25 Functional Equations Problems

Problem 1 Find all functions \(f:\mathbb{Q}\rightarrow\mathbb{Q}\) such that \(f(1)=2\) and \(f(xy)=f(x)f(y)-f(x+y)+1\).

Problem 2 (Belarus 1997) Find all functions \(g:\mathbb{R}\rightarrow\mathbb{R}\) such that for arbitrary real numbers \(x\) and \(y\): \[g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y).\]

Problem 3 The function \(f:\mathbb{R}\rightarrow\mathbb{R}\) satisfies \(x+f(x)=f(f(x))\) for every \(x\in\mathbb{R}\). Find all solutions of the equation \(f(f(x))=0\).

Problem 4 Find all injective functions \(f:\mathbb{N}\rightarrow\mathbb{N}\) that satisfy: \[(a)\ f(f(m)+f(n))=f(f(m))+f(n), \quad (b)\ f(1)=2,\ f(2)=4.\]

Problem 5 (BMO 1997, 2000) Solve the functional equation \[f(xf(x)+f(y))=y+f(x)^2,\ x,y\in\mathbb{R}.\]

Problem 6 (IMO 1979, shortlist) Given a function \(f:\mathbb{R}\rightarrow\mathbb{R}\), if for every two real numbers \(x\) and \(y\) the equality \(f(xy+x+y)=f(xy)+f(x)+f(y)\) holds, prove that \(f(x+y)=f(x)+f(y)\) for every two real numbers \(x\) and \(y\).

Problem 7 Does there exist a function \(f:\mathbb{R}\rightarrow\mathbb{R}\) such that \(f(f(x))=x^2-2\) for every real number \(x\)?

Problem 8 Find all functions \(f:\mathbb{R}^+\rightarrow\mathbb{R}^+\) such that \(f(x)f(yf(x))=f(x+y)\) for every two positive real numbers \(x,y\).

Problem 9 (IMO 2000, shortlist) Find all pairs of functions \(f:\mathbb{R}\rightarrow\mathbb{R}\) and \(g:\mathbb{R}\rightarrow\mathbb{R}\) such that for every two real numbers \(x,y\) the following relation holds: \[f(x+g(y))=xf(y)-yf(x)+g(x).\]

Problem 10 (IMO 1992, shortlist) Find all functions \(f:\mathbb{R}^+\rightarrow\mathbb{R}^+\) which satisfy \[f(f(x))+af(x)=b(a+b)x.\]

Problem 11 (Vietnam 2003) Let \(F\) be the set of all functions \(f:\mathbb{R}^+\rightarrow\mathbb{R}^+\) which satisfy the inequality \(f(3x)\geq f(f(2x))+x\), for every positive real number \(x\). Find the largest real number \(\alpha\) such that for all functions \(f\in F\): \(f(x)\geq \alpha\cdot x\).

Problem 12 Find all functions \(f,g,h:\mathbb{R}\rightarrow\mathbb{R}\) that satisfy \[f(x+y)+g(x-y)=2h(x)+2h(y).\]

Problem 13 Find all functions \(f:\mathbb{Q}\rightarrow\mathbb{Q}\) for which \[f(xy)=f(x)f(y)-f(x+y)+1.\] Solve the same problem for the case \(f:\mathbb{R}\rightarrow\mathbb{R}\).

Problem 14 (IMO 2003, shortlist) Let \(\mathbb{R}^+\) denote the set of positive real numbers. Find all functions \(f:\mathbb{R}^+\rightarrow \mathbb{R}^+\) that satisfy the following conditions:

• (i) \(f(xyz)+f(x)+f(y)+f(z)=f(\sqrt{xy})f(\sqrt{yz})f(\sqrt{zx})\)
  
  
• (ii) \(f(x) < f(y)\) for all \(1\leq x < y\).

First notice that the solution of this functional equation is not one of the common solutions that we are used to work with. Namely one of the solutions is \(f(x)=x+ \frac 1x\) which tells us that this equality is unlikely to be shown reducing to the Cauchy equation. First, setting \(x=y=z=1\) we get \(f(1)=2\) (since \(f(1) > 0\)). One of the properties of the solution suggested above is \(f(x)=f(1/x)\), and proving this equality will be our next step. Putting \(x=ts\), \(y=\frac ts\), \(z=\frac st\) in (i) gives \[ f(t)f(s)=f(ts)+f(t/s).\quad\quad\quad\quad\quad (1)\] In particular, for \(s=1\) the last equality yields \(f(t)=f(1/t)\); hence \(f(t)\geq f(1)=2\) for each \(t\). It follows that there exists \(g(t)\geq1\) such that \( f(t)=g(t)+ \frac1{g(t)}\). Now it follows by induction from (1) that \(g(t^n)=g(t)^n\) for every integer \(n\), and therefore \(g(t^q)= g(t)^q\) for every rational \(q\). Consequently, if \(t > 1\) is fixed, we have \(f(t^q)=a^q+a^{-q}\), where \(a=g(t)\). But since the set of \(a^q\) (\(q\in\mathbb{Q}\)) is dense in \(\mathbb{R}^+\) and \(f\) is monotone on \((0,1]\) and \([1,\infty)\), it follows that \(f(t^r)= a^r+a^{-r}\) for every real \(r\). Therefore, if \(k\) is such that \(t^k=a\), we have \[f(x)=x^k+x^{-k} \hspace{5mm}\mbox{for every }x \in\mathbb{R}. \]

Problem 15 Find all functions \(f:[1,\infty)\rightarrow [1,\infty)\) that satisfy:

• (i) \(f(x)\leq 2(1+x)\) for every \(x\in [1,\infty)\);
  
  
• (ii)\(xf(x+1)=f(x)^2-1\) for every \(x\in [1,\infty)\).

It is not hard to see that \(f(x)=x+1\) is a solution. Let us prove that this is the only solution. Using the given conditions we get \[f(x)^2=xf(x+1)+1\leq x(2(x+1))+1 < 2(1+x)^2,\] i.e. \(f(x)\leq \sqrt{2}(1+x)\). With this we have found the upper bound for \(f(x)\). Since our goal is to prove \(f(x)=x+1\) we will use the same method for lowering the upper bound. Similarly we get \[f(x)^2=xf(x+1)+1\leq x(\sqrt{2}(x+1))+1 < 2^{1/4}(1+x)^2.\] Now it is clear that we should use induction to prove \[f(x) < 2^{1/2^k}(1+x),\] for every \(k\). However this is shown in the same way as the previous two inequalities. Since \(2^{1/2^k}\rightarrow 1\) as \(k\rightarrow+\infty\), hence for fixed \(x\) we can't have \(f(x) > x+1\). This implies \(f(x)\leq x+1\) for every real number \(x\geq 1\). It remains to show that \(f(x)\geq x+1\), for \(x\geq 1\). We will use the similar argument. From the fact that the range is \([1,+\infty)\) we get \( \frac{f(x)^2-1}{x}=f(x+1)\geq 1\), i.e. \(f(x)\geq \sqrt{x+1} > x^{1/2}\). We further have \(f(x)^2=1+xf(x+1) > 1+x\sqrt{x+2} > x^{3/2}\) and similarly by induction \[f(x) > x^{1-1/2^k}.\] Passing to the limit we further have \(f(x)\geq x\). Now again from the given equality we get \(f(x)^2=1+xf(x+1)\geq (x+1/2)^2\), i.el \(f(x)\geq x+1/2\). Using the induction we get \(f(x)\geq x+1- \frac{1}{2^k}\), and passing to the limit we get the required inequality \(f(x)\geq x+1\).

Problem 16 (IMO 1999, probelm 6) Find all functions \(f:\mathbb{R}\rightarrow \mathbb{R}\) such that \[f(x-f(y))=f(f(y))+xf(y)+f(x)-1.\]

Let \(A=\{f(x)\,|\, x\in\mathbb{R}\}\), i.e. \(A=f(\mathbb{R})\). We will determine the value of the function on \(A\). Let \(x=f(y)\in A\), for some \(y\). From the given equality we have \(f(0)=f(x)+x^2+f(x)-1\), i.e. \[f(x)= \frac{c+1}{2}-\frac{x^2}{2},\] where \(f(0)=c\). Now it is clear that we have to analyze set \(A\) further. Setting \(x=y=0\) in the original equation we get \(f(-c)=f(c)+c-1\), hence \(c\neq 0\). Furthermore, plugging \(y=0\) in the original equation we get \(f(x-c)-f(x)=cx+f(c)-1\). Since the range of the function (on \(x\)) on the right-hand side is entire \(\mathbb{R}\), we get \(\{f(x-c)-f(x)\,|\, x\in\mathbb{R}\} =\mathbb R\), i.e. \(A-A=\mathbb{R}\). Hence for every real number \(x\) there are real numbers \(y_1,y_2\in A\) such that \(x=y_1-y_2\). Now we have \[f(x) = f(y_1-y_2)= f(y_1-f(z))= f(f(z))+y_1f(z)+f(y_1)-1\] \[= f(y_1)+f(y_2)+y_1y_2-1=c-\frac{x^2}{2}.\] From the original equation we easily get \(c=1\). It is easy to show that the function \(f(x)=1- \frac{x^2}{2}\) satisfies the given equation.

Problem 17 Given an integer \(n\), let \(f:\mathbb{R}\rightarrow\mathbb{R}\) be a continuous function satisfying \(f(0)=0\), \(f(1)=1\), and \(f^{(n)}(x)=x\), for every \(x\in[0,1]\). Prove that \(f(x)=x\) for each \(x\in[0,1]\).

Problem 18 Find all functions \(f: (0,+\infty)\rightarrow(0,+\infty)\) that satisfy \(f(f(x)+y)=xf(1+xy)\) for all \(x,y\in(0,+\infty)\).

Problem 19 (Bulgaria 1998) Prove that there is no function \(f:\mathbb{R}^+\rightarrow\mathbb{R}^+\) such that \(f(x)^2\geq f(x+y)(f(x)+y)\) for every two positive real numbers \(x\) and \(y\).

Problem 20 Let \(f:\mathbb{N}\rightarrow\mathbb{N}\) be a function satisfying \[f(1)=2,\quad f(2)=1,\quad f(3n)=3f(n),\quad f(3n+1)=3f(n)+2,\quad f(3n+2)=3f(n)+1.\] Find the number of integers \(n\leq 2006\) for which \(f(n)=2n\).

Problem 21 (BMO 2003, shortlist) Find all possible values for \(f\Big( \frac{2004}{2003}\Big)\) if \(f:\mathbb{Q}\rightarrow[0,+\infty)\) is the function satisfying the conditions:

• (i) \(f(xy)=f(x)f(y)\) for all \(x,y\in\mathbb{Q}\);
  
  
• (ii) \(f(x)\leq 1\Rightarrow f(x+1)\leq 1\) for all \(x\in\mathbb{Q}\);
  
  
• (iii) \(f\Big( \frac{2003}{2002}\Big)=2\).

Notice that from (i) and (ii) we conclude that \(f(x) > 0\), for every rational \(x\). Now (i) implies that for \(x=y=1\) we get \(f(1)=0\) and similarly for \(x=y=-1\) we get \(f(-1)=1\). By induction \(f(x)\leq 1\) for every integer \(x\). For \(f(x)\leq f(y)\) from \(f\Big( \frac yx\Big)f(y)=f(x)\) we have that \(f\Big( \frac yx\Big)\leq 1\), and according to (ii) \(f\Big( \frac yx+1\Big)\leq 1.\) This implies \[f(x+y)=f\Big( \frac yx+1\Big)f(x)\leq f(x),\] hence \(f(x+y)\leq \max \{f(x),f(y)\}\), for every \(x,y\in\mathbb Q\). Now you might wonder how did we get this idea. There is one often neglected fact that for every two relatively prime numbers \(u\) and \(v\), there are integers \(a\) and \(b\) such that \(au+bv=1\). What is all of this good for? We got that \(f(1)=1\), and we know that \(f(x)\leq 1\) for all \(x\in\mathbb{Z}\) and since \(1\) is the maximum of the function on \(\mathbb{Z}\) and since we have the previous inequality our goal is to show that the value of the function is \(1\) for a bigger class of integers. We will do this for prime numbers. If for every prime \(p\) we have \(f(p)=1\) then \(f(x)=1\) for every integer implying \(f(x)\equiv 1\) which contradicts (iii). Assume therefore that \(f(p)\neq 1\) for some \(p\in\mathbb{P}\). There are \(a\) and \(b\) such that \(ap+bq=1\) implying \(f(1)=f(ap+bq)\leq\max\{f(ap),f(bq)\}.\) Now we must have \(f(bq)=1\) implying that \(f(q)=1\) for every other prime number \(q\). From (iii) we have \[f\Big(\frac{2003}{2002}\Big)=\frac{f(2003)}{f(2)f(7)f(11)f(13)}=2,\] hence only one of the numbers \(f(2),f(7),f(11),f(13)\) is equal to \(1/2\). Thus \(f(3)=f(167)=f(2003)\) giving: \[f\Big(\frac{2004}{2003}\Big)=\frac{f(2)^2f(3)f(167)} {f(2003)}=f(2)^2.\] If \(f(2)=1/2\) then \( f\Big(\frac{2003}{2002}\Big)= \frac 14\), otherwise it is \(1\).

It remains to construct one function for each of the given values. For the first value it is the multiplicative function taking the value \(1/2\) at the point \(2\), and \(1\) for all other prime numbers; in the second case it is a the multiplicative function that takes the value \(1/2\) at, for example, \(7\) and takes \(1\) at all other prime numbers. For these functions we only need to verify the condition (ii), but that is also very easy to verify.

Problem 22 Let \(I=[0,1]\), \(G=I\times I\) and \(k\in\mathbb{N}\). Find all \(f:G\rightarrow I\) such that for all \(x,y,z\in I\) the following statements hold:

• (i) \(f(f(x,y),z)=f(x,f(y,z))\);
  
  
• (ii) \(f(x,1)=x\), \(f(x,y)=f(y,x)\);
  
  
• (iii) \(f(zx,zy)=z^kf(x,y)\) for every \(x,y,z\in I\), where \(k\) is a fixed real number.

The function of several variables appears in this problem. In most cases we use the same methods as in the case of a single-variable functions. From the condition (ii) we get \(f(1,0)=f(0,1)=0\), and from (iii) we get \(f(0,x)=f(x,0)=x^kf(1,0)=0\). This means that \(f\) is entirely defined on the edge of the region \(G\). Assume therefore that \(0 < x\leq y < 1\). Notice that the condition (ii) gives the value for one class of pairs from \(G\) and that each pair in \(G\) can be reduced to one of the members of the class. This implies \[f(x,y)=f(y,x)=y^kf\Big(1,\frac xy\Big)=y^{k-1}x.\] This can be written as \(f(x,y)=\min(x,y)(\max(u,v))^{k-1}\) for all \(0 < x,y < 1\). Let us find all possible values for \(k\). Let \(0 < x\leq \frac 12\leq y < 1\). From the condition (i), and the already obtained results we get \[f\Big(f\Big(x, \frac 12\Big),y\Big)=f\Big(x\Big( \frac 12\Big)^{k-1},y\Big)=f\Big(x,f\Big( \frac 12\Big)\Big)=f\Big(x,\frac 12 y^{k-1}\Big).\] Let us now consider \(x\leq 2^{k-1}y\) in order to simplify the expression to the form \(f\Big(x, \frac 12 y^{k-1}\Big)=x\Big(\frac y2\Big)^{k-1}\), and if we take \(x\) for which \(2x\leq y^{k-1}\) we get \(k-1=(k-1)^2\), i.e. \(k=1\) or \(k=2\). For \(k=1\) the solution is \(f(x,y)=\min(x,y)\), and for \(k=2\) the solution is \(f(x,y)=xy\). It is easy to verify that both solutions satisfy the given conditions.

Problem 23 (APMO 1989) Find all strictly increasing functions \(f:\mathbb{R}\rightarrow\mathbb{R}\) such that \[f(x)+g(x)=2x,\] where \(g\) is the inverse of \(f\).

Clearly every function of the form \(x+d\) is the solution of the given equation. Another useful idea appears in this problem. Namely denote by \(S_d\) the the set of all numbers \(x\) for which \(f(x)=x+d\). Our goal is to prove that \(S_d=\mathbb{R}\). Assume that \(S_d\) is non-empty. Let us prove that for \(x\in S_d\) we have \(x+d\in S_d\) as well. Since \(f(x)=x+d\), according to the definition of the inverse function we have \(g(x+d)=x\), and the given equation implies \(f(x+d)=x+2d\), i.e. \(x+d\in S_d\). Let us prove that the sets \(S_{d^{\prime}}\) are empty, where \(d^{\prime} < d\). From the above we have that each of those sets is infinite, i.e. if \(x\) belongs to some of them, then each \(x+kd\) belongs to it as well. Let us use this to get the contradiction. More precisely we want to prove that if \(x\in S_d\) and \(x\leq y\leq x+(d-d^{\prime})\), then \(y\not\in S_{d^{\prime}}\). Assume the contrary. From the monotonicity we have \(y+d^{\prime}=f(y)\geq f(x)=x+d\), which is a contradiction to our assumption. By further induction we prove that every \(y\) satisfying \[x+k(d-d^{\prime})\leq y < x+(k+1)(d-d^{\prime}),\] can't be a member of \(S_{d^{\prime}}\). However this is a contradiction with the previously established properties of the sets \(S_d\) and \(S_{d^{\prime}}\). Similarly if \(d^{\prime} > d\) switching the roles of \(d\) and \(d^{\prime}\) gives a contradiction. Simple verification shows that each \(f(x)=x+d\) satisfies the given functional equation.

Problem 24 Find all functions \(h:\mathbb{N}\rightarrow\mathbb{N}\) that satisfy \[h(h(n))+h(n+1)=n+2.\]

Notice that we have both \(h(h(n))\) and \(h(n+1)\), hence it is not possible to form a recurrent equation. We have to use another approach to this problem. Let us first calculate \(h(1)\) and \(h(2)\). Setting \(n=1\) gives \(h(h(1))+h(2)=3\), therefore \(h(h(1))\leq 2\) and \(h(2)\leq 2\). Let us consider the two cases:

• \(1^{\circ}\) \(h(2)=1\). Then \(h(h(1))=2\). Plugging \(n=2\) in the given equality gives \(4=h(h(2))+h(3)=h(1)+h(3)\). Let \(h(1)=k\). It is clear that \(k\neq 1\) and \(k\neq 2\), and that \(k\leq 3\). This means that \(k=3\), hence \(h(3)=1\). However from \(2=h(h(1))=h(3)=1\) we get a contradiction. This means that there are no solutions in this case.
  
  
• \(2^{\circ}\) \(h(2)=2\). Then \(h(h(1))=1\). From the equation for \(n=2\) we get \(h(3)=2\). Setting \(n=3,4,5\) we get \(h(4)=3,h(5)=4,h(6)=4\), and by induction we easily prove that \(h(n)\geq 2\), for \(n\geq 2\). This means that \(h(1)=1\). Clearly there is at most one function satisfying the given equality. Hence it is enough to guess some function and prove that it indeed solves the equation (induction or something similar sounds fine). The solution is \[h(n)=\lfloor n\alpha\rfloor+1,\] where \(\alpha=\frac{-1+\sqrt{5}}{2}\) (this constant can be easily found \(\alpha^2+\alpha=1\)). Proof that this is a solution uses some properties of the integer part (although it is not completely trivial).

Problem 25 (IMO 2004, shortlist) Find all functions \(f:\mathbb{R}\rightarrow\mathbb{R}\) satisfying the equality \[f(x^2+y^2+2f(xy))=f(x+y)^2.\]

Πηγή: imomath

Let us make the substitution \(z=x+y\), \(t=xy\). Given \(z,t\in \mathbb{R}\), \(x,y\) are real if and only if \(4t\leq z^2\). Define \(g(x)=2(f(x)-x)\). Now the given functional equation transforms into \[ f\left(z^2+g(t)\right)=\left(f(z)\right)^2\;\; \mbox{for all }\; t,z \in\mathbb{R}\;\mbox{ with }\;z^2\geq 4t.\quad\quad\quad\quad\quad (1) \] Let us set \(c=g(0)=2f(0)\). Substituting \(t=0\) into (1) gives us \[f(z^{2}+c)=\left(f(z)\right)^2\;\mbox{ for all }\; z\in\mathbb{R}. \quad\quad\quad\quad\quad (2)\] If \(c < 0\), then taking \(z\) such that \(z^2+c=0\), we obtain from (2) that \(f(z)^2=c/2\), which is impossible; hence \(c\geq0\). We also observe that \[ x > c\;\;\mbox{ implies }\;\; f(x)\geq0. \quad\quad\quad\quad\quad (3)\] If \(g\) is a constant function, we easily find that \(c=0\) and therefore \(f(x)=x\), which is indeed a solution.

Suppose \(g\) is nonconstant, and let \(a,b\in\mathbb{R}\) be such that \(g(a)-g(b)=d > 0\). For some sufficiently large \(K\) and each \(u,v\geq K\) with \(v^2-u^2=d\) the equality \(u^2+g(a)=v^2+g(b)\) by (1) and (2) implies \(f(u)=f(v)\). This further leads to \(\ g(u)-g(v)=2(v-u)= \frac d{u+\sqrt{u^2+d}}\). Therefore every value from some suitably chosen segment \([\delta,2\delta]\) can be expressed as \(g(u)-g(v)\), with \(u\) and \(v\) bounded from above by some \(M\).

Consider any \(x,y\) with \(y > x\geq 2\sqrt{M}\) and \(\delta < y^2-x^2 < 2\delta\). By the above considerations, there exist \(u,v\leq M\) such that \(g(u)-g(v)=y^2-x^2\), i.e., \(x^2+g(u)=y^2+g(v)\). Since \(x^2\geq4u\) and \(y^2\geq 4v\), (1) leads to \(f(x)^2=f(y)^2\). Moreover, if we assume w.l.o.g. that \(4M\geq c^2\), we conclude from (3) that \(f(x)=f(y)\). Since this holds for any \(x,y\geq 2\sqrt{M}\) with \(y^2-x^2\in[\delta, 2\delta]\), it follows that \(f(x)\) is eventually constant, say \(f(x)= k\) for \(x\geq N=2\sqrt{M}\). Setting \(x > N\) in (2) we obtain \(k^2=k\), so \(k=0\) or \(k=1\).

By (1) we have \(f(-z)=\pm f(z)\), and thus \(|f(z)|\leq 1\) for all \(z\leq-N\). Hence \(g(u)=2f(u)-2u\geq-2-2u\) for \(u\leq-N\), which implies that \(g\) is unbounded. Hence for each \(z\) there exists \(t\) such that \(z^2+g(t) > N\), and consequently \(f(z)^2=f(z^2+g(t))=k=k^2\). Therefore \(f(z)=\pm k\) for each \(z\).

If \(k=0\), then \(f(x)\equiv0\), which is clearly a solution. Assume \(k=1\). Then \(c=2f(0)=2\) (because \(c\geq 0\)), which together with (3) implies \(f(x)=1\) for all \(x\geq2\). Suppose that \(f(t)=-1\) for some \(t < 2\). Then \(t-g(t)=3t+2 > 4t\). If also \(t-g(t)\geq0\), then for some \(z\in\mathbb{R}\) we have \(z^2=t-g(t) > 4t\), which by (1) leads to \(f(z)^2=f(z^2+g(t))=f(t) =-1\), which is impossible. Hence \(t-g(t) < 0\), giving us \(t < -2/3\). On the other hand, if \(X\) is any subset of \((-\infty,-2/3)\), the function \(f\) defined by \(f(x)=-1\) for \(x\in X\) and \(f(x)=1\) satisfies the requirements of the problem.

To sum up, the solutions are \(f(x)=x\), \(f(x)=0\) and all functions of the form \[f(x)=\left\{\begin{array}{ll} 1, &x\not \in X,\newline -1,& x\in X, \end{array}\right.\] where \(X\subset(-\infty,-2/3)\).