Tintarn wrote: Among all triples $(a,b,c)$ of natural numbers satisfying \[(a+14\sqrt{3})(b-14c\sqrt{3})=2024,\]determine the one with the maximal value of $a$. So $b=ac$ and $ab-588c=2024$ and so $a^2=588+\frac{2024}c$ and we are looking for the smallest $ c$ $c=1$ does not fit $c=2$ fits giving $\boxed{\text{greatest a }=40}$ and $\boxed{(a,b,c)=(40,80,2)}$ and indeed $(40+14\sqrt 3)(80-28\sqrt 3)=2024$