Book 3. Functional Equations

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.

With \(x=0\), \(f(y)=f(0)+2y\). Hence \(f(x)=2x+c\). Check: \(f(x+y)=2x+2y+c=f(x)+2y\). Answer: \(f(x)=2x+c\), where \(c\in\mathbb R\).

With \(y=0\), \(f(x)=f(x)+f(0)+x\), so \(f(0)=-x\) for all \(x\). This is impossible. Therefore there are no solutions.

Then \(g(x+y)=g(x)+g(y)\), and \(g(1)=3\). On \(\mathbb Q\), \(g(q)=3q\). Hence \(f(q)=3q+5\). Verification is direct.

With \(x=y\), \(f(0)=0\). With \(x=2\), \(y=1\): \(f(1)=f(2)-f(1)\), so \(f(1)=5\). The equation gives additivity on \(\mathbb Z\), hence \(f(n)=5n\).

The recurrence gives \(f(n)=c+2(n-1)\). Then \(f(f(1))=f(c)=c+2(c-1)=3c-2\). The condition gives \(3c-2=5\), so \(c=7/3\), not a positive integer. Therefore no such function exists.

With \(y=0\), \(f(x)+f(0)=f(x)+f(x)\), so \(f(x)=f(0)\) for all \(x\). Every constant function works.

With \(x=y\), \(f(2y)=f(0)+4y\). Thus for every \(t\), \(f(t)=f(0)+2t\). Check: \(f(x+y)=f(0)+2x+2y=f(0)+2x-2y+4y=f(x-y)+4y\).

Let \(g(x)=f(x)-x^2\). Then \(g(x+y)=g(x)+g(y)\), and \(g(1)=3\). On \(\mathbb Q\), \(g(q)=3q\). Hence \(f(q)=q^2+3q\). Verification follows from \((x+y)^2=x^2+y^2+2xy\).

Put \(g(x)=f(x)-x^2\). Then \(g(x+y)=g(x-y)\) for all \(x,y\). Choosing \(u=x+y\), \(v=x-y\), we can represent any rational \(u,v\) by \(x=(u+v)/2\), \(y=(u-v)/2\). Hence \(g(u)=g(v)\) for all \(u,v\), so \(g\) is constant. Since \(g(0)=0\), \(g\equiv0\), and \(f(q)=q^2\).

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, otherwise \(f(f(n))=n\). Hence \(f(n)=n+1\), and this function works.

With \(x=0\), \(f(f(0)+y)=f(y)\). Thus shifting by \(f(0)\) does not change \(f\). With \(y=0\), \(f(f(x))=f(0)+x\). Comparing these relations gives \(f(0)=0\), hence \(f(f(x))=x\). Now replace \(x\) by \(f(t)\): \(f(t+y)=f(y)+f(t)\). Thus \(f\) is additive. On \(\mathbb Q\), \(f(q)=cq\), and \(f(f(q))=q\) gives \(c^2=1\). Answer: \(f(q)=q\) and \(f(q)=-q\).

Let \(g(x)=f(x)-x^3\). Since \((x+y)^3-x^3-y^3=3xy(x+y)\), the function \(g\) is additive. Also, \(g(1)=3\), so \(g(q)=3q\) on \(\mathbb Q\). Hence \(f(q)=q^3+3q\).

Put \(g(x)=f(x)-2x^2\). Then \(g(x+y)+g(x-y)=2g(x)\). With \(x=0\), \(g(y)+g(-y)=0\). With \(x=(u+v)/2\), \(y=(u-v)/2\), \(g(u)+g(v)=2g((u+v)/2)\). On \(\mathbb Q\), Jensen's equation with \(g(0)=0\), \(g(1)=0\) gives \(g\equiv0\). Hence \(f(q)=2q^2\).

Let \(A(x)=f(x)-\frac{x^2}{2}\). Then \(A(x+y)=A(x)+A(y)\). Therefore all solutions are \(f(x)=\frac{x^2}{2}+A(x)\), where \(A:\mathbb R\to\mathbb R\) is any additive function. Verification follows from \((x+y)^2/2=x^2/2+y^2/2+xy\).

For \(t\in[0,1]\), \(f(t)\ge-1\), and \(f(t)=f(1)-f(1-t)\le f(1)+1\). Hence \(f\) is bounded on \([0,1]\). An additive function bounded on an interval is continuous at zero, and therefore linear: \(f(x)=xf(1)\).

With \(y=1\), \(f(x)=x f(1)+f(x)\), so \(f(1)=0\). Put \(f(x)=xD(x)\) for \(x\ne0\). Then the equation becomes \(xyD(xy)=xyD(y)+xyD(x)\), hence \(D(xy)=D(x)+D(y)\), and \(D(2)=0\). Conversely, every multiplicatively additive \(D:\mathbb Q^{*}\to\mathbb Q\) with \(D(2)=0\), together with \(f(0)=0\), gives a solution by \(f(x)=xD(x)\). Thus the answer is a family, not a single formula.

Since \((x+y)^3-x^3-y^3=3x^2y+3xy^2\), put \(g(x)=f(x)-\frac{x^3}{3}\). Then \(g\) is additive. We have \(g(1)=1-\frac13=\frac23\), so on \(\mathbb Q\), \(g(q)=\frac{2q}{3}\). Answer: \(f(q)=\frac{q^3+2q}{3}\).

With \(x=0\), \(f(f(y))=2y\). Since \(f\) is surjective, for every \(t\) there is \(y\) with \(t=f(y)\), and then \(f(t)=2y\). Hence \(f(x+t)=x+f(t)\) for all \(x,t\). Thus \(f(u)-u\) is constant: \(f(u)=u+c\). Then \(f(f(y))=y+2c\), but it must be \(2y\), impossible for all \(y\). Therefore there are no solutions.

For \(u,v\), \(\Phi(u)+\Phi(v)=2\Phi((u+v)/2)\Phi((u-v)/2)\). For \(u,-v\), \[\Phi(u)+\Phi(-v)=2\Phi\left(\frac{u-v}{2}\right)\Phi\left(\frac{u+v}{2}\right).\] The right sides are equal, so \(\Phi(u)+\Phi(v)=\Phi(u)+\Phi(-v)\). Hence \(\Phi(v)=\Phi(-v)\) for all \(v\).

For any \(s\), since \(s0\). Fix \(r1+2^n a\) for all \(n\). For \(n=1\), \(H(r)\ge H(t_0)^2=(1+a)^2>1+2a\). For the induction step, take \(m=(r+t_0)/2\). It lies between \(r\) and \(t_0\); applying the same midpoint argument gives \(H(m)>1+2^n a\), hence \(H(r)\ge H(m)^2>1+2^{n+1}a\). Thus the fixed number \(H(r)\) exceeds \(1+2^n a\) for all \(n\), impossible. Therefore \(H(t)\le1\) for all \(t\), and with nonnegativity, \(0\le H(t)\le1\).

Let \(a_1,a_2\), \(b_1,b_2\), \(c_1,c_2\) be the roots of \(A,B,C\). Consider \(S=A+B+C\). At \(c_1,c_2\), \(C(c_1)=C(c_2)=0\), so equal values of \(A+B\) give \(S(c_1)=S(c_2)\). Similarly, \(S(a_1)=S(a_2)\) and \(S(b_1)=S(b_2)\). Since \(S\) is a quadratic polynomial with positive leading coefficient, two distinct points with equal values are symmetric about the same axis \(x=d\). Therefore \(a_1+a_2=b_1+b_2=c_1+c_2=2d\).

Put \(a=u\), \(b=2v\). The condition becomes \(p(a^2+b^2)\ge p(2ab)\). For any \(0\le r\le q\), there exist nonnegative \(a,b\) with \(2ab=r\), \(a^2+b^2=q\): take \(a=\frac{\sqrt{q+r}+\sqrt{q-r}}{2}\), \(b=\frac{\sqrt{q+r}-\sqrt{q-r}}{2}\). Hence \(p(q)\ge p(r)\) whenever \(0\le r\le q\), so \(p\) is nondecreasing on \([0,\infty)\). If both roots were nonnegative, a quadratic with two distinct roots would change sign between and after them, and could not be nondecreasing on all of \([0,\infty)\). Thus at least one root is negative.

Assume the contrary: among the seven sums there are at most two distinct values. Then one value, say \(s\), occurs in at least four cycles. For every number \(x\) in such a cycle, \(x+F(x)+F(F(x))=s\). Hence all \(12\) numbers from these four cycles are roots of \(G(x)=x+F(x)+F(F(x))-s\). Since \(F\) is cubic, \(F(F(x))\) has degree \(9\), so \(G\) has degree at most \(9\). A nonzero polynomial of degree at most \(9\) cannot have \(12\) distinct roots. Contradiction. Therefore at least three distinct sums occur.

Let \(00\). If \(c=1\), then \(R_2-R_1=\sqrt{b^2-4a}\), the positive square root of an integer; for distinct roots, \(R_2-R_1\ge1\). Hence \(R_1\ge50\), \(R_2\ge51\), and \(a=R_1R_2\ge50\cdot51=2550\). If \(c\ge2\), then \(a=cR_1R_2\ge2\cdot50^2>2550\). Thus \(a\ge2550\). This bound is attained because \(2550x^2-101x+1\) has roots \(\frac1{50}\) and \(\frac1{51}\).

For integers \(n\), \(f(n)=5n\). If \(q=m/n\), then \(n f(q)=f(m)=5m\), so \(f(q)=5q\).

Then \(g\) is additive and \(g(2)=6\), so \(g(1)=3\). Hence \(g(q)=3q\), and \(f(q)=3q+1\).

With \(x=0\), \(f(y)=f(0)\) for all \(y\). Every constant function works.

From \(x=y=0\), \(f(0)=0\). For any \(u,v\in\mathbb Q\), take \(x=u/2\), \(y=v/3\): then \(f(u+v)=f(u)+f(v)\). Hence \(f(q)=7q\).

Let \(g(x)=f(x)-4\). Then \(g(x+y)=g(x)+g(y)\), and \(g(1)=5\). On \(\mathbb Q\), \(g(q)=5q\), so \(f(q)=5q+4\).

From \(x=0\) in the second condition, \(f(1)=5\). Now \(g=f-2\) is additive and \(g(1)=3\). Hence \(g(q)=3q\), so \(f(q)=3q+2\).

For \(g=f-1\), \(g((x+y)/2)=(g(x)+g(y))/2\), \(g(0)=0\), \(g(1)=3\). On rationals, the midpoint relation gives additivity: \(g(u+v)=g(u)+g(v)\) after taking \(x=2u\), \(y=2v\) and using \(g(2t)=2g(t)\). Hence \(g(q)=3q\), and \(f(q)=3q+1\).

We have \(f(f(x))=a(ax+b)+b=a^2x+b(a+1)\). Thus \(a^2=4\), \(b(a+1)=3\). If \(a=2\), then \(b=1\); if \(a=-2\), then \(b=-3\). Answer: \(2x+1\) and \(-2x-3\).

From \(f(x+1)=f(x)+2\) at \(x=1\), \(f(2)=f(1)+2\), so \(f(1)=5\). From the first equation at \(x=y=1\), \(f(2)=2f(1)-f(0)\), hence \(7=10-f(0)\), so \(f(0)=3\). Then \(g=f-3\) is additive and \(g(1)=2\), so \(g(q)=2q\). Answer: \(f(q)=2q+3\). Direct verification checks all three conditions.

On rationals, \(f(q)=qf(1)\). Let rational sequences \(r_n\uparrow x\), \(s_n\downarrow x\). By monotonicity, \(f(r_n)\le f(x)\le f(s_n)\), so \(r_nf(1)\le f(x)\le s_nf(1)\), with the sign handled by two-sided approximation. Passing to the limit gives \(f(x)=xf(1)\).

If \(h\) is small, choose \(n\) so that \(0\le nh\le1\). Then \(|f(h)|=|f(nh)|/n\le10/n\), which tends to zero as \(h\to0\). Thus \(f\) is continuous at zero. A continuous additive function is linear: \(f(x)=xf(1)\).

If \(x\le y\), then \(y-x\ge0\), so \(f(y)-f(x)=f(y-x)\ge0\). Hence \(f\) is nondecreasing. By the previous result, \(f(x)=cx\), with \(c=f(1)\ge0\).

Additivity gives \(f(q)=cq\). Substitute into the second condition: \(cxy=x\cdot cy+y\cdot cx=2cxy\). With \(x=y=1\), \(c=2c\), so \(c=0\). Answer: \(f\equiv0\).

If \(x>0\), then \(x=t^2\) for some \(t\ne0\). Thus \(f(x)=f(t)^2\ge0\). Moreover \(f(t)\ne0\), otherwise \(1=f(1)=f(t\cdot1/t)=0\). Hence \(f(x)>0\) for \(x>0\), so \(f\) is increasing. An increasing additive function is linear: \(f(x)=cx\). From \(f(1)=1\), \(c=1\).

We have \(f(f(x)+1)=a(ax+b+1)+b=a^2x+ab+a+b\). Thus \(a^2=9\) and \(b(a+1)+a=5\). If \(a=3\), then \(4b+3=5\), so \(b=1/2\). If \(a=-3\), then \(-2b-3=5\), so \(b=-4\). Both work.

Let \(c=f(0)\), \(g=f-c\). Then \(g\) is additive, and \(f(3x)=g(3x)+c=3g(x)+c\). The right side of the second condition is \(3(g(x)+c)-4=3g(x)+3c-4\). Hence \(c=3c-4\), so \(c=2\). The additive part on \(\mathbb Q\) is \(g(q)=kq\). Answer: \(f(q)=kq+2\), \(k\in\mathbb Q\).

Let \(g(x)=f(x)-3\). Then \(g\) is additive. On \([0,1]\), \(g(x)\ge -8\). Also \(g(x)=g(1)-g(1-x)\le g(1)+8\), so \(g\) is bounded on \([0,1]\). Therefore \(g(x)=cx\), and \(f(x)=cx+3\).

From \(f(0)=0\), \(b=0\), so \(f(x)=ax\). Then \(f(f(x))=a^2x\), which cannot equal \(x+1\) for all \(x\), because the right side has constant term \(1\). No solutions.

By monotonicity, the additive function is linear: \(f(x)=cx\). Then \(f(f(x))=c^2x=x\), so \(c^2=1\). Since \(f\) is nondecreasing, \(c\ge0\), hence \(c=1\). Answer: \(f(x)=x\).

Let \(A(x)=f(x)-2\). Then \(A(x+y)=A(x)+A(y)\). Without extra conditions, all solutions are \(f(x)=A(x)+2\), where \(A\) is arbitrary additive. If \(f\) is nondecreasing, then \(A\) is nondecreasing, so \(A(x)=cx\). The condition \(f(1)=5\) gives \(c+2=5\), hence \(c=3\). Thus \(f(x)=3x+2\).

Since \(f\) is injective, the equality gives \(f(x)+y=f(y)+x\). Hence \(f(x)-x=f(y)-y\) for all \(x,y\). Fixing \(y=0\), we get \(f(x)-x=f(0)\), so \(f(x)=x+f(0)\). Take \(c=f(0)\).

By surjectivity, for every \(t\in\mathbb R\) there is \(y\) such that \(t=f(y)\). Hence \(f(x+t)=f(x)+t\) for all \(x,t\). Taking \(x=0\), we get \(f(t)=f(0)+t\). Thus \(f(x)=x+c\), where \(c=f(0)\). Checking: \(f(x+f(y))=x+f(y)+c=f(x)+f(y)\). The answer is all \(f(x)=x+c\), \(c\in\mathbb R\).

If \(f(a)=f(b)\), then \(f(f(a))=f(f(b))\). By the condition, \(a+1=b+1\), so \(a=b\). Thus \(f\) is injective.

Now let \(T\in\mathbb Q\). Then \(f(f(T-1))=T\). Hence \(T\) lies in the image of \(f\). Since \(T\) was arbitrary, \(f\) is surjective. Therefore \(f\) is bijective.

Since \(f\) is surjective, \(t=f(y)\) runs through all real numbers. Hence \(f(x+t)=x+t+1\) for all \(x,t\). Every number \(u\) can be written as \(u=x+t\), so \(f(u)=u+1\). The check is immediate: \(f(x+f(y))=x+f(y)+1\). The answer is \(f(x)=x+1\).

By injectivity, \(f(x)+f(y)=x+y\). Taking \(y=0\), we get \(f(x)+f(0)=x\), so \(f(x)=x-c\), where \(c=f(0)\). Substitute this into \(f(x)+f(y)=x+y\): \((x-c)+(y-c)=x+y\), hence \(c=0\). Therefore \(f(x)=x\). The check is immediate.

With \(x=0\), \(f(f(y))=f(0)+y\). If \(f(a)=f(b)\), then \(f(f(a))=f(f(b))\), hence \(a=b\). Thus \(f\) is injective. With \(y=0\), \(f(x+f(0))=f(x)\), so \(f(0)=0\). Hence \(f(f(y))=y\).

Put \(y=f(t)\): \(f(x+f(f(t)))=f(x)+f(t)\), so \(f(x+t)=f(x)+f(t)\). On \(\mathbb Q\), an additive function has the form \(f(x)=cx\). From \(f(f(y))=y\), we get \(c^2=1\). The answers are \(f(x)=x\) and \(f(x)=-x\). Direct checking works.

With \(y=0\), \(f(f(x))=x+f(0)\), so \(f\) is injective and surjective. With \(x=0\), \(f(f(0)+y)=f(y)\). By injectivity, \(f(0)+y=y\), hence \(f(0)=0\). Thus \(f(f(x))=x\).

Put \(y=f(t)\): \(f(f(x)+f(t))=x+t=f(f(x+t))\). By injectivity, \(f(x)+f(t)=f(x+t)\). Hence \(f\) is additive on \(\mathbb Q\), so \(f(x)=cx\). The condition \(f(f(x))=x\) gives \(c^2=1\). The answer is \(f(x)=x\) or \(f(x)=-x\).

By surjectivity, \(t=f(y)\) is arbitrary. Hence \(f(x+t)=f(x)+2t\) for all \(x,t\). Taking \(x=0\), we get \(f(t)=f(0)+2t\). Therefore \(f(x)=2x+c\). Checking: \(f(x+f(y))=2x+2f(y)+c=f(x)+2f(y)\). All functions \(f(x)=2x+c\) work.

If \(f(a)=f(b)\), comparing the original equality for \(y=a\) and \(y=b\) gives \(3a=3b\), so \(a=b\). Thus \(f\) is injective. With \(y=0\), \(f(x+f(0))=f(x)\), hence \(f(0)=0\). With \(x=0\), \(f(f(y))=3y\), so \(f\) is surjective.

Let \(t=f(y)\). Then \(f(t)=3y\), and the original equation becomes \(f(x+t)=f(x)+f(t)\). Hence \(f\) is additive on \(\mathbb Q\), so \(f(x)=cx\) for some rational \(c\). The condition \(f(f(y))=3y\) gives \(c^2=3\), impossible for rational \(c\). Contradiction.

By surjectivity, \(t=f(y)\) is arbitrary, so \(f(x+t)=f(x)+t+1\). Taking \(x=0\), we get \(f(t)=f(0)+t+1\). Then the left-hand side of \(f(x+t)=f(x)+t+1\) equals \(f(0)+x+t+1\), while the right-hand side equals \(f(0)+x+1+t+1\). This gives \(0=1\), a contradiction. No such function exists.

If \(f(a)=f(b)\), comparing the equations for \(y=a\) and \(y=b\) gives \(a=b\), so \(f\) is injective. Let \(c=f(0)\). With \(x=0\), \(f(f(y))=c+y+1\). Put \(t=f(y)\). Then \(f(t)=c+y+1\), so \(y+1=f(t)-c\). Hence the original equation gives \(f(x+t)=f(x)+f(t)-c\) for all \(x,t\) in the image. The function is surjective because \(f(f(T-c-1))=T\), so this holds for all \(t\).

Let \(g(x)=f(x)-c\). Then \(g(x+t)=g(x)+g(t)\), so on \(\mathbb Q\) we have \(g(x)=ax\). Also, from \(f(f(y))=c+y+1\) we get \(g(g(y)+c)=y+1\). Since \(g\) is additive, \(g(g(y))+g(c)=y+1\). At \(y=0\), \(g(c)=1\), hence \(g(g(y))=y\). Therefore \(a^2=1\) and \(ac=1\). If \(a=1\), then \(c=1\); if \(a=-1\), then \(c=-1\). The answer is \(f(x)=x+1\) and \(f(x)=-x-1\). Direct checking works.

If \(f(a)=f(b)\), comparing the equations for \(y=a\) and \(y=b\) gives \(a=b\), so \(f\) is injective. Let \(c=f(0)\). With \(x=0\), \(f(f(y))=c+f(y)+y\). If \(t=f(y)\), then \(f(t)=c+t+y\), so \(t+y=f(t)-c\). Hence \(f(x+t)=f(x)+f(t)-c\). Thus \(g(x)=f(x)-c\) is additive, and on \(\mathbb Q\), \(g(x)=ax\).

Therefore \(f(x)=ax+c\). Substitute into the original equation: \(a(x+ay+c)+c=ax+(a+1)y+2c\). Comparing coefficients gives \(a^2=a+1\) and \(c(a-1)=0\). But \(a^2-a-1=0\) has no rational roots. Contradiction.

If \(f(a)=f(b)\), comparing the equalities for \(y=a\) and \(y=b\) gives \(2a=2b\), so \(a=b\). The function 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 \(f(t)=2y\), hence \(f(x+t)=f(x)+f(t)\). The function is additive and increasing, so \(f(x)=cx\). From \(f(f(y))=2y\), we get \(c^2=2\), and increasing gives \(c>0\). The answer is \(f(x)=\sqrt{2}x\). Direct checking works.

With \(y=0\), \(f(f(x))=x+f(0)\). This implies injectivity: if \(f(a)=f(b)\), then \(a+f(0)=b+f(0)\). It also implies surjectivity. With \(x=0\), \(f(f(0)+y)=f(y)\), so injectivity gives \(f(0)=0\). Hence \(f(f(x))=x\).

Put \(y=f(t)\): \(f(f(x)+f(t))=x+t=f(f(x+t))\). By injectivity, \(f(x)+f(t)=f(x+t)\). The function is continuous and additive, so \(f(x)=cx\). The condition \(f(f(x))=x\) gives \(c^2=1\). The answer is \(f(x)=x\) and \(f(x)=-x\).

If \(f(a)=f(b)\), comparing the equations for \(y=a\) and \(y=b\) gives \(a=b\), so \(f\) is injective. Let \(c=f(0)\). With \(x=0\), \(f(f(y))=c+f(y)+y\). If \(t=f(y)\), then \(f(t)=c+t+y\), so \(t+y=f(t)-c\). Hence \(f(x+t)=f(x)+f(t)-c\). Therefore \(g(x)=f(x)-c\) is additive and continuous, so \(g(x)=ax\).

Thus \(f(x)=ax+c\). Substituting into the original equation gives \(a(x+ay+c)+c=ax+(a+1)y+2c\). Hence \(a^2=a+1\) and \(c(a-1)=0\). Since the roots of the first equation are not \(1\), we get \(c=0\). Thus \(a=\frac{1+\sqrt{5}}{2}\) or \(a=\frac{1-\sqrt{5}}{2}\). Both functions \(f(x)=ax\) work.

Since \(f\) is surjective, \(t=f(x)\) runs through all real numbers. Then \(f(t+y)=t+f(y)\) for all \(t,y\). Taking \(y=0\), we obtain \(f(t)=t+f(0)\). Hence \(f(x)=x+c\). Checking: \(f(f(x)+y)=f(x)+y+c=x+c+y+c=f(x)+f(y)\). All functions \(f(x)=x+c\) work.

Since \(f\) is surjective, \(t=f(y)\) is arbitrary. Therefore \(f(x+t)+f(x-t)=2f(x)\) for all \(x,t\). Taking \(u=x+t\), \(v=x-t\), we get \(f\left(\frac{u+v}{2}\right)=\frac{f(u)+f(v)}{2}\). A continuous function preserving midpoints is affine: \(f(x)=ax+b\).

Surjectivity requires \(a\ne0\). Checking shows that every function \(f(x)=ax+b\), \(a\ne0\), satisfies the original equation. The answer is \(f(x)=ax+b\), where \(a\ne0\).

If \(f(a)=f(b)\), then using the equations with the same \(y\) gives \(a+y=b+y\), so \(a=b\). Thus \(f\) is injective. Also, every \(T\) equals \(f(f(T)+f(0))\), so \(f\) is surjective. Let \(g=f^{-1}\).

Putting \(y=0\), we get \(f(f(x)+f(0))=x\), hence \(g(x)=f(x)+f(0)\). Let \(c=f(0)\). Then \(g(x)=f(x)+c\). From the original equation with \(x=y=g(0)\), we obtain \(f(0)=2g(0)\). But from \(g(0)=f(0)+c\), we get \(g(0)=2c\). Hence \(c=4c\), so \(c=0\). Therefore \(g=f\) and \(f(f(x))=x\).

Now for arbitrary \(u=f(x)\), \(v=f(y)\), we have \(f(u+v)=x+y=f(u)+f(v)\), because \(g=f\). Thus \(f\) is additive. An increasing additive function has the form \(f(x)=ax\), where \(a>0\). The condition \(f(f(x))=x\) gives \(a^2=1\), hence \(a=1\). The answer is \(f(x)=x\).

With \(x=0\), \(f(f(y))=f(0)+y\). This implies injectivity in the usual way. With \(y=0\), \(f(x+f(0))=f(x)\), hence \(f(0)=0\). Thus \(f(f(y))=y\).

Putting \(y=f(t)\), we get \(f(x+t)=f(x)+f(t)\), so \(f\) is additive. A continuous additive function has the form \(f(x)=cx\). From \(f(f(y))=y\), we get \(c^2=1\), hence \(f(x)=x\) or \(f(x)=-x\). In the first case \(f(2)=2\), in the second \(f(2)=-2\). Neither gives \(3\). Contradiction.

If \(f(a)=f(b)\), comparing the equations for \(y=a\) and \(y=b\) gives \(a=b\), so \(f\) is injective. Let \(c=f(0)\). With \(x=0\), \(f(f(y))=c+f(y)+y+1\). Put \(t=f(y)\). Then \(f(t)=c+t+y+1\), so \(t+y+1=f(t)-c\). Hence \(f(x+t)=f(x)+f(t)-c\). Therefore \(g(x)=f(x)-c\) is additive and continuous, so \(g(x)=ax\).

From \(f(c)=f(0+f(0))=c+c+0+1=2c+1\), we get \(g(c)=c+1\). Also, \(f(f(y))=c+f(y)+y+1\) becomes \(g(g(y)+c)+c=g(y)+y+2c+1\). Since \(g\) is additive and \(g(c)=c+1\), this gives \(g(g(y))=g(y)+y\). For \(g(x)=ax\), we get \(a^2=a+1\).

The remaining condition is \(g(c)=c+1\), that is \(ac=c+1\), or \(c(a-1)=1\). Thus \(a\) is one of the roots of \(a^2-a-1=0\), and \(c=\frac{1}{a-1}\). Hence the two functions are \(f(x)=ax+\frac{1}{a-1}\), where \(a=\frac{1+\sqrt{5}}{2}\) or \(a=\frac{1-\sqrt{5}}{2}\). Direct substitution confirms both.

The statement is true for \(n=1\). If \(f(n)=7n\), then \(f(n+1)=f(n)+f(1)=7n+7=7(n+1)\). Hence \(f(n)=7n\) for all \(n\in\mathbb N\).

From \(f(2)=f(1)+f(1)\), we get \(f(1)=5\). For positive \(n\), \(f(n)=5n\). Also \(0=f(0)=f(n+(-n))=f(n)+f(-n)\), so \(f(-n)=-5n\). The answer is \(f(n)=5n\) for all \(n\in\mathbb Z\).

By additivity on \(\mathbb Q\), \(f(q)=qf(1)\). From \(f(3)=3f(1)=12\), we get \(f(1)=4\). The answer is \(f(q)=4q\).

Let \(g(x)=f(x)+6\). Then \(g(x+y)=f(x+y)+6=f(x)+f(y)+12=g(x)+g(y)\). Hence \(g(x)=ax\), where \(a\in\mathbb Q\). Therefore \(f(x)=ax-6\). All such functions work.

If \(n=p_1^{a_1}\cdots p_k^{a_k}\), then by multiplicativity \(f(n)=f(p_1)^{a_1}\cdots f(p_k)^{a_k}=p_1^{2a_1}\cdots p_k^{2a_k}=n^2\). The answer is \(f(n)=n^2\).

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

Since \(f\) is continuous and additive, \(f(x)=cx\). From \(c\sqrt{3}=6\), we get \(c=\frac{6}{\sqrt{3}}=2\sqrt{3}\). The answer is \(f(x)=2\sqrt{3}x\).

By the standard fact, an additive function bounded on an interval is continuous at zero and hence everywhere. Therefore \(f(x)=cx\). Since \(f(1)=2\), we get \(c=2\). Thus \(f(x)=2x\).

As in the theory, continuous solutions have the form \(f(x)=ax+b\). From \(f(0)=3\), we get \(b=3\). From \(f(2)=9\), \(2a+3=9\), so \(a=3\). The answer is \(f(x)=3x+3\).

The recursion gives \(f(n+1)-f(n)=2n+1=(n+1)^2-n^2\). Since \(f(1)=1=1^2\), telescoping yields \(f(n)=n^2\). Multiplicativity checks: \(f(mn)=m^2n^2=f(m)f(n)\). The answer is \(f(n)=n^2\).

By additivity, \(f((x+y)^2)=f(x^2)+2f(xy)+f(y^2)\). On the other hand, by the condition this equals \((x+y)f(x+y)=(x+y)(f(x)+f(y))\). After cancelling \(x f(x)+y f(y)\), we get \(2f(xy)=x f(y)+y f(x)\). Put \(y=1\). Then \(2f(x)=x f(1)+f(x)\), hence \(f(x)=x f(1)\). Take \(c=f(1)\).

Let \(g(x)=f(x)-x^2\). Then \(g(x+y)+g(x-y)=2g(x)\). On \(\mathbb Q\), this midpoint equation gives \(g(x)=ax+b\). Hence \(f(x)=x^2+ax+b\), where \(a,b\in\mathbb Q\). Direct checking works.

Since \(f\) is additive on \(\mathbb Q\), \(f(x)=cx\), where \(c\in\mathbb Q\). Then \(f(f(x))=c^2x\). The condition gives \(c^2=4\), hence \(c=2\) or \(c=-2\). The answers are \(f(x)=2x\) and \(f(x)=-2x\).

Let \(g(x)=f(x)-1\). The first condition gives \(g(x+y)=g(x)+g(y)\). The second condition, after substituting \(f=g+1\), becomes \(g(xy)=g(x)g(y)\). On \(\mathbb Q\), an additive function has the form \(g(x)=cx\), and multiplicativity gives \(c=0\) or \(c=1\). Thus \(f(x)=1\) or \(f(x)=x+1\). Both functions work.

From the midpoint equation and continuity, \(f(x)=ax+b\). Then \(f(f(x))=a(ax+b)+b=a^2x+b(a+1)\). This must equal \(x\) for all \(x\). Hence \(a^2=1\) and \(b(a+1)=0\). If \(a=1\), then \(b=0\), giving \(f(x)=x\). If \(a=-1\), then \(b\) is arbitrary, giving \(f(x)=-x+b\). All these functions work.

With \(y=0\), we get \(f(0)=0\). Continuous solutions of the quadratic Cauchy equation have the form \(f(x)=cx^2\). From \(f(2)=12\), \(4c=12\), so \(c=3\). The answer is \(f(x)=3x^2\). Checking follows by expanding squares.

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

Continuous additivity gives \(f(x)=cx\). The second condition becomes \(c^2xy=cxy+xy\) for all \(x,y\). Hence \(c^2=c+1\). We get \(c=\frac{1+\sqrt{5}}{2}\) or \(c=\frac{1-\sqrt{5}}{2}\). Both functions work.

Let \(g(x)=f(x)-1\). Then \(g(x+y)=g(x)+g(y)\). The second condition gives \(g(xy)+1=(g(x)+1)(g(y)+1)-(g(x)+1)-(g(y)+1)+2=g(x)g(y)+1\), hence \(g(xy)=g(x)g(y)\). On \(\mathbb Q\), additivity gives \(g(x)=cx\). Multiplicativity gives \(c=c^2\), so \(c=0\) or \(c=1\). The answer is \(f(x)=1\) and \(f(x)=x+1\).

Putting \(x=y\), we get \(f(2x)+f(0)=4f(x)\), so \(f(2x)=4f(x)\). A standard induction using the quadratic Cauchy formula gives \(f(nx)=n^2f(x)\) for integers \(n\). With \(x=1\), this gives \(f(n)=n^2\). Then from \(f(n\cdot \frac{m}{n})=n^2 f(\frac{m}{n})\), we obtain \(f(\frac{m}{n})=\frac{m^2}{n^2}\). Hence \(f(q)=q^2\) for all rational \(q\).

For an arbitrary real \(x\), choose rationals \(q_k\to x\). By continuity, \(f(x)=\lim f(q_k)=\lim q_k^2=x^2\). Therefore \(f(x)=x^2\) for all \(x\).

For \(n>0\), induction gives \(f(n)=4n-1\). For \(n<0\), from \(f(n)=f(n+1)-4\) we get the same formula. The answer is \(f(n)=4n-1\).

If \(n=2k\), then \(f(n)=2+6k=3n+2\). If \(n=2k+1\), then \(f(n)=5+6k=3n+2\). The answer is \(f(n)=3n+2\). The formula also works for negative \(k\), since the recurrence is reversible.

Then \(g(m+n)=f(m+n)+1=f(m)+f(n)+2=g(m)+g(n)\). On \(\mathbb N\), \(g(n)=ng(1)=5n\). Hence \(f(n)=5n-1\).

Let \(c=f(0)\). By induction, \(f(1)\equiv c+1\), \(f(2)\equiv c+2\), and so on. Since all values are taken modulo \(7\), we get \(f(x)\equiv c+x\pmod 7\) for every residue \(x\).

The condition is equivalent to \(f(n+2)-f(n+1)=f(n+1)-f(n)\). Thus first differences are constant. Since \(f(1)-f(0)=2\), we have \(f(n+1)-f(n)=2\) for all \(n\). Hence \(f(n)=2n\).

For \(n>0\), \(f(n)=nf(1)=-3n\). Also \(f(0)=0\), and from \(0=f(n+(-n))\), we get \(f(-n)=-f(n)=3n\). Thus for all \(n\in\mathbb Z\), \(f(n)=-3n\).

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

Let \(B(n)=\binom{n}{2}=\frac{n(n-1)}{2}\). Then \(B(m+n)=B(m)+B(n)+mn\). Hence \(g(n)=f(n)-B(n)\) is additive on \(\mathbb Z\), so \(g(n)=cn\). The answer is \(f(n)=\binom{n}{2}+cn\), where \(c\in\mathbb Z\).

Put \(n=1\): \(f(m+1)+f(m-1)=2f(m)\). This means \(f(m+1)-f(m)=f(m)-f(m-1)\), so first differences are constant. Therefore \(f(n)=an+b\). All such functions indeed work.

For every \(n\), the number \(2n+1\) is odd. Hence \(f(n+1)\) and \(f(n)\) have opposite parity. Therefore all values of the function cannot have the same parity. Contradiction.

With \(n=1\), \(f(m+1)+f(m-1)=2f(m)+2\). Hence \(f(m+1)-f(m)=f(m)-f(m-1)+2\). Let \(d_m=f(m)-f(m-1)\). Then \(d_{m+1}=d_m+2\), and \(d_1=f(1)-f(0)=1\). Thus \(d_m=2m-1\) for positive \(m\), and summing gives \(f(n)=1+3+\cdots+(2n-1)=n^2\). For negative \(n\), the equation with \(m=0\) gives \(f(-n)=f(n)\). Therefore \(f(n)=n^2\) for all integers \(n\).

The condition gives \(g(mn)=g(m)+g(n)\). For a prime \(p\), \(g(p)=1\). If \(n=p_1^{a_1}\cdots p_k^{a_k}\), then \(g(n)=a_1+\cdots+a_k\). Hence \(f(n)=n(a_1+\cdots+a_k)\), that is, \(f(n)=n\Omega(n)\), where \(\Omega(n)\) is the number of prime factors counted with multiplicity.

Let \(a=f(1)\). For a residue \(x\), we have \(x=1+\cdots+1\) modulo \(p\). Then \(f(x)=xf(1)=ax\). This also holds for \(x=0\), since \(f(0)=0\).

With \(n=0\), \(f(f(m))=f(0)+m\). If \(f(a)=f(b)\), then \(f(f(a))=f(f(b))\), hence \(a=b\); so \(f\) is injective. With \(m=0\), \(f(n+f(0))=f(n)\), hence \(f(0)=0\). Thus \(f(f(m))=m\). Put \(m=f(t)\): \(f(n+t)=f(n)+f(t)\). Therefore \(f\) is additive on \(\mathbb Z\), so \(f(n)=cn\). From \(f(f(n))=n\), we get \(c^2=1\). The answer is \(f(n)=n\) and \(f(n)=-n\).

For \(g(n)=f(n)+2\), we get \(g(m+n)=g(m)+g(n)\). Thus \(g(n)=an\), where \(a\in\mathbb Z\), and \(f(n)=an-2\). Then \(f(f(n))=a(an-2)-2=a^2n-2a-2\). This equals \(n\) for all \(n\), so \(a^2=1\) and \(-2a-2=0\). Therefore \(a=-1\). The answer is \(f(n)=-n-2\).

By additivity on residues, \(f(x)=ax\), where \(a=f(1)\). Then \(f(f(x))=a^2x\). The condition gives \(a^2\equiv1\pmod p\). Since \(p\) is an odd prime, \((a-1)(a+1)\equiv0\pmod p\), so \(a\equiv1\) or \(a\equiv-1\). The answer is \(f(x)=x\) and \(f(x)=-x\) modulo \(p\).

For \(n>0\), \(f(n)=\sum_{k=0}^{n-1}(2k+1)=n^2\). For negative \(n\), move backwards: \(f(n)=f(n+1)-(2n+1)\), and the same formula \(n^2\) satisfies the recurrence and the initial value. The answer is \(f(n)=n^2\).

Notice that \(4mn=(m+n)^2-(m-n)^2\). Let \(u=m+n\), \(v=m-n\). Then \(u\) and \(v\) have the same parity, and the condition says \(f(u)-u^2=f(v)-v^2\). Hence \(f(n)-n^2\) is constant separately on even and on odd \(n\). Thus all solutions are \(f(n)=n^2+c_0\) for even \(n\) and \(f(n)=n^2+c_1\) for odd \(n\), where \(c_0,c_1\in\mathbb Z\). The check is immediate because \(m+n\) and \(m-n\) always have the same parity.

The function \(n^2\) satisfies the recurrence and the three initial values. A step-three recurrence determines values separately on the classes \(0,1,2\pmod3\). The initial values on these three classes agree with \(n^2\), so by induction forward and backward \(f(n)=n^2\) for all integers \(n\).

With \(x=0\), \(f(f(y))=f(0)+y\). If \(f(a)=f(b)\), then \(f(f(a))=f(f(b))\), hence \(a=b\); the function is injective. On a finite set it is bijective. With \(y=0\), \(f(x+f(0))=f(x)\), so \(f(0)=0\). Thus \(f(f(y))=y\).

Put \(y=f(t)\): \(f(x+t)=f(x)+f(t)\). Hence \(f\) is additive on \(\mathbb Z/p\mathbb Z\), so \(f(x)=ax\). From \(f(f(x))=x\), we get \(a^2\equiv1\pmod p\). Since \(p\) is an odd prime, \(a\equiv1\) or \(a\equiv-1\). The answer is \(f(x)=x\) and \(f(x)=-x\) modulo \(p\).

With \(y=0\), \(P(0)=0\). If the degree of \(P\) is at least \(2\), then \(P(x+y)\) has a mixed term of highest degree, absent from \(P(x)+P(y)\). Hence \(P\) is linear or zero. Thus \(P(x)=ax+b\), and \(P(0)=0\) gives \(b=0\). The answer is \(P(x)=ax\).

Let \(P\) have degree \(d\ge2\) and leading coefficient \(a\). The leading part on the left is \(a((x+y)^d+(x-y)^d)\), while on the right it is \(2ax^d\). Expanding produces a term with \(y^2x^{d-2}\) when \(d\ge2\), impossible. Hence \(d\le1\). All linear polynomials \(P(x)=ax+b\) work.

Putting \(y=0\), we get \(P(0)=0\). Comparing leading parts shows that the degree is at most \(2\). Let \(P(x)=ax^2+bx\). Substitution gives \(2ax^2+2ay^2+2bx=2ax^2+2bx+2ay^2+2by\), hence \(b=0\). From \(P(1)=2\), \(a=2\). The answer is \(P(x)=2x^2\).

The polynomial \(Q(x)=P(x)-5x\) satisfies \(Q(x+1)=Q(x)\). A periodic polynomial is constant. Hence \(Q(x)=c\), and \(P(x)=5x+c\). All such polynomials work.

Since \((x+1)^2-x^2=2x+1\), the polynomial \(Q(x)=P(x)-x^2\) is periodic with period \(1\). Therefore \(Q\) is constant. The answer is \(P(x)=x^2+c\).

The polynomial \(F(x)=\frac{x(x-1)(2x-1)}{6}\) has difference \(F(x+1)-F(x)=x^2\). Then \(Q=P-F\) is periodic with period \(1\), hence constant. The answer is \(P(x)=\frac{x(x-1)(2x-1)}{6}+c\).

The constant solutions are \(P=0\), \(P=1\). Let \(P\) be nonconstant. Then \(P(1)=1\). If \(r\ne0\) and \(P(r)=0\), then \(P(ry)=0\) for all \(y\), so \(P\equiv0\), a contradiction. Hence all roots, if any, are \(0\), and \(P(x)=ax^n\). Substitution gives \(a=a^2\), so \(a=1\). The answer is \(0\), \(1\), and \(x^n\), \(n\ge1\).

If \(P\) is constant, impossible. Let \(\deg P=d\). Then \(\deg P(P(x))=d^2\), while the right-hand side has degree \(1\). Thus \(d=1\). Let \(P(x)=ax+b\). Then \(P(P(x))=a^2x+b(a+1)\). We get \(a^2=1\), \(b(a+1)=0\). The answer is \(P(x)=x\) and \(P(x)=-x+b\), where \(b\in\mathbb R\).

A constant \(P\) is impossible. If \(\deg P=d\ge1\), then \(\deg P(P(x))=d^2\). The right-hand side has degree \(2\). We would need \(d^2=2\), impossible for an integer \(d\). Hence no such polynomial exists.

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

If \(P\) is constant, then \(P=0\). Let \(d\ge1\). The degree of the left-hand side in \(y\) is \(d^2\), while the right-hand side has degree \(d\), so \(d=1\). Let \(P(x)=ax+b\), \(a\ne0\). Substitution gives \(a(x+ay+b)+b=ax+b+ay+b\). Hence \(a^2=a\), so \(a=1\); any constant works. The answer is \(P=0\) and \(P(x)=x+b\).

The constant solutions are \(0\), \(1\). Let \(P\) be nonconstant. If \(r\ne0\) is a root, then \(P(r^2)=P(r)^2=0\), and then \(P(r^{2^k})=0\) for infinitely many \(k\), impossible. Thus the only root is \(0\), and \(P(x)=ax^n\). Substitution gives \(a^2=a\), hence \(a=1\). The answer is \(0\), \(1\), and \(x^n\), \(n\ge1\).

Constant solutions satisfy \(c^3=c\), so \(c=0,1,-1\). Let \(P\) be nonconstant. A nonzero root would generate an infinite chain of roots \(r,r^3,r^9,\ldots\), so the only root is \(0\). Hence \(P(x)=ax^n\). Substitution gives \(a^3=a\), so \(a=1\) or \(a=-1\). The answer is \(0\), \(1\), \(-1\), \(x^n\), and \(-x^n\), \(n\ge1\).

Applying the condition twice gives \(P(x+2)=-P(x+1)=P(x)\). Thus \(P\) is periodic with period \(2\), so it is constant. Then \(c=-c\), hence \(c=0\). The answer is \(P\equiv0\).

For \(x^2\), \((x+1)^2-(x-1)^2=4x\). Therefore \(Q=P-x^2\) satisfies \(Q(x+1)=Q(x-1)\), i.e. \(Q(x+2)=Q(x)\). A periodic polynomial is constant. The answer is \(P(x)=x^2+c\).

The polynomial \(R(x)=P(x+1)-P(x)-x^2\) has infinitely many roots, so \(R\equiv0\). Thus the problem reduces to \(P(x+1)-P(x)=x^2\). As before, \(P(x)=\frac{x(x-1)(2x-1)}{6}+c\).

Let \(Q(x)=P(x)-x^2\). Then \(Q(x+y)=P(x+y)-(x+y)^2=P(x)+P(y)+2xy-x^2-2xy-y^2=Q(x)+Q(y)\). Hence \(Q(x)=ax\). The answer is \(P(x)=x^2+ax\).

The constant solutions are \(0\), \(1\). Let \(P\) be nonconstant. If \(r\ne0\) is a root, then \(P(r^2)=P(r)P(-r)=0\), again producing an infinite chain of roots. Thus the only root is \(0\), so \(P(x)=ax^n\). Substitution gives \(a x^{2n}=a^2(-1)^n x^{2n}\). Hence \(a=a^2(-1)^n\). Since \(a e0\), \(a=(-1)^n\). The answer is \(P(x)=x^{2k}\) and \(P(x)=-x^{2k+1}\), together with \(0\), \(1\).

If \(P\) is constant, say \(P=c\), then \(c=c^2\), so \(c=0\) or \(c=1\). Now let \(P\) be nonconstant. Then the image of \(P\) is infinite. For every number \(t\) in this image, there is \(x\) such that \(t=P(x)\), and the condition gives \(P(t)=t^2\). Hence the polynomial \(P(T)-T^2\) has infinitely many roots, so it is identically zero. Therefore \(P(x)=x^2\). Checking shows that \(0\), \(1\), and \(x^2\) all work.

Let the degree of \(P\) be \(d\), with leading coefficient \(a e0\). The leading part of the equality is \(a(x^d+y^d+(-x-y)^d)\). Put \(y=x\): we get \(a(2+(-2)^d)x^d\). If \(d\ge2\), this coefficient is nonzero: for even \(d\) it is positive, and for odd \(d\ge3\) it is \(2-2^d\). Hence \(d\le1\). Let \(P(x)=ax+b\). Substitution gives \(a(x+y-x-y)+3b=0\), so \(b=0\). The answer is \(P(x)=ax\), including \(a=0\).

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

Let \(y\) be in the image. Then \(y=f(t)\) for some \(t\). Therefore \(f(y)=f(f(t))=f(t)=y\). Hence \(y\) is fixed.

If \(f(a)>a\), increasingness gives \(f(f(a))>f(a)\), so \(a>f(a)\), contradiction. If \(f(a)

We have \(f(f(x))=a(ax+b)+b=a^2x+b(a+1)\). Thus \(a^2=1\) and \(b(a+1)=0\). If \(a=1\), then \(b=0\), giving \(f(x)=x\). If \(a=-1\), then \(b\) is arbitrary, giving \(f(x)=-x+b\).

The cycles of the permutation have lengths \(1\) or \(3\). If there are no fixed points, all cycles have length \(3\), so the number of elements is divisible by \(3\). But \(10\) is not divisible by \(3\). Hence there is a cycle of length \(1\), i.e. a fixed point.

Let \(f(x)=cx\). Then \(f(f(x))=c^2x\). We get \(c^2=4\), so \(c=2\) or \(c=-2\). The answer is \(f(x)=2x\), \(f(x)=-2x\).

Additivity on \(\mathbb Q\) gives \(f(x)=cx\). The condition \(f^3(x)=8x\) gives \(c^3=8\), hence \(c=2\). The answer is \(f(x)=2x\).

We have \(f^2(x)=a^2x+b(a+1)\). Thus \(a^2=1\). If \(a=1\), then \(2b=6\), so \(b=3\). If \(a=-1\), the shift is \(0\), not \(6\). The answer is \(f(x)=x+3\).

The coefficient of \(x\) in \(f^3(x)\) is \(a^3\), so \(a^3=1\), hence \(a=1\). Then \(f(x)=x+b\), and \(f^3(x)=x+3b\). We get \(3b=6\), so \(b=2\). The answer is \(f(x)=x+2\).

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

Every constant polynomial works. Let \(P\) be nonconstant. Then the image of \(P\) is infinite. For every \(t=P(x)\), we have \(P(t)=t\). Thus the polynomial \(P(T)-T\) has infinitely many roots, so \(P(T)=T\). The answer is all constant polynomials and \(P(x)=x\).

Cycle lengths can be \(1,2,4\). If there are no cycles of length \(1\) or \(2\), then all cycles have length \(4\). Hence the number of elements must be divisible by \(4\), but \(6\) is not divisible by \(4\). Contradiction.

From \(f(n+1)=f(n)+1\), the value \(f(n)-n\) is constant on \(\mathbb Z\). Hence \(f(n)=n+c\). Then \(f^2(n)=n+2c\). The condition gives \(2c=4\), so \(c=2\). The answer is \(f(n)=n+2\).

Let \(m=f(n)\). Then \(m>n\), and \(f(m)=n+2\). Applying the condition to \(m\), we get \(f(m)>m\). Hence \(n+2>m\). Thus \(n

For all \(x\), \(a^2x\equiv x\pmod p\), so \(a^2\equiv1\pmod p\). Thus \((a-1)(a+1)\equiv0\pmod p\). Since \(p\) is prime, \(a\equiv1\) or \(a\equiv-1\pmod p\).

If \(P\) is constant, impossible. Let \(\deg P=d\). Then \(\deg P^3=d^3\), while the right-hand side has degree \(1\), so \(d=1\). Let \(P(x)=ax+b\). The coefficient of \(x\) in the third iterate is \(a^3\), so \(a^3=1\), hence over \(\mathbb R\), \(a=1\). Then \(P(x)=x+b\), and \(P^3(x)=x+3b\). Thus \(b=0\). The answer is \(P(x)=x\).

Additivity on \(\mathbb Q\) gives \(f(x)=cx\) with rational \(c\). Then \(f(f(x))=c^2x\). The condition requires \(c^2=-1\), impossible for a rational number. Hence no such function exists.

If \(P=c\) is constant, then \(c=c^2\), so \(c=0\) or \(c=1\). Let \(P\) be nonconstant. For every \(t\) in the image of \(P\), we have \(t=P(x)\), hence \(P(t)=t^2\). The image of a nonconstant polynomial is infinite, so \(P(T)-T^2\) has infinitely many roots. Therefore \(P(T)=T^2\). The answer is \(P=0\), \(P=1\), \(P(x)=x^2\).

If \(f(a)>a\), increasingness gives \(f^2(a)>f(a)>a\), and then \(f^3(a)>f^2(a)>a\), contradicting \(f^3(a)=a\). If \(f(a)

Fix \(a\). If \(f(a)>a\), then strict increasingness gives \(f^2(a)>f(a)\), then \(f^3(a)>f^2(a)\), and so on. Thus \(f^k(a)>a\), a contradiction. If \(f(a)

Since \(f\) is increasing and additive, \(f(x)=cx\). From \(f(1)=4\), \(c=4\). The answer is \(f(x)=4x\).

An additive function bounded on an interval is continuous. Therefore \(f(x)=cx\). From \(f(1)=3\), we get \(f(x)=3x\).

If \(x

If \(f(a)>0\), then \(f(na)=nf(a)\) exceeds \(100\) for large \(n\), contradiction. If \(f(a)<0\), then \(f((-n)a)=-nf(a)\) exceeds \(100\). Hence \(f(a)=0\) for every \(a\).

If \(f(a)>a\), then \(f(f(a))>f(a)\), so \(a>f(a)\), contradiction. If \(f(a)

Boundedness on an interval implies linearity: \(f(x)=cx\). Since \(f(2)=10\), \(2c=10\), so \(c=5\). The answer is \(f(x)=5x\).

For \(x\), \(f(x)\ge x\). For \(-x\), \(f(-x)\ge -x\), i.e. \(-f(x)\ge -x\), so \(f(x)\le x\). Therefore \(f(x)=x\) for all \(x\).

Let \(g(x)=f(x)-f(0)\). Then \(g(0)=0\), \(g\) satisfies the same Jensen equation and is bounded above on an interval. The standard result gives continuity and linearity \(g(x)=ax\). Hence \(f(x)=ax+b\).

The function is strictly increasing and additive, hence linear: \(f(x)=cx\). From \(f(1)=1\), \(c=1\). The answer is \(f(x)=x\).

Nonnegativity on the positive half-line gives nondecreasingness. Hence the additive function is linear: \(f(x)=cx\). From \(0\le f(1)\le1\), we get \(0\le c\le1\).

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

Let \(x e0\). For \(q>0\), \(qf(x)=f(qx)\ge q^2x^2\), so \(f(x)\ge qx^2\). As \(q o\infty\), the right-hand side grows without bound, impossible for fixed \(f(x)\). Contradiction.

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

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

Let \(g(x)=f(x)-f(0)\). Then \(g(0)=0\), \(g\) satisfies Jensen's equation and is nondecreasing. By the standard fact, a Jensen function with monotonicity is linear: \(g(x)=ax\). Hence \(f(x)=ax+b\).

The estimate \(|f(x)|\le2|x|\) gives boundedness on \([-1,1]\), hence \(f(x)=cx\). The condition becomes \(|c||x|\le2|x|\), so \(|c|\le2\). The answer is all \(f(x)=cx\), where \(|c|\le2\).

If \(x0\), and \(f(y)-f(x)=f(y-x)>0\). Thus \(f\) is strictly increasing. A strictly increasing additive function is linear: \(f(x)=cx\). From \(f(1)=1\), we get \(c=1\).