First Substitutions

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 \(\frac1{51}\).