Find all $x,y\in\mathbb{Z},$ such that $x^{2008}+{2008}!=21^{y}$

Question

Find all $x,y\in\mathbb{Z},$ such that $x^{2008}+{2008}!=21^{y}$

My try:

So,I want to solve this problem without 'Lifting the exponent lemma'.Here is what I have tried: $x^{2008}+{2008}!=21^{y}\implies 21\big|(x^{2008}+{2008}!)$

But I don't know how to proceed from here.Any help would be highly appreciated.Thank you!

Answer 1

First,

• $y$ cannot $< 0$ because in that case LHS is an integer while RHS isn't.
• $y$ cannot $= 0$ because in that case LHS is $> 1$ while RHS $= 1$.

So $y > 0$. It is also clear $x$ cannot be $0$. Notice if $(x,y)$ is a solution, so does $(-x,y)$. In the search of a solution, we only need to consider the case $x > 0$.

For any prime $p$ and integer $n \ne 0$, let $\nu_p(n)$ be the highest power of $p$ divides $n$. More precisely, we define $$\nu_p(n) = \max\{ k \in \mathbb{N} : p^k | n \}$$

There are two properties we need:

1. $\nu_p(n^k) = k\nu_p(n)$
2. $\nu_p(u + v) = \min(\nu_p(u),\nu_p(v))$ whenever $\nu_p(u) \ne \nu_p(v)$.

Notice $\nu_3(2008!) = 1000$ and $\nu_7(2008!) = 331$.

• If $y \ne 331$, then $$2008\nu_7(x) = \nu_7(x^{2008}) = \nu_7(21^y - 2008!) = \min(y,331)$$ There are no integer solution of $\nu_7(x)$ because $0 < \min(y,331) \le 331 < 2008$.

• Otherwise, $y = 331$ and $$2008\nu_3(x) = \nu_3(x^{2008}) = \nu_3(21^{331} - 2008!) = \min(331,1000) = 331$$ Once again, there are no integer solution for $\nu_3(x)$.

Combine these, we can conclude the equation $x^{2008} + 2008! = 21^y$ has no integer solutions.