# If $B(x+y)-B(x)-B(y)\in\mathbb Z$ can we add an integer function to $B$ to make it additive?

Given a function $$B:\mathbb R\to\mathbb R$$ satisfying $$B(x+y)-B(x)-B(y)\in\mathbb Z$$ for all real numbers $$x$$ and $$y$$, is there a function $$Z:\mathbb R\to\mathbb Z$$ such that $$B+Z$$ is an additive function? In other words, is there a function $$A:\mathbb R\to\mathbb R$$ satisfying $$A(x+y)=A(x)+A(y)$$ for all real numbers $$x$$ and $$y$$, such that $$A(x)-B(x)\in\mathbb Z$$ for every real number $$x$$?

My motivation: I was thinking about real solutions of the d'Alembert functional equation, $$f(x+y)+f(x-y)=2f(x)f(y)$$, without assuming continuity. There was a case where I could show that for some real function $$B$$, $$f(x)=\cos\big(2\pi B(x)\big)$$ and $$\cos\Big(2\pi\big(B(x+y)-B(x)-B(y)\big)\Big)=1$$ so for every $$x$$ and $$y$$ we have $$B(x+y)-B(x)-B(y)\in\mathbb Z$$. I was wondering if there's an additive function $$A$$ such that $$A(x)-B(x)\in\mathbb Z$$ for every $$x$$. In that case, I could show that the solution is of the form $$f(x)=\cos(2\pi A(x))$$ where $$A$$ is additive. It's easy to verify that every function of this form is indeed a solution to the functional equation. (In other cases I could show that $$f$$ is the constant zero function or is of the form $$f(x)=\cosh\big(A(x)\big)$$ for some additive function $$A$$. But they're not related to my question here.)

My attempt: I defined $$n(x,y)=B(x+y)-B(x)-B(y)$$. So $$n(x,y)=n(y,x)$$ and $$n(x,0)=-B(0)$$. Without loss of generality, we can assume that $$B(0)=0$$ (otherwise we can subtract $$B(0)$$ from $$B(x)$$ and continue). Because $$x+(y+z)=(x+y)+z$$, therefore I could conclude that $$n(x,y+z)+n(y,z)=n(x+y,z)+n(x,y)$$. But this doesn't seem to help much.

There is a similar problem here: USA January TST 2015

Firstly, we show that there is a function $$B:\mathbb Q\to\mathbb R$$ such that for every rational $$x$$ and $$y$$, $$B(x+y)-B(x)-B(y)\in\mathbb Z$$, but there's no additive function $$A$$ such that $$B(x)-A(x)\in\mathbb Z$$ for every rational $$x$$.

It's well known that if $$A$$ is an additive function then there is a constant $$c$$ such that $$A(x)=cx$$ for every rational $$x$$. See here for a proof.

Now construct $$B\left(\frac pq\right)=\frac pqK(q)$$ where $$\gcd(p,q)=1$$, $$q>0$$ and $$K(q)=\sum\limits_{i=0}^{q-1} i!$$.

I claim that we have $$B(x+y)-B(x)-B(y) \in \mathbb{Z}$$, for every rational $$x$$ and $$y$$.

Let $$x=\frac{a}{b}$$, $$y=\frac{c}{d}$$ and $$\frac{p}{q}=\frac{a}{b}+\frac{c}{d}$$ where $$\gcd(a,b)=\gcd(c,d)=\gcd(p,q)=1$$. Then, in mod $$1$$, we have:

$$B\left(\frac pq\right)-B\left(\frac ab\right)-B\left(\frac cd\right)=\frac pqK(q)-\frac abK(b)-\frac cdK(d)\equiv\left(\frac pq-\frac ab-\frac cd\right)K(bd)=0$$

Notice that for $$m\ge n$$, we have $$m!\equiv0\pmod n$$, so $$K(q+m)\equiv K(q)\pmod q$$ for all $$m\ge0$$.

Now I claim that there is no $$c$$ such that $$B(x)-cx\in\mathbb{Z}$$ for all $$x\in\mathbb{Q}$$.

Suppose that there is such $$c$$. If $$q$$ is a positive integer, then $$B(\frac 1q)-\frac cq=\frac1q(K(q)-c)\in\mathbb Z$$. So $$c$$ is an integer such that for every postive integer $$q$$, we have $$K(q)\equiv c\pmod q$$. for instance, we have $$K(q!)\equiv c\pmod{q!}$$. But by definition of $$K$$, we know that $$K(q!)\equiv K(q)\pmod{q!}$$ which leads to $$K(q)\equiv c\pmod{q!}$$. Hence there is a sequence of integers like $$(k_q)_{q\in\mathbb Z^+}$$ such that $$c=k_q\cdot q!+K(q)$$. Now for every positive integer $$q$$:

$$0=c-c=k_{q+1}\cdot (q+1)!+K(q+1)-k_q\cdot q!-K(q)=\left((q+1)k_{q+1}-k_q+1\right)q!$$ $$\therefore\quad k_{q+1}=\dfrac{k_q-1}{q+1}$$

Now we show that for every natural number $$n$$, we must have $$|k_q|\ge q^n$$. For the base case, we note that if $$k_q=0$$, then $$k_{q+1}$$ can't be an integer, so $$|k_q|\ge1=q^0$$. For the induction step, we have:

$$\dfrac{|k_q|+1}{q+1}\ge\dfrac{|k_q-1|}{q+1}=|k_{q+1}|\ge (q+1)^n$$ $$\therefore\quad|k_q|\ge(q+1)^{n+1}-1\ge q^{n+1}$$

But this leads to an obvious contradiction. So $$c$$ doesn't exist.

Finally, we show that there is a function $$B:\mathbb R\to\mathbb R$$ such that for every real $$x$$ and $$y$$, $$B(x+y)-B(x)-B(y)\in\mathbb Z$$, but there's no additive function $$A$$ such that $$B(x)-A(x)\in\mathbb Z$$ for every real $$x$$. So the answer to the original question is negative.

Let $$({\bf e}_i)_{i\in I}$$ be a Hamel Basis. So for every real number $$x$$, there is a finite set of indices $$I_x\subseteq I$$ and rational numbers $$\left(\frac{p_i}{q_i}\right)_{i\in I_x}$$ such that $$\gcd(p_i,q_i)=1$$, $$q_i>0$$ and $$x=\sum\limits_{i\in I_x}\dfrac{p_i}{q_i}{\bf e}_i$$. Define:

$$B(x)=\sum_{i\in I_x}\frac{p_i}{q_i}K(q_i)$$

Note that $$I_x$$ and $$\left(\frac{p_i}{q_i}\right)_{i\in I_x}$$ are uniquely determined and thus $$B$$ is well defined. We can check that $$B$$ satisfies the desired conditions, similar to what we did before.

Let $$B(2^{-2n})=\frac13+2^{-2n},\\B(2^{-2n-1})=\frac23+2^{-2n-1}$$ If that can be completed to $B(\mathbb{R})$,
then $A(1)=2^{2n}A(2^{-2n})$ has no finite value.

Let $B= (e_i)_{i\in I}$ a base of $\bf R$ as a $\bf Q$ vector space. For each $i$ choose an integer $n_i$. Let $f(x)=\sum _i (x_i+n_i)$. Then $f$ satisfies your property

• $f$ satisfies which property? Commented Dec 25, 2015 at 16:05