7. Problem: Let $a, b, c, d$ be integers such that $a>b>c>d \geq-2021$ and
$$
\frac{a+b}{b+c}=\frac{c+d}{d+a}
$$
(and $b+c \neq 0 \neq d+a)$. What is the maximum possible value of $a c$ ?
Solution: We claim that the maximum value of $a c$ is $2 \times 505^{2}=510050$, and this is uniquely achieved by $(a, b, c, d)=(1010,506,505,-2021)$.
To prove this we start by rearranging the expression $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ to get $(c+b)(c+d)=$ $(a+b)(a+d)$. Now expand the brackets and move all the terms to one side to get $(c-a)(a+b+c+d)=0$. Since $c>a$ we can divide by $c-a$ to get
$$
a+b+c+d=0 .
$$

Note that this is equivalent to the $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ condition. Now assume that $(a, b, c, d)$ is the tuple such that $a c$ is maximised. Since $4 a>a+b+c+d=0$ we trivially have $a>0$. Furthermore if $c \leq 0$ then we would have $a c \leq 0$ which would contradict the definition that $a c$ is maximal. Therefore
$$
a>b>c>0 \geq d \geq-2021 .
$$

Now we perform two little proofs by contradiction:
- If $d>-2021$ then we could consider $\left(a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime}\right)=(a+1, b, c, d-1)$ so that
$$
a^{\prime} c^{\prime}=(a+1) c>a c .
$$

However this would contradict the definition that $a c$ was maximal.
Therefore $d=-2021$.
- If $b>c+1$ then we could consider $\left(a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime}\right)=(a+1, b-1, c, d)$ so that
$$
a^{\prime} c^{\prime}=(a+1) c>a c .
$$

However this would contradict the definition that $a c$ was maximal.
Therefore $b=c+1$.
Therefore $(a, b, c, d)=(a, c+1, c,-2021)$, and since $a+b+c+d=0$, this means $a=2020-2 c$. So the final expression we are trying to maximise is
$$
\begin{aligned}
a c & =(2020-2 c) c \\
& =2 \times 505^{2}-2(c-505)^{2} \\
& \geq 2 \times 505^{2} .
\end{aligned}
$$
with equality iff $c=505$. Therefore the maximum value of $a c$ is $2 \times 505^{2}=510050$ which is achieved by
$$
(a, b, c, d)=(1010,506,505,-2021) .
$$