Find all positive integers a,b,c,d such that $2019^a = b^3+c^3+d^3-5$

Question

Working on a problem...

Find all positive integers a,b,c,d such that $2019^a = b^3+c^3+d^3-5$

Using $(a+b+c)^3$ yields far too many terms, and I cannot come up with any solutions other than guessing and checking.

Answer 1

HINT: If $a \ge 2$, then $2019^a \equiv 0 \pmod{9}$, while $b^3 \equiv -1, 0, \ \text{or} \ 1 \pmod{9}$ and similarly for $c^3$ and $d^3$.

Do you see a contradiction? After ruling out $a \ge 2$, you only need to solve the $a = 1$ case, i.e. $b^3+c^3+d^3 = 2024$. Since $b,c,d$ are positive integers, none of them can be larger than $\sqrt[3]{2024}$, so there aren't too many possibilities.