now the functional equation becomes that: f(f(x)+y2+1))=x+(y+1)f(y)Take x=−(y+1)f(y) then f(f(−(y+1)f(y))+y2+1)=0⇒f(−(y+1)f(y))=−y2 by (1) we have f(−f(y2))=−y2 and applies f and we obtain −f(y2)+2=f(−y2)...(3). Now, solving the system equation with (1),(2),(3) we have f(y)=y−1 for all real y and that is a solution. Case 2, f(0)=1 We have f(f(x))=x P(f(x),1)⇒f(x+1)=f(x)−1f(f(x)+y2)+1=f(f(x)+y2−1)=x+(y+1)f(y)Now take x=−(y+1)f(y)+1 f(−(y+1)f(y)+1)+y2−1=0⇒f(−(y+1)f(y)+2)=−y2⇒−(y+1)f(y)+2=f(−y2)⇒f(−y2)+f(y2)=2And one more time, solve the system equation and we have f(y)=1−y for all real y.
12.08.2021 15:28
jasperE3 wrote: ... Case 1: f(0)=1 What we previously obtained becomes f(f(x))=x and f(x+1)=f(x)−1. P(f−1(x−y2),y+1)−P(f−1(x−y2),y)⇒Q(x,y):f(x+2y)=f(x)+f(y)−y−1 Q(0,x2)⇒f(x)+x2=f(x2) Q(x,y2)⇒f(x+y)=f(x)+f(y)−1 Let g(x)=f(x)−1, then g(x+y)=g(x)+g(y) ... I think from here there is a shorter way to finish it: Q(0,x/2) yields g(2x)+x=g(x). By additivity, g(2x)=2g(x). Hence, g(x)=−x and f(x)=1−x (and similarly in the second case).
12.08.2021 18:02
Gryphos wrote: jasperE3 wrote: ... Case 1: f(0)=1 What we previously obtained becomes f(f(x))=x and f(x+1)=f(x)−1. P(f−1(x−y2),y+1)−P(f−1(x−y2),y)⇒Q(x,y):f(x+2y)=f(x)+f(y)−y−1 Q(0,x2)⇒f(x)+x2=f(x2) Q(x,y2)⇒f(x+y)=f(x)+f(y)−1 Let g(x)=f(x)−1, then g(x+y)=g(x)+g(y) ... I think from here there is a shorter way to finish it: Q(0,x/2) yields g(2x)+x=g(x). By additivity, g(2x)=2g(x). Hence, g(x)=−x and f(x)=1−x (and similarly in the second case). Thank you.
31.12.2021 15:40
(y+1)f(y)=(1−y)f(−y)f(x+y2−1)=f(x)+f(y2)f(x2)=ax2+bRemaining easy
18.06.2022 20:10
Find all functions f:R→R such that f(f(x)+y2)=x−1+(y+1)f(y) holds for all real numbers x,y.
23.03.2024 21:18
jasperE3 wrote: Find all functions f:R→R such that f(f(x)+y2)=x−1+(y+1)f(y) holds for all real numbers x,y. Let P(x,y) be the assertion of the problem. It's obvious from this equation that f is both injective and surjective. By P(1,−1) we have there is t such that f(t)=0. Then P(t,0) infers that f(1)=0. P(1,y) gives that f(y2)=(y+1)f(y). We name this relation as A. Also, by P(x,0) we have f(f(x))=x−1+f(0). We name this relation as B. We know from A and B that f(y2+f(x))=f(y2)+f(f(x))−f(0). This with f being surjective plus some calculations give that f is semi - additive. i.e. for all x,y∈R f(x+y)=f(x)+f(y)−f(0).Thus we have f(2x)=2f(x)−f(0) and f(4x)=4f(x)−3f(0). Plugging y→2y in relation A using these two equations we have f(4y2)=4yf(y)+4f(y)−3f(0),(2y+1)f(2y)=(2y+1)(2f(y)−f(0))=4yf(y)+2f(y)−(2y+1)f(0).Then 4yf(y)+4f(y)−3f(0)=4yf(y)+2f(y)−(2y+1)f(0)→f(y)=(1−y)f(0).Plugging into P(x,y) we get that f(0)=±1 and then answers are f(x)=x−1:∀x and f(x)=1−x:∀x.