Book 3. Functional Equations

Putting \(x=0\), we get \(f(y)=f(0)+y\). Hence \(f(x)=x+c\). Direct verification shows that all such functions work.

Put \(g(x)=f(x)-2\). Then \(g(x+y)=g(x)+g(y)\), so on \(\mathbb Q\), \(g(q)=cq\). Thus \(f(q)=cq+2\).

Put \(g(x)=f(x)-2x^2\). Then \(g(x+y)=g(x)+g(y)\), and \(g(1)=0\). Hence \(g(q)=0\), so \(f(q)=2q^2\).

Put \(A(x)=f(x)+1\). Then \(A\) is additive. Therefore \(f(x)=A(x)-1\), where \(A:\mathbb R\to\mathbb R\) is any additive function.

If \(f(u)=f(v)\), then \(u=v\), because \(f(x+f(u))=x+u\). Thus \(f\) is injective. With \(x=0\), \(f(f(y))=y\). Substituting \(y=f(t)\), we get \(f(x+t)=x+f(t)\), so \(f(s)=s+c\). From \(f(f(y))=y\), \(c=0\), hence \(f(x)=x\).

The function is strictly increasing, so \(f(n)\ge f(1)+n-1\ge n\). If \(f(k)>k\) for some \(k\), then the value \(k\) could not appear in the image. Therefore \(f(n)=n\).

Because in this course \(\mathbb N\) means the positive integers. A substitution is allowed only when the value belongs to the domain.

The left side is \(2(x+y)^2=2x^2+4xy+2y^2\). The right side is \(2x^2+2y^2+4xy\). They are equal.

With \(x=0\), \(f(y)=f(0)+2y\). Hence \(f(x)=2x+c\). Verification is direct.

With \(y=0\), \(f(x)=f(x)+f(0)+x\), impossible for all \(x\). There are no solutions.

Let \(g(x)=f(x)-4\). Then \(g\) is additive and \(g(1)=3\). Hence \(g(q)=3q\), so \(f(q)=3q+4\).

Put \(g(x)=f(x)-x^2\). Then \(g\) is additive and \(g(1)=0\), hence \(g\equiv0\). Answer: \(f(q)=q^2\).

Subtract the common term \(F(x)\) to get \(F(y)=F(-y)\) for all \(y\).

With \(y=0\), \(f(f(x))=f(0)+x\). With \(x=0\), \(f(f(0)+y)=f(y)\), so \(f(0)=0\). Hence \(f(f(x))=x\). Substituting \(x=f(t)\), we get \(f(t+y)=t+f(y)\), so \(f\) is additive and \(f(t)=t\) or \(f(t)=-t\). Both work.

They are symmetric about the axis of the parabola. This is often used after substituting roots into another quadratic function.

Use \((x+y)^3=x^3+y^3+3xy(x+y)\). Thus the formula works.

For integers \(n\), \(f(n)=6n\). If \(q=m/n\), then \(n f(q)=f(m)=6m\), so \(f(q)=6q\).

Let \(g=f-3\). Then \(g\) is additive and \(g(1)=2\), so \(f(q)=2q+3\).

First \(f(0)=0\). For any \(u,v\), take \(x=u/2\), \(y=v/3\), obtaining \(f(u+v)=f(u)+f(v)\). Hence \(f(q)=4q\).

Take \(x=0\): \(f(y)=f(0)\). Hence the function is constant.

On rationals, \(f(q)=qf(1)\). Monotonicity gives continuity, and rational approximation gives the formula for all \(x\).

Boundedness gives continuity at zero: multiply small \(h\) by large \(n\) so that \(nh\in[0,1]\). A continuous additive function is linear.

\(a^2=4\), \(b(a+1)=3\). Thus \(a=2,b=1\) or \(a=-2,b=-3\).

The left side is \(A(x+y)+2=A(x)+A(y)+2\), and the right side is \(A(x)+2+A(y)+2-2\). They are equal.

Since \(f\) is injective, from \(f(f(x)+y)=f(f(y)+x)\) we get \(f(x)+y=f(y)+x\). Rearranging gives \(f(x)-x=f(y)-y\) for all \(x,y\). Hence \(f(x)-x\) does not depend on \(x\).

Comment. The key point is not to expand the function, but to remove the outside \(f\).

Since \(f\) is surjective, the number \(t=f(y)\) is arbitrary. Thus \(f(x+t)=f(x)+t\) for all \(x,t\). Taking \(x=0\), we get \(f(t)=f(0)+t\). Hence \(f(x)=x+c\), where \(c=f(0)\). Checking gives \(f(x+f(y))=x+f(y)+c=f(x)+f(y)\). All such functions work.

Comment. Surjectivity replaces a special value of the function by a free variable.

If \(f(a)=f(b)\), then \(f(f(a))=f(f(b))\), hence \(a+1=b+1\), so \(a=b\). Thus \(f\) is injective. Now take any \(T\in\mathbb Q\). We have \(f(f(T-1))=T\), so \(T\) is a value of \(f\). Therefore \(f\) is surjective and hence bijective.

Comment. An iteration \(f(f(x))\) often gives both injectivity and surjectivity.

Putting \(x=0\), we get \(f(f(y))=f(0)+y\). If \(f(a)=f(b)\), then \(f(f(a))=f(f(b))\), hence \(a=b\); so \(f\) is injective. With \(y=0\), \(f(x+f(0))=f(x)\), and injectivity gives \(f(0)=0\). Therefore \(f(f(y))=y\). Put \(y=f(t)\): then \(f(x+t)=f(x)+f(t)\). On \(\mathbb Q\), this gives \(f(x)=cx\). From \(f(f(y))=y\), we get \(c^2=1\). The answers are \(f(x)=x\) and \(f(x)=-x\), and both work.

Comment. On \(\mathbb Q\), additivity immediately becomes linearity.

By surjectivity, \(t=f(y)\) is arbitrary. Hence \(f(x+t)=f(x)+t+1\) for all \(x,t\). At \(x=0\), this gives \(f(t)=f(0)+t+1\). Substituting back, the left-hand side is \(f(0)+x+t+1\), while the right-hand side is \(f(0)+x+1+t+1\). This gives \(0=1\), a contradiction.

Comment. An extra constant often appears only during the final check.

If \(f(a)=f(b)\), comparing the equations with \(y=a\) and \(y=b\) gives \(a=b\), so \(f\) is injective. With \(y=0\): \(f(x+f(0))=f(x)\), hence \(f(0)=0\). With \(x=0\): \(f(f(y))=2y\), so \(f\) is surjective. Let \(t=f(y)\). Then \(f(t)=2y\), and the original equation becomes \(f(x+t)=f(x)+f(t)\). Thus \(f\) is additive. An increasing additive function is linear: \(f(x)=cx\). From \(f(f(y))=2y\), we obtain \(c^2=2\), and increasing gives \(c>0\). Therefore \(f(x)=\sqrt{2}x\).

Comment. Without regularity on \(\mathbb R\), the additive part could be more complicated.

With \(y=0\), we get \(f(f(x))=x+f(0)\), which implies injectivity. The same formula also implies surjectivity. With \(x=0\): \(f(f(0)+y)=f(y)\), so \(f(0)=0\). Hence \(f(f(x))=x\). Put \(y=f(t)\): then \(f(f(x)+f(t))=x+t=f(f(x+t))\). By injectivity, \(f(x)+f(t)=f(x+t)\). A continuous additive function has the form \(f(x)=cx\), and \(f(f(x))=x\) gives \(c^2=1\). Therefore \(f(x)=x\) or \(f(x)=-x\).

Comment. Continuity is used only at the end to exclude nonlinear additive functions.

Since \(f\) is surjective, the value \(t=f(y)\) is arbitrary. Therefore \(f(x+t)+f(x-t)=2f(x)\) for all \(x,t\). This means that the value at the midpoint equals the arithmetic mean of the values at the endpoints: \(f\left(\frac{u+v}{2}\right)=\frac{f(u)+f(v)}{2}\). A continuous function with this property is affine: \(f(x)=ax+b\). Surjectivity requires \(a\ne0\). Checking shows that all \(f(x)=ax+b\), \(a\ne0\), work.

Comment. Surjectivity turns \(f(y)\) into any distance from the midpoint.

From \(f(0)=f(0)+f(0)\), we get \(f(0)=0\). For natural \(n\), \(f(n)=nf(1)\) by induction. Also \(0=f(n+(-n))=f(n)+f(-n)\), hence \(f(-n)=-nf(1)\). If \(q=\frac{m}{n}\), then \(n f(q)=f(nq)=f(m)=mf(1)\), so \(f(q)=qf(1)\).

Comment. This is the main fact that cannot be transferred to \(\mathbb R\) without regularity.

Let \(g(x)=f(x)-4\). Then \(g(x+y)=f(x+y)-4=f(x)+f(y)-8=g(x)+g(y)\). Thus \(g(x)=ax\) on \(\mathbb Q\). Hence \(f(x)=ax+4\), where \(a\in\mathbb Q\). Direct checking works.

Comment. First choose the shift, then solve the ordinary Cauchy equation.

By additivity, \(f(x)=cx\). Multiplicativity gives \(cxy=c^2xy\) for all \(x,y\). Hence \(c=0\) or \(c=1\). The answers are \(f(x)=0\) and \(f(x)=x\). Both work.

Comment. In multiplicative problems, the zero function must be checked separately.

A continuous additive function has the form \(f(x)=cx\). Therefore \(c\sqrt{2}=5\), so \(c=\frac{5}{\sqrt{2}}\). The answer is \(f(x)=\frac{5}{\sqrt{2}}x\).

Comment. An olympiad solution should explicitly state where continuity is used.

Put \(u=x+y\), \(v=x-y\). Then \(x=\frac{u+v}{2}\), and the condition becomes \(f\left(\frac{u+v}{2}\right)=\frac{f(u)+f(v)}{2}\). A continuous function preserving midpoints is affine: \(f(x)=ax+b\). Checking shows that all such functions work.

Comment. This is not the quadratic equation: the right-hand side does not contain \(2f(y)\).

With \(y=0\), we get \(2f(x)=2f(x)+2f(0)\), so \(f(0)=0\). Continuous solutions of the quadratic Cauchy equation have the form \(f(x)=cx^2\). From \(f(1)=3\), we get \(c=3\). The answer is \(f(x)=3x^2\). Checking: \(3(x+y)^2+3(x-y)^2=6x^2+6y^2\).

Comment. It is important to distinguish the affine and quadratic cases.

Notice that \((x+y)^3-x^3-y^3=3xy(x+y)\). Let \(g(x)=f(x)-\frac{x^3}{3}\). Then \(g(x+y)=g(x)+g(y)\). On \(\mathbb Q\), \(g(x)=ax\). Hence \(f(x)=\frac{x^3}{3}+ax\), where \(a\in\mathbb Q\). The check follows from the same cubic identity.

Comment. This is an olympiad version of Cauchy's method: remove the correction first.

Continuous additivity gives \(f(x)=cx\). Substituting into the second condition gives \(c^2xy=cxy+xy\) for all \(x,y\). Hence \(c^2=c+1\). Therefore \(c=\frac{1+\sqrt{5}}{2}\) or \(c=\frac{1-\sqrt{5}}{2}\). Both answers work.

Comment. The second condition does not replace the first; it fixes the coefficient.

For \(n>0\), induction gives \(f(n)=2+5n\). For negative integers, move backwards: \(f(n)=f(n+1)-5\), so the same formula remains true. The answer is \(f(n)=5n+2\).

Comment. On \(\mathbb Z\), always check the backward direction.

On even numbers, \(f(2k)=1+4k=2(2k)+1\). On odd numbers, \(f(2k+1)=3+4k=2(2k+1)+1\). Hence \(f(n)=2n+1\) for all integers \(n\).

Comment. If the initial values were not compatible, the formula could differ on even and odd numbers.

Let \(g(n)=f(n)-n^2\). Then \(g(m+n)=g(m)+g(n)\). On \(\mathbb Z\), \(g(n)=cn\), where \(c\in\mathbb Z\). The answer is \(f(n)=n^2+cn\). Direct checking works.

Comment. Polynomial corrections are often recognised through square identities.

Use \(\binom{m+n}{2}=\binom{m}{2}+\binom{n}{2}+mn\). Let \(g(n)=f(n)-\binom{n}{2}\). Then \(g(m+n)=g(m)+g(n)\), so \(g(n)=cn\). The answer is \(f(n)=\binom{n}{2}+cn\).

Comment. This correction remains integral for all integers \(n\).

Every element is \(x\cdot1\) in the additive sense modulo \(p\). Therefore \(f(x)=f(1+\cdots+1)=xf(1)\). All equalities are modulo \(p\).

Comment. This is the finite version of additivity on \(\mathbb Q\), but the proof is shorter.

The difference \(f(n+1)-f(n)=2n+1\) is always odd. Hence neighbouring values have opposite parity. Therefore all values cannot have the same parity.

Comment. This is short, but it is an important olympiad check.

With \(n=0\), \(f(f(m))=f(0)+m\). This implies injectivity. With \(m=0\), \(f(n+f(0))=f(n)\), so \(f(0)=0\). Thus \(f(f(m))=m\). Put \(m=f(t)\): \(f(n+t)=f(n)+f(t)\). Therefore \(f(n)=cn\) on \(\mathbb Z\). From \(f(f(n))=n\), we get \(c^2=1\). The answer is \(f(n)=n\) and \(f(n)=-n\).

Comment. The method resembles the real-domain one, but linearity on \(\mathbb Z\) is simpler.

With \(x=0\), \(f(f(y))=f(0)+y\), so \(f\) is injective, and on a finite set it is bijective. With \(y=0\), \(f(x+f(0))=f(x)\), hence \(f(0)=0\). Thus \(f(f(y))=y\). Substituting \(y=f(t)\), we get \(f(x+t)=f(x)+f(t)\), so \(f(x)=ax\). From \(f(f(y))=y\), \(a^2=1\), hence \(a=1\) or \(a=-1\). The answer is \(f(x)=x\) and \(f(x)=-x\).

Comment. Finiteness replaces the proof of surjectivity.

With \(y=0\), \(P(0)=0\). Let \(\deg P=d\). If \(d\ge2\), then \(P(x+y)\) contains a mixed term of highest degree, while \(P(x)+P(y)\) contains no mixed terms. Hence \(d\le1\). Thus \(P(x)=ax+b\), and \(P(0)=0\) gives \(b=0\). The answer is \(P(x)=ax\).

Comment. One may also use additivity on \(\mathbb Q\), but degree comparison is faster.

Notice that \((x+1)^2-x^2=2x+1\). Thus \(P(x)-x^2\) has zero finite difference: \(Q(x+1)=Q(x)\). A nonconstant polynomial cannot be periodic, so \(Q\) is constant. The answer is \(P(x)=x^2+c\).

Comment. The difference formula is often guessed from the expected degree.

If \(P\) is constant, the left-hand side is constant, impossible. If \(\deg P=d\ge1\), then \(\deg P(P(x))=d^2\). The right-hand side has degree \(2\). We would need \(d^2=2\), but no integer degree \(d\) has this property. Contradiction.

Comment. Always check the constant case first.

The constant solutions are \(P=0\) and \(P=1\). Let \(P\) be nonconstant. Then \(P(1)=1\). If \(P\) has a nonzero root \(r\), then \(P(ry)=P(r)P(y)=0\) for all \(y\), so \(P\equiv0\), a contradiction. Therefore the only possible root is \(0\), and \(P(x)=ax^n\). Substitution gives \(a=a^2\), hence \(a=1\). The answer is \(0\), \(1\), and \(x^n\) for \(n\ge1\).

Comment. Roots cannot be ignored: they often give the whole structure.

The constant solutions are \(0\) and \(1\). If \(P\) is nonconstant and \(r\ne0\) is a root, then \(P(r)^2=P(r^2)=0\), then \(P(r^{2^k})=0\) for infinitely many values, impossible for a nonzero polynomial. Hence the only root is \(0\), so \(P(x)=ax^n\). Comparing leading coefficients gives \(a^2=a\), so \(a=1\). The answer is \(0\), \(1\), and \(x^n\).

Comment. Here an infinite chain of roots replaces division.

Let \(\deg P=d\). If \(d\ge2\), the degree of the left-hand side in \(y\) is \(d^2\), while the degree of the right-hand side is \(1\), impossible. Hence \(P(x)=ax+b\). Substitution gives \(a(x+ay+b)+b=ax+b+y\). Therefore \(a^2=1\) and \(ab=0\). Since \(a\ne0\), \(b=0\). The answer is \(P(x)=x\) and \(P(x)=-x\).

Comment. First degree, then coefficients.

The polynomial \(P(x+1)-P(x)-x^2\) has infinitely many roots, so it is identically zero. Since \(\frac{x(x-1)(2x-1)}{6}\) has finite difference \(x^2\), all solutions are \(P(x)=\frac{x(x-1)(2x-1)}{6}+c\).

Comment. Integer points give an identity when there are infinitely many of them.

Let the degree of \(P\) be \(d\), with leading coefficient \(a\). The leading homogeneous part gives \(a(x^d+y^d+(-x-y)^d)=0\). Put \(y=x\): \(a(2+(-2)^d)x^d=0\). For \(d\ge2\), this is not identically zero; for even \(d\) the coefficient is positive, and for odd \(d\ge3\) it is \(2-2^d\ne0\). Hence \(d\le1\). Substituting \(P(x)=ax+b\) gives \(3b=0\). The answer is \(P(x)=ax\).

Comment. There is no need to expand the entire polynomial.

If \(f(a)=a\), then \(f(f(a))=f(a)=a\). But the condition gives \(f(f(a))=a+1\). Thus \(a=a+1\), a contradiction.

Comment. Always check what the condition gives at a fixed point.

Suppose \(f(a)>a\). Then by increasingness, \(f(f(a))>f(a)\), so \(a>f(a)\), a contradiction. Similarly, if \(f(a)

Comment. The whole chain \(a, f(a), f^2(a)\) is compared.

We have \(f(f(x))=a(ax+b)+b=a^2x+b(a+1)\). Thus \(a^2=1\). If \(a=1\), then \(2b=6\), so \(b=3\). If \(a=-1\), then \(b(a+1)=0\), but we need \(6\), impossible. The answer is \(f(x)=x+3\).

Comment. Affine tasks often reduce to a system for \(a,b\).

Additivity on \(\mathbb Q\) gives \(f(x)=cx\), \(c\in\mathbb Q\). Then \(f(f(x))=c^2x\). Hence \(c^2=9\), so \(c=3\) or \(c=-3\). The answer is \(f(x)=3x\), \(f(x)=-3x\).

Comment. This is the iterative version of the linear Cauchy fact.

The cycles of the permutation have lengths dividing \(3\), so they have length \(1\) or \(3\). If there are no fixed points, all cycles have length \(3\). Then the number of elements must be divisible by \(3\), but \(10\) is not divisible by \(3\). Contradiction.

Comment. This reasoning appears often in olympiad combinatorics and algebra.

If \(P\) is constant, impossible. Let \(\deg P=d\). Then \(d^2=1\), hence \(d=1\). Let \(P(x)=ax+b\). We get \(P(P(x))=a^2x+b(a+1)=x\). Thus \(a^2=1\), \(b(a+1)=0\). The answer is \(P(x)=x\) or \(P(x)=-x+b\).

Comment. This is a polynomial involution.

If \(y\) lies in the image, then \(y=f(x)\) for some \(x\). Hence \(f(y)=f(f(x))=f(x)=y\). Therefore every image point is a fixed point.

Comment. This is useful when the function cannot be fully determined.

If \(f(a)>a\), then increasingness gives \(f^2(a)>f(a)>a\), then \(f^3(a)>f^2(a)\), and so on. Thus \(f^k(a)>a\), contradicting \(f^k(a)=a\). The case \(f(a)

Comment. This is the final template for ordered cycles.

An increasing additive function is linear: \(f(x)=cx\). From \(f(1)=3\), we get \(c=3\). The answer is \(f(x)=3x\).

Comment. The solution must state exactly what monotonicity gives.

An additive function bounded on an interval is continuous. Hence \(f(x)=cx\). From \(f(1)=2\), we get \(f(x)=2x\).

Comment. Values outside the interval are not needed.

If \(x0\), so \(f(y)-f(x)=f(y-x)\ge0\). Thus \(f\) is nondecreasing. A monotone additive function is linear: \(f(x)=cx\). The condition for \(x>0\) gives \(c\ge0\).

Comment. Positivity becomes order.

If \(f(a)>0\), then for large negative \(n\), \(f(na)=nf(a)<-1\), contradiction. If \(f(a)<0\), then for large positive \(n\), again \(f(na)<-1\). Hence \(f(a)=0\) for every \(a\).

Comment. A global estimate is stronger than a local one.

For every \(q>0\), \(f(qx)=qf(x)\le q^2x^2\). Divide by \(q\): \(f(x)\le qx^2\). Letting \(q\to0+\), we get \(f(x)\le0\). Applying this to \(-x\), we get \(-f(x)\le0\), so \(f(x)\ge0\). Therefore \(f(x)=0\).

Comment. This is one of the most useful tricks in the module.

Let \(g(x)=f(x)-f(0)\). Then \(g\) satisfies Jensen's equation and is bounded on an interval. Standard result: such a function is continuous and Jensen-linear, so \(g(x)=cx\). Hence \(f(x)=cx+b\).

Comment. For this course, it can be used as a theoretical fact.

A nondecreasing additive function is linear: \(f(x)=cx\), where \(c\ge0\). Then \(cxy=c^2xy\) for all \(x,y\). Hence \(c=0\) or \(c=1\). The answers are \(f\equiv0\) and \(f(x)=x\).

Comment. The zero function remains valid because it is nondecreasing.

Taking \(y=x\), we get \(f(x)^2\le x^2\), so \(f\) is bounded on \([-1,1]\). Hence \(f(x)=cx\). The condition becomes \(c^2xy\le xy\) for all \(x,y\). If \(xy>0\), then \(c^2\le1\); if \(xy<0\), then \(c^2\ge1\). Therefore \(c^2=1\). The answer is \(f(x)=x\), \(f(x)=-x\).

Comment. It is essential to consider both signs of \(xy\).

Let \(g(x)=f(x)-\frac a2 x^2\). Then \(g(x+y)=g(x)+g(y)\). By continuity, \(g(x)=cx\). Hence \(f(x)=\frac a2 x^2+cx\). Direct checking works.

Comment. The parameter remains free, and so does \(c\).

From the previous example, \(f(x)=x^2+cx\). The condition \(f(1)=0\) gives \(1+c=0\), so \(c=-1\). The answer is \(f(x)=x^2-x\).

Comment. Do not start with \(f(1)=0\); first get the structure.

Additivity gives \(f(x)=cx\). Then \(cx^2=c^2x^2\) for all \(x\). Hence \(c=0\) or \(c=1\). The answers are \(f=0\) and \(f(x)=x\).

Comment. Do not forget the zero answer.

By additivity, \(f(x)=cx\). Then \(cxy=2cxy\) for all \(x,y\), so \(c=0\). The answer is \(f\equiv0\).

Comment. This is a rational-domain version of a derivation-type condition.

Continuous additivity gives \(f(x)=cx\). Then \(c^2xy=cxy+xy\), hence \(c^2=c+1\). The answer is \(f(x)=\frac{1+\sqrt{5}}{2}x\) or \(f(x)=\frac{1-\sqrt{5}}{2}x\).

Comment. The coefficient may be irrational because the domain is \(\mathbb R\).

Let \(f(x)=cx\), \(c\in\mathbb Q\). Then \(f(f(x))=c^2x\), so \(c^2=2\). No rational \(c\) has this property. Contradiction.

Comment. The domain changes existence.

As in earlier modules, comparing values of \(y\) gives injectivity, then \(f(0)=0\), \(f(f(y))=2y\), and \(f(x+t)=f(x)+f(t)\). An increasing additive function is linear: \(f(x)=cx\). Then \(c^2=2\), and increasing gives \(c>0\). The answer is \(f(x)=\sqrt{2}x\).

Comment. This combines injectivity, surjectivity, and order.

From the first condition, \(f(x)=\frac a2x^2+cx\). If \(a e0\), then \(f(f(x))\) has degree \(4\), impossible since it equals \(x\). Hence \(a=0\). Then \(f(x)=cx\), and \(c^2=1\). The answer is \(a=0\), \(f(x)=x\) or \(f(x)=-x\).

Comment. This is a typical advanced problem: the parameter is checked by degree.

Let \(g(x)=f(x)-2x^2\). Then \(g(x+y)=g(x)+g(y)\), so \(g(x)=cx\). From \(f(1)=3\), \(2+c=3\), hence \(c=1\). The answer is \(f(x)=2x^2+x\).

Comment. The method is recognised by \(xy\).

Let \(f(x)=cx\). Then \(c^2=9\). The answers are \(f(x)=3x\) and \(f(x)=-3x\).

Comment. The domain immediately gives linearity.

The right-hand side is \((n+1)^3-n^3\). Thus \(P(x)-x^3\) has zero difference at all integers, so that difference is identically zero. Hence \(P(x)=x^3+c\).

Comment. Infinitely many integer points are enough for an identity.

Cycles have length \(1\) or \(3\). If there were no fixed points, all cycles would have length \(3\), so the number of elements would be divisible by \(3\). But \(14\) is not divisible by \(3\). Therefore a fixed point exists.

Comment. The method is combinatorial, not analytic.

For \(q>0\), \(qf(x)=f(qx)\le q^2x^2\). Divide by \(q\) and let \(q o0+\), obtaining \(f(x)\le0\). Applying this to \(-x\), we get \(f(x)\ge0\). Hence \(f=0\).

Comment. Scaling is the main move.

All constant polynomials work. If \(P\) is nonconstant, its image is infinite. For every \(t=P(x)\), we have \(P(t)=t\), so \(P(T)-T\) has infinitely many roots. Therefore \(P(x)=x\).

Comment. The image, not degree comparison, is the key.

Comparing values with equal \(f(y)\) gives injectivity. Then \(f(0)=0\), \(f(f(y))=3y\), and with \(t=f(y)\) we get \(f(x+t)=f(x)+f(t)\). An increasing additive function is linear: \(f(x)=cx\). Then \(c^2=3\), \(c>0\). The answer is \(f(x)=\sqrt{3}x\).

Comment. Three techniques are mixed here.

The general form is \(f(x)=\frac a2x^2+cx\). If \(a e0\), the composition has degree \(4\), impossible. Hence \(a=0\), \(f(x)=cx\), and \(c^2=1\). The answer is \(a=0\), \(f(x)=x\) or \(f(x)=-x\).

Comment. The parameter is also part of the answer.