Apply the <Szemerédi theorem> with density $\varepsilon$ and progression length five. For all sufficiently large $n$, the set $A$ contains a nonconstant five-term <arithmetic progression>
$$
e-2r,\quad e-r,\quad e,\quad e+r,\quad e+2r,
$$
where $r\ne0$. Set
$$
a=e-2r,
\quad b=e-r,
\quad c=e+r,
\quad d=e+2r.
$$
These four elements and $e$ are distinct, and direct addition gives
$$
a+b+c+d
=(e-2r)+(e-r)+(e+r)+(e+2r)
=4e.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 122 / 3 / b / Solution

{c}
{wiki=Szemerédi's_theorem}

For every positive density $\alpha$ and positive integer $r$, every sufficiently large subset of $[n]$ of size at least $\alpha n$ contains a nonconstant arithmetic progression of length $r$.


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 Incoming links: Szemerédi theorem