moldovan wrote: Find all monotone (not necessarily strictly) functions $ f: \mathbb{R}^{ + }_0\rightarrow \mathbb{R}$ such that: $ f(x + y) - f(x) - f(y) = f(xy + 1) - f(xy) - f(1) \forall x,y \ge 0$; $ f(3) + 3f(1) = 3f(2) + f(0).$ 1) Let us solve the easier equation $ (E1)$ : "Find all functions $ g(x)$ from $ \mathbb N\to\mathbb R$ such that : $ g(2x+y)-g(2x)-g(y)=g(2y+x)-g(2y)-g(x)$ $ \forall x,y\in\mathbb N$" The set $ \mathbb S$ of solutions is a $ \mathbb R$-vector space. Setting $ y=1$, we get $ g(2x+1)=g(2x)+g(1)+g(x+2)-g(2)-g(x)$ Setting $ y=2$, we get $ g(2x+2)=g(2x)+g(2)+g(x+4)-g(4)-g(x)$ From these two equations, we see that knowledge of $ g(1),g(2),g(3),g(4)$ and $ g(6)$ gives knowledge of $ g(x)$ $ \forall x\in\mathbb N$ and so dimension of $ \mathbb S$ is at most $ 5$. But the $ 5$ functions below are independant solutions : $ g_1(x)=1$ $ g_2(x)=x$ $ g_3(x)=x^2$ $ g_4(x)=1$ if $ x=0\pmod 2$ and $ g_4(x)=0$ if $ x\neq 0\pmod 2$ $ g_5(x)=1$ if $ x=0\pmod 3$ and $ g_5(x)=0$ if $ x\neq 0\pmod 3$ And the general solution of $ (E1)$ is $ g(x)=a\cdot x^2+b\cdot x+c+d\cdot g_4(x)+e\cdot g_5(x)$ 2) Solutions of the original equation : Let $ P(x,y)$ be the assertion $ f(x+y)-f(x)-f(y)=f(xy+1)-f(xy)-f(1)$ In a first time, let us forget the monotonous constraint and the equation about $ f(0)$ Comparing $ P(xy,z)$ and $ P(xz,y)$, we get $ Q(x,y,z)$ : $ f(xy+z)-f(xy)-f(z)=f(xz+y)-f(xz)-f(y)$ Let then $ p$ a positive integer. $ Q(2,\frac mp,\frac np)$ $ \implies$ $ f(\frac{2m+n}{p})-f(\frac{2m}{p})-f(\frac np)=f(\frac{2n+m}{p})-f(\frac{2n}{p})-f(\frac mp)$ So $ f(\frac xp)$ is a solution of $ (E1)$ and so $ f(\frac xp)=a_p\cdot x^2+b_p\cdot x+c_p+d_p\cdot g_4(x)+e_p\cdot g_5(x)$ $ \forall x\in\mathbb N$ Choosing $ x=kp$, it's easy to see that $ a_p=\frac{a}{p^2}$, then that $ b_p=\frac bp$ Choosing $ x=2kp$, $ x=3kp$ and $ x=6kp$, it's easy to see that $ c_p=c$ and $ d_p=e_p=0$ And so $ f(\frac xp)=a(\frac xp)^2+b(\frac xp)+c$ $ \forall x,p\in\mathbb N$ And so $ f(x)=ax^2+bx+c$ $ \forall x\in\mathbb Q^{+*}$ Now, we must remember that $ f(x)$ is monotonous, and so $ f(x)=ax^2+bx+c$ $ \forall x\in\mathbb R^{+*}$ The condition "monotonous" implies too that $ ab\geq 0$ The condition $ f(0)=f(3)+3f(1)-3f(2)$ implies $ f(0)=c$ And so the necessary conditions are : $ \boxed{f(x)=ax^2+bx+c}$ $ \forall x\in\mathbb R_0^+$ with $ ab\geq 0$ And it's immediate to check that these necessary conditions are indeed sufficient.