Find all functions f:R→R such that for all real numbers x and y, f(x+f(y))+xy=f(x)f(y)+f(x)+y. Andrew Carratu
Problem
Source: ELMO Shortlist 2024/A3
Tags: algebra, ELMO Shortlist, functional equation
22.06.2024 18:42
22.06.2024 22:37
22.06.2024 22:50
Let P(x,y) denote the relation f(x+f(y))+xy=f(x)f(y)+f(x)+y.
23.06.2024 02:33
The solutions are f(x)≡x and f(x)≡−x, which work. Let P(x,y) denote the assertion that f(x+f(y))+xy=f(x)f(y)+f(x)+y.P(0,x) gives f(f(x))=f(0)f(x)+f(0)+x. Notice that P(f(x),y) gives f(f(x)+f(y))+yf(x)=f(f(x))f(y)+f(f(x))+y,and subtracting this by its symmetric variant with x and y swapped gives yf(x)−xf(y)=f(f(x))f(y)+f(f(x))+y−f(f(y))f(x)−f(f(y))−x.Using P(0,x) to rewrite all the nested f's, the RHS simplifies to xf(y)−yf(x), so we have xf(y)=yf(x). Plugging in y=1, we get f(x)=xf(1). If f(1)=c, then the functional equation simplifies to c(x+cy)+xy=c2xy+cx+y, which is only true for c=1,−1, as desired. ◻
23.06.2024 05:23
Solved with Orthogonal.. We claim the only functions are f(x)≡x and f(x)≡−x. It is easy to check that these work. Let P(x,y) denote the given assertion. P(0,x) gives f(f(x))=f(0)f(x)+f(0)+x. Comparing P(f(x),1) and P(f(1),x) gives f(x)≡f(1)x. We can show that f(1)=±1, as desired. ◻
23.06.2024 05:35
The only solutions are f(x)=x and f(x)=−x, which work. Let P(x,y) denote the given assertion. Case 1: f(0)=0 Then P(0,x):f(f(x))=x, so f is an involution. P(1,f(x)):f(x+1)=xf(1)+f(1), so f is linear, meaning f(x)=cx for some constant c. Since f is an involution, either c=1 or c=−1. Case 2: f(0)≠0. Claim: f is injective. Proof: If f(a)=f(b), then P(0,a) compared with P(0,b) gives a=b. ◻ Claim: x+f(x) is injective. Proof: Suppose a+f(a)=b+f(b). Then P(x,x):f(x+f(x))−(x+f(x))=(f(x)−x)(x+f(x)), so comparing x=a and x=b gives that (a+f(a))(f(a)−a))=(a+f(a))(f(b)−b). If a+f(a)=0, then P(a,a):f(0)=0, absurd. Hence a+f(a)≠0, so f(a)−a=f(b)−b. Hence (f(a)−a)+(f(a)+a)=(f(b)−b)+(f(b)+b)⟹f(a)=f(b)⟹a=b◻ Now, the equation gives f(x+f(y))−f(x)−y=f(x)f(y)−xy,so swapping x,y here gives f(x+f(y))−f(x)−y=f(y+f(x))−f(y)−x, so f(x+f(y))+(x+f(y))=f(y+f(x))+(y+f(x)), which by our earlier claim implies x+f(y)=y+f(x)⟹f(x)−x=f(y)−y, so f(x)=x+c for some constant c. P(0,0):f(f(0))=f(0)2+f(0), so 2c=c2+c⟹c∈{0,1}. Checking, we see that c=1 fails, so c=0, but this is absurd since f(0)≠0.
23.06.2024 14:12
IAmTheHazard wrote: Find all functions f:R→R such that for all real numbers x and y, f(x+f(y))+xy=f(x)f(y)+f(x)+y. Andrew Carratu f(x)f(y)f(z)=f(x+f(y))f(z)+xyf(z)−f(z)f(x)−yf(z)=f(x+f(y)+f(z))+(x+f(y))f(z)−f(x+f(y))−z+xyf(z)−f(x+f(z))−xz+f(x)+z−yf(z) Now by the symmetry of y,z we get that: f(x+f(y)+f(z))+(x+f(y))f(z)−f(x+f(y))−z+xyf(z)−f(x+f(z))−xz+f(x)+z−yf(z)= f(x+f(z)+f(y))+(x+f(z))f(y)−f(x+f(z))−y+xzf(y)−f(x+f(y))−xy+f(x)+y−zf(y) ⇒xf(z)+xyf(z)−xz−yf(z)=xf(y)+xzf(y)−xy−zf(y) In this for x=0 we get that: −yf(z)=−zf(y)⇒f(z)z=f(y)y⇒f(x)=cx Now in the start we have that: cx+c2y+xy=c2xy+cx+y⇒c2y=y⇒c=+−1 So we get that: f(x)≡x and f(x)≡−x. In the same way we can solve the problem if f:R+→R+
23.06.2024 15:44
Cute one. P(0,x) -> f(f(x))=f(x)f(0)+f(0)+x (1) By looking at P(f(x),y) and P(f(y),x) we have f(f(x)+f(y))=f(f(x))f(y)+f(f(x))+y−f(x)y=f(f(y))f(x)+f(f(y))+x−f(y)x, and by using (1) we observe that xf(y)=yf(x), which means that f(x)=cx. And by plugging it into equation one can prove that only solutions are f(x)≡x and f(x)≡−x.
02.07.2024 23:41
I am not sure about the surjectivety part Answer: f(x)=±x∀x∈R Solution: Let P(x,y)-denote the given assertion. Claim: f-bijective Proof: P(0,x)⟹f(f(x))=f(x)f(0)+f(0)+x⟹f-is surjective
Since f is both injective and surjective we get that f-is bijective. ◻ Claim: f(0)=0 Proof: Since f-is surjective there exists an α such that f(α)=0⟺∃α∈R:f(α)=0 P(α,0)⟹f(α+f(0))=0=f(α)⟹f(α+f(0))=f(α)f−injective⟹α+f(0)=α⟹f(0)=0. ◻ Claim: f(x)=±x∀∈R Proof: P(0,x)⟹f(f(x))=x ...(∗) P(x,x)⟹f(x+f(x))+x2=f(x)2+f(x)+x⟹f(x+f(x))−x−f(x)=f(x)2−x2 ...(3) P(f(x),f(x))(∗)⟹f(x+f(x))+f(x)2=x2+x+f(x)⟹f(x+f(x))−x−f(x)=x2−f(x)2 ...(4) (3)−(4)⟹f(x)2−x2=x2−f(x)2⟹f(x)2=x2⟹f(x)=±x∀∈R ◼ POINTWISE TRAP!
01.08.2024 20:13
This solution might be one of the lamest ever : P(x,y):f(x+f(y))+xy=f(x)f(y)+f(x)+y We will just simplify P(f(y),1) to find the answer : f(f(y)+f(1))+f(y)−1=f(f(y))(f(1)+1) Now : P(0,y) , gives f(f(y))=f(0)(f(y)+1)+y So for the right side we get that : f(f(y))(f(1)+1)=f(y).f(0)(f(1)+1)+f(0)(f(1)+1)+y(f(1)+1) For the left side using P(f(1),y) , we get that : f(f(y)+f(1))=f(f(1))(f(y)+1)+y(1−f(1))=[f(0)(f(1)+1)+1](f(y)+1)+y(1−f(1)) So :f(f(y)+f(1))+f(y)−1=f(0)(f(1)+1).f(y)+f(0)(f(1)+1)+f(y)+1+f(y)−1+y(1−f(1)) Identifying both sides will be left with : 2f(y)+y(1−f(1))=y(f(1)+1) , thus f(y)=yf(1) , the rest is obvious .
01.08.2024 20:26
For above that equation doesn't imply that f is surjective as changing x to some value will also change f(x)
01.08.2024 20:39
For above that equation doesn't imply that f is surjective as changing x to some value will also change f(x) an example is f(x)=(x+1)2 ,will get on the right : (x+1)2+(x+1) , which doesn't take all values in R such as −4
09.08.2024 18:42
The solutions are f(x)=x and f(x)=−x, which are both easy to verify. Claim: f(0)=0. Proof: Assume that f(0)≠0 FTSOC. Taking P(x,0) gives us f(x+f(0))=f(x)⋅(1+f(0)),so incrementing x by f(0) gives some geometric sequence. But the original equation can be rewritten as f(x+f(y))−f(x)(1+f(y))=y(1−x),and if we increment x by f(0) here, the LHS will multiply by 1+f(0). However, the RHS cannot grow exponentially since it is negative for all x>1 and positive for all x<1, something no exponential function does. This gives us a contradiction. Now, taking P(0,x) gives us f(f(x))=x, implying surjectivity, and P(1,x) gives us f(1+f(x))=f(1)(1+f(x)),which is enough to imply that f is linear. Plugging this into our original equation gives the claimed solutions.
18.09.2024 03:23
f(x+f(y))+x.y=f(x).f(y)+f(x)+y Let' s call this P(x,y) P(0,0) gives us f(f(0))=f(0)^2+f(0) this call * P(x, 0) gives us f(x+f(0))=f(x).f(0)+f(x) x->-f(0) gives us f(0)=f(-f(0)).f(0)+f(-f(0)) from this, we can get that: f(0)/f(f(0))=f(0)+1 from this, * gives us f(0)^2=f(-f(0)).f(f(0))=A P(-f(0), f(0)) f(-f(0)+f(f(0)))-f(0)^2=A+f(-f(0))+f(0) f(-f(0)+f(0)^2+f(0))=2.f(0)^2+f(0)+f(-f(0)) this give f(f(0)^2)=f(0)^2+f(f(0))+f(-f(0)) P(f(0), f(0)^2) f(f(0)+f(f(0)^2))+f(0)^3=f(f(0)).f(f(0)^2)+f(f(0))+f(0)^2 f(f(0)+f(0)^2+f(f(0))+f(-f(0)))+f(0)^3=f(f(0)).f(f(0)^2)+f(f(0))+f(0)^2 f(2f(f(0))+f(-f(0)))+f(0)^3=f(f(0)).f(f(0)^2)+f(f(0))+f(0)^2 Now, P(2f(f(0)), -f(0)), which give us f(2f(f(0))+f(-f(0)))-2.f(f(0)).f(0)=f(2f(f(0))).f(-f(0))+f(2f(f(0)))-f(0) Now P(x, f(0)) W f(x+f(f(0)))+f(0).x=f(f(0)).f(x)+f(x)+f(0) x->f(f(0)), I don' t know why I didn't put this here W f(2f(f(0)))+f(0).f(f(0))=f(f(0)).f(f(f(0)))+f(f(f(0)))+f(0) Now I want to find f(f(f(0))) if we take this P(0, y) we will get this f(f(y))=f(0).f(y)+f(0)+y y->f(0), f(f(f(0)))=f(f(0)).f(0)+2.f(0) if we call f(0) to be (a) and f(f(0)) to be (b), where b=a^2+a( from the beginning) Then, f(2f(f(0)))+f(0).f(f(0))=f(f(0)).(f(f(0)).f(0)+2.f(0))+f(f(0)).f(0)+2.f(0)+f(0) we will get this f(2f(f(0)))+a.b=b.(b.a+2.a)+b.a+2.a+a, which is equivalent to f(2f(f(0)))=b(b.a+2.a)+3.a f(2f(f(0))+f(-f(0)))-2.f(f(0)).f(0)=f(2f(f(0))).f(-f(0))+f(2f(f(0)))-f(0), Now, from this, we have: f(2f(f(0))+f(-f(0)))-2.a.b=(b.(b.a+2.a)+3.a).f(-f(0))+b.(b.a+2.a)+3.a-a, call this T From this, f(2f(f(0))+f(-f(0)))+f(0)^3=f(f(0)).f(f(0)^2)+f(f(0))+f(0)^2, we have that the left side is equal to, also I want to say that f(f(0)^2)=f(0)^2+f(f(0))+f(-f(0))=a^2+b+f(-f(0)) Now, nightmare is coming The left side=b(a^2+b+f(-f(0)))+b+a^2-a^3 But we have that: f(0)^2=f(f(0)).f(-f(0)), which is equal to a^2=b.f(-f(0)), i.e the left side=a^2.b+b^2+a^2+b+a^2-a^3=a^2.b+b^2+2.a^2+b-a^3, call D Now-if we multiply whole equation with b, we will get this (The left side).b=2.b^2.a+(b^2.a+2.a.b+3.a).a^2+b^3.a+2.a.b^2+2.a.b From D multiply with (b), we get this a^2.b^2+b^3+2.a^2.b+b^2-a^3.b to be equal to 2.b^2.a+b^2.a^3+2.a^3.b+3.a^3+b^3.a+2.a.b^2+2.a.b Now, if this above wasn' t nightmare, after we will replaceable b with a^2+a- this will be a real bad nightmare, let's start (teeth) a^2.(a^2+a)^2+(a^2+a)^3+2.a^2.(a^2+a)+(a^2+a)^2-a^3.(a^2+a)=2.(a^2+a)^2.a+(a^2+a)^2.a^3+2.a^3.(a^2+a)+3.a^3+(a^2+a)^3.a+2.a.(a^2+a)^2+2.a.(a^2+a) After we use Wolfram Alpha, oops, just a joke, I figured it all myself (sad) , we will get this +2.a^7+3.a^6+6.a^5+5.a^4+4.a^3+a^2=0, the solution of this two in Real numbers, the one of them is approximately to some digit-negative and stranger, the other solution is 0 Then we have a=0=f(0) Then, if we put this in this: f(f(y)=f(0).f(y)+f(0)+y=y, i.e the function is involution Now, let P(f(x), f(y)) gives us f(f(x)+y)+f(x).f(y)=x.y+f(y)+x But P(y,x) gives us f(f(x)+y)+x.y=f(x).f(y)+f(y)+x, call G Let sum this two, then this gives us f(f(x)+y)=f(y)+x Put this in G gives x.y=f(x).f(y), y->x gives f(x)^2=x^2 f(x)=+x, f(x)=-x
02.01.2025 16:35
P(f(x),y) ,change x,y after this and y=0 implies that f(f(x))×(f(0)+1)=f(x)×f(f(0))+f(f(0))+x-x×f(0). Plug x=0 we get f(f(y))=f(0)×f(y)+f(0)+y we get f is injective. Also we use this equation above and we get f(x)×(f(0)²+f(0)-f(f(0))) =-2x×f(0)-(f(0)²+f(0)-f((0))) if f(0)²+f(0)-f(f(0)) isn't equal 0 then f(x)=ax+b we check and we get f(x)=x or f(x)=-x. Other case implies that f(0)=0.In common equation we plug x=0 and we get f(f(x))=x.In next steps P(x,f(y)) and we change x,y we get x×f(y)=y×f(x) y=1 and we get f(x)=cx.It is easy to check and find f(x)=x and f(x)=-x answers.We are done.
02.01.2025 16:49
P(0,x)⟹f(f(x))=f(0)(f(x)+1) P(f(x),y)⟹f(f(x)+f(y))+yf(x)=f(0)(f(x)+1)(f(y)+1)+xf(y)+x+y P(f(x),y)−P(x,f(y))⟹xf(y)=yf(x)⟹f(x)=±x