Mixed Functional Equations

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

Then \(c^2x=9x\), so \(c=3\) or \(c=-3\). The answers are \(f(x)=3x\), \(f(x)=-3x\).

If \(f(a)>a\), then \(f(f(a))>f(a)\), i.e. \(a>f(a)\), contradiction. The case \(f(a)

The polynomial \(P(x+1)-P(x)-((x+1)^3-x^3)\) has infinitely many roots, so it is identically zero. Thus \(P(x)-x^3\) is periodic with period \(1\), hence constant. The answer is \(P(x)=x^3+c\).

For \(n=2k\), \(f(n)=1+8k=4n+1\). For \(n=2k+1\), \(f(n)=5+8k=4n+1\). The answer is \(f(n)=4n+1\).

Cycles have length \(1\) or \(3\). If there are no fixed points, all cycles have length \(3\), so \(14\) is divisible by \(3\), false. Therefore a fixed point exists.

For \(q>0\), \(qf(x)=f(qx)\le q^2x^2\), so \(f(x)\le qx^2\). Letting \(q o0+\), \(f(x)\le0\). For \(-x\), we get \(-f(x)\le0\), hence \(f(x)=0\).

The constant solutions are \(0\), \(1\). If \(P\) is nonconstant and has a nonzero root \(r\), then \(P(ry)=0\) for all \(y\), impossible. Hence \(P(x)=ax^n\). Substitution gives \(a=a^2\), so \(a=1\). The answer is \(0\), \(1\), and \(x^n\), \(n\ge1\).

Continuous solutions have the form \(f(x)=ax+b\). From \(f(0)=1\), \(b=1\). From \(f(2)=5\): \(2a+1=5\), so \(a=2\). The answer is \(f(x)=2x+1\).

We get \(cx^2=2c x^2\) for all \(x\), hence \(c=0\). The answer is \(f\equiv0\).

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

If \(f(a)=f(b)\), comparing for \(y=a,b\) gives \(a=b\). With \(y=0\), \(f(0)=0\); with \(x=0\), \(f(f(y))=3y\). For \(t=f(y)\), we have \(f(t)=3y\), so \(f(x+t)=f(x)+f(t)\). An increasing additive function is linear: \(f(x)=cx\). Then \(c^2=3\), \(c>0\). The answer is \(f(x)=\sqrt{3}x\).

If \(f(x)=cx\), then \(f(f(x))=c^2x\). We need \(c^2=3\), impossible for rational \(c\). Contradiction.

Constant polynomials work. If \(P\) is nonconstant, then for infinitely many \(t=P(x)\), \(P(t)=t\). Hence \(P(T)-T\equiv0\). The answer is all constant polynomials and \(P(x)=x\).

We have \(m>n\) and \(f(m)=n+2\). But by the condition, \(f(m)>m\), so \(n+2>m\). Hence \(n

With \(y=x\), \(f(x)^2\le x^2\), so \(f\) is bounded on \([-1,1]\), and \(f(x)=cx\). Then \(c^2xy\le xy\). If \(xy>0\), \(c^2\le1\); if \(xy<0\), \(c^2\ge1\). Hence \(c^2=1\). The answer is \(f(x)=x\), \(f(x)=-x\).

The general form is \(f(x)=\frac a2x^2+cx\). If \(a\ne0\), then \(f(f(x))\) has degree \(4\), impossible. Hence \(a=0\). Then \(f(x)=cx\), and \(c^2x=x\), so \(c=1\) or \(c=-1\). The answer is \(a=0\), \(f(x)=x\), \(f(x)=-x\).

With \(x=0\), \(f(f(y))=f(0)+y\), so \(f\) is injective, hence bijective on a finite set. With \(y=0\), \(f(x+f(0))=f(x)\), so \(f(0)=0\). Thus \(f(f(y))=y\). Putting \(y=f(t)\), we get \(f(x+t)=f(x)+f(t)\), hence \(f(x)=ax\). From \(a^2=1\) modulo \(p\), \(a=\pm1\). The answer is \(f(x)=x\), \(f(x)=-x\).

From \(f(2)=6\), \(2a=6\), so \(a=3\). The condition \(f(g(x))=x\) gives \(ab=1\), hence \(b=\frac13\). The answer is \(f(x)=3x\), \(g(x)=\frac{x}{3}\).

From the first condition, after subtracting \(a x^2\), we get \(f(x)=a x^2+cx+d\). If \(a\ne0\), then the composition \(f(f(x))\) has degree \(4\), impossible. Hence \(a=0\), \(f(x)=cx+d\). The condition \(f(f(x))=x\) gives \(c^2=1\), \(d(c+1)=0\). The answer is \(a=0\), \(f(x)=x\), and \(f(x)=-x+d\) for any \(d\in\mathbb R\).