4. Find all functions $f: \mathbb{R}^{+} \mapsto \mathbb{R}^{+}$, where $\mathbb{R}^{+}$denotes the positive reals, such that whenever $a>b>c>d>0$ are reel numbers with $a d=b c$,
$$
f(a+d)+f(b-c)=f(a-d)+f(b+c) .
$$
Solution: Since $f(a+d)-f(a-d)$ only depends on $a d$, we can have a function $g$ mapping positive reals to reals such that whenever $a>d$,
$$
g(a d)=f(a+d)-f(a-d) .
$$

Also,
$$
\begin{aligned}
g\left(k a d+k(k+1) d^{2}\right) & =g((a+(k+1) d)(k d)) \\
& =f(a+(2 k+1) d)-f(a+d) \\
& \text { and } \\
g\left((k+1) a d+k(k+1) d^{2}\right) & =g((a+k d)((k+1) d)) \\
& =f(a+(2 k+1) d)-f(a-d) \\
& =g(a d)+g\left(k a d+k(k+1) d^{2}\right)
\end{aligned}
$$
and
$$
\begin{aligned}
g\left((k+1) a d+k(k+1) d^{2}\right) & =g((a+k d)((k+1) d)) \\
& =f(a+(2 k+1) d)-f(a-d) \\
& =g(a d)+g\left(k a d+k(k+1) d^{2}\right)
\end{aligned}
$$
for any constant $k>0$.
Let $a=2 d$, and let $x=d^{2}$. Then we have the following:
$$
g\left(\left(k^{2}+3 k+2\right) x\right)=g(2 x)+g\left(\left(k^{2}+3 k\right) x\right)
$$

However, $\left(k^{2}+3 k\right) x$ is surjective over the positive reals as $k>0$, so if we let $y=\left(k^{2}+3 k\right) x$, we obtain
$$
g(x+y)=g(x)+g(y)
$$
for all positive real numbers $x$ and $y$. Consequently, for any positive real number $x$, we can always find a unique $\lambda>0$ such that $\lambda(\lambda+1)=x$. Thus,
$$
g(x)=g(\lambda(\lambda+1))=f(2 \lambda+1)-f(1) \geq-f(1)
$$

Because $g$ is bounded below and satisfies Cauchy's Functional Equation, there exists a real number $a$ such that $g(x)=a x$ for all $x>0$. That gives, for $u>1$,
$$
f(u)=f(1)+g\left(\left(u^{2}-1\right) / 4\right)=\frac{a}{4} u^{2}+\frac{-a+4 f(1)}{4}
$$
and for $u0$ are the only possible solutions. It is not hard to see that this is a solution to the functional equation if and only if $c$ and $d$ are nonnegative real numbers which are not both zero.

This problem was proposed by Calvin Deng.