#1 What Is a Functional Equation?

With \(x=0\), \(f(y)=f(0)+y\). Hence \(f(x)=x+c\), where \(c=f(0)\). Verification: \(f(x+y)=x+y+c=f(x)+y\). Answer: \(f(x)=x+c\), \(c\in\mathbb R\).

Then \(g(x+y)=g(x)+g(y)\). On \(\mathbb Q\), such a function has the form \(g(q)=cq\), where \(c=g(1)\in\mathbb Q\). Hence \(f(q)=cq+2\). Verification is direct.

We have \(f(n)=1+\sum_{k=0}^{n-1}(2k+3)=1+n(n-1)+3n=(n+1)^2\). The recurrence check is immediate.

With \(y=0\), \(f(x)=f(x)f(0)=0\). Hence \(f\equiv0\). This function indeed satisfies the equation.

Additivity gives \(f(3)=3f(1)\), so \(f(1)=4\). For integers \(n\), \(f(n)=4n\), and from \(m f(m/n)=f(m)=4m\), \(f(m/n)=4m/n\). Answer: \(f(q)=4q\).

Put \(g(x)=f(x)-2x^2\). Then \(g(x+y)=g(x)+g(y)\), and \(g(1)=0\). On \(\mathbb Q\), this gives \(g\equiv0\). Hence \(f(q)=2q^2\).

\(f(n)=\sum_{k=0}^{n-1}(3k+1)=3n(n-1)/2+n=\frac{n(3n-1)}2\).

Put \(A(x)=f(x)+1\). Then \(A(x+y)=A(x)+A(y)\). Thus all solutions are \(f(x)=A(x)-1\), where \(A:\mathbb R\to\mathbb R\) is any additive function. Verification by substitution is direct.

Then \(g\) is additive and \(g(1)=1\). On \(\mathbb Q\), \(g(q)=q\). Hence \(f(q)=q-1\). Check: \(x+y-1=(x-1)+(y-1)+1\).

Since \(f\) is strictly increasing and takes positive integer values, \(f(n)\ge n\). If \(f(k)\ge k+2\), then \(f(f(k))\ge f(k+2)\ge f(k)+2\ge k+4\), contradiction. Thus \(f(n)\le n+1\). Equality \(f(n)=n\) is impossible, since then \(f(f(n))=n\). Hence \(f(n)=n+1\), and verification gives \(f(f(n))=n+2\).

Put \(g(x)=f(x)-x^2\). Then \(g(x+y)=g(x)+g(y)\). Also, \(g(2)=f(2)-4=0\), so \(g(1)=0\). On \(\mathbb Q\), \(g(q)=0\), hence \(f(q)=q^2\). Check: \((x+y)^2=x^2+y^2+2xy\).

With \(y=1\), \(f(x)=x f(1)\). Thus \(f(x)=cx\), where \(c\in\mathbb Q\). Check: \(f(xy)=cxy=x\cdot cy=x f(y)\).

For integers \(n\), \(f(n)=n f(1)=0\). If \(q=m/n\), then \(n f(q)=f(m)=0\), so \(f(q)=0\). But on \(\mathbb R\), without regularity, additive functions exist that are not determined by their values on \(\mathbb Q\); therefore one cannot conclude \(f\equiv0\) without extra assumptions.

Let \(g(x)=f(x)-\frac{x^2}{2}\). Then \(g(x+y)=g(x)+g(y)\). Also, \(g(1)=f(1)-1/2=-1/2\). On \(\mathbb Q\), \(g(q)=-q/2\), hence \(f(q)=\frac{q^2-q}{2}\).

If \(f(u)=f(v)\), then \(f(x+f(u))=x+u\) and \(f(x+f(v))=x+v\) imply \(u=v\). Thus \(f\) is injective. With \(x=0\), \(f(f(y))=y\). Substitute \(y=f(t)\): \(f(x+t)=x+f(t)\). For fixed \(t\), this gives \(f(s)=s+f(t)-t\), so \(f(s)=s+c\). Then \(f(f(y))=y+2c\), hence \(c=0\). Answer: \(f(x)=x\).

Since \(f(1)\ge1\), induction gives \(f(n)\ge n\). If \(f(k)>k\) for some \(k\), then monotonicity would give \(f(n)>k\) for all \(n\ge k\), and for \(n

Let \(|f(t)|\le M\) on \([0,1]\). For small \(h\), choose \(n\) so that \(|nh|\le1\). Then \(|f(h)|=|f(nh)|/n\le M/n\), so \(f(h)\to0\) as \(h\to0\). An additive function continuous at zero is linear: \(f(x)=xf(1)\) first for rationals, then by continuity for all reals.

On one hand, \(f((x+y)^2)=(x+y)(f(x)+f(y))\). On the other hand, by additivity, \(f(x^2+2xy+y^2)=xf(x)+2f(xy)+yf(y)\). Hence \(2f(xy)=xf(y)+yf(x)\). With \(y=1\), \(2f(x)=xf(1)+f(x)\), so \(f(x)=cx\), where \(c=f(1)\). Every function \(f(x)=cx\) works.

Put \(g(x)=f(x)-x^3\). Then \(g(x+y)=g(x)+g(y)\), because the cubic extra term is completely removed. Also, \(g(1)=0\), so on \(\mathbb Q\), \(g(q)=0\). Therefore \(f(q)=q^3\). Verification follows from the identity above.

We have \(f(x+y)=\frac{(x+y)^2}{2}+A(x+y)=\frac{x^2}{2}+\frac{y^2}{2}+xy+A(x)+A(y)=f(x)+f(y)+xy\). If \(A\) is an arbitrary additive function on \(\mathbb R\), it need not be linear without regularity assumptions. Thus the solution set is broader than just quadratic polynomials.