any hints?

Well, it is quite a classical trick to consider what happens when you add the conjugate number $(a-b\sqrt{2024})^p$ (and hope that this is small, say that $a-b\sqrt{2024} \in (0,1)$). The rest is not very hard to work out.

Nice probleme

Bluecloud123 wrote: any hints? Exactly that was the solution

let $a=\lfloor b\sqrt{2024} \rfloor+1$ we'll get that $a-b\sqrt{2024}\in(0,1)$ and $\lfloor (a+b\sqrt{2024})^p+(a-b\sqrt{2024})^p\rfloor\in\mathbb{N}$ thus $\lfloor (a+b\sqrt{2024})^p+(a-b\sqrt{2024})^p\rfloor=\lfloor(a+b\sqrt{2024})^p\rfloor+1$ $\Rightarrow \lfloor(a+b\sqrt{2024})^p\rfloor=2\sum_{k=0}^{\frac{p-1}{2}} {{p}\choose{2k}} a^{p-2k}b^{2k}2024^k -1$ $\equiv 2a^p-1\equiv 2a-1(mod~p)$ every prime $p$ thus if $(a,b,c)=(\lfloor b\sqrt{2024} \rfloor+1,b,2\lfloor b\sqrt{2024} \rfloor+1),~~~\forall b \in\mathbb{N}$ every prime p divides $\lfloor(a+b\sqrt{2024})^p\rfloor-c$ $\therefore$ we can conclude that there exist infinitly many triples $(a,b,c)$ of positive integers such that every prime $p$ divides$\left\lfloor\left(a+b\sqrt{2024}\right)^p\right\rfloor-c.~~~~~\square$

Tintarn wrote: Well, it is quite a classical trick to consider what happens when you add the conjugate number $(a-b\sqrt{2024})^p$ (and hope that this is small, say that $a-b\sqrt{2024} \in (0,1)$). The rest is not very hard to work out. thanks for the hints its very easy indeed when u figure out the trick