Seems to simple... It is well known that since nice numbers have positive density then there exists arbitrarily large arithmetic progressions, all of whose terms are nice. Take $a$, $a+r$, $a+2r$, $a+3r$ nice arithmetic progression. Then $$(a+r)(a+2r)-(a)(a+3r)=2r^2$$As $r$ can take on arbitrarily large values, we are done.

Well, it might seem simple, but all Belarusian and Iranian contestants got zero on it. In the official solution your "well-known fact" (I myself actually didn't know it before) is replaced by "color every number by the color with number equals to the distance to the nearest bigger nice number, by Van Der Waerden find arithmetic progressions and shift it to nice numbers"

The problem condition implies that the sequence has positive [upper] density in $\mathbb{N}$, so therefore by Szemerédi's theorem, there exists arbitararily large arithmetic progressions and now by $a(a+3d)-(a+d)(a+2d)=2d^2$, we're done.

I am sorry, I don't want to start some disputes or raise tensions, but what is the point of writing exactly the same post as someone has already written? I expect people to make new interesting posts with solutions or ideas that no one has showed yet, but not posts to show that you can solve the problem too or increase the post counter.