8

this is a problem from the 2024 National Ukrainian Mathematics Olympiad. None of contestants got more than 0 out of 7 points. Here it is:

Show that there are an infinite number of positive integers that cannot be represented in the form of $a^{bc} - b^{ad}$, where $a$, $b$, $c$, $d$ are positive integers, $a$, $b$ $> 1$.

My attempt:

Let's rewrite the problem. Prove that there are infinitely many c $\in \mathbb{N}$, such that for any solution $(x, y)$ to the equation $c = a^x - b^y$ either $b \nmid x$, $a \nmid y$, or both. I think we also can use the fact that exponential diophantine equations have at max 2 solutions, but I don't know how.

Any help would be appreciated, thanks in advance!

Edit: I found a solution (see my nnswer) but it has absolutely no motivation behind it, I'm sure there is more general solution.

Why $8k+3$? The choice makes absolutely no sense. So, again, I'm sure there is a better solution, any help would be appreciated.

Bill Dubuque
  • 282,220
Pashtet671
  • 151
  • 7
  • 2
    I think that's an excellent solution (+1). Using modular arithmetic arguments with small moduli is a very standard way to attack diophantine equations. Specifically, it's a great technique when, as here, you want to find an obstruction to a solution. I'd be surprised if there is a simpler argument out there. – lulu Aug 15 '24 at 13:03
  • Note: you might want to post your solution below. That way, users here can check your computations and the site can close the question. Of course, others will still be able to post alternate methods of proof. – lulu Aug 15 '24 at 13:04
  • Hey, @lulu , thanks for the answer. Yeah, I see now, mod 8 is really standard given that one number is odd and bigger than 3 and the other one is even. And the motivation behind mod 8 is also $(2u)^{3v} = 8u^{3v}$. I will post the solution below now, thanks! – Pashtet671 Aug 15 '24 at 14:07

1 Answers1

7

Let's show that no number of the form $8k+3$ can be represented in this form. Suppose that $a^{bc}−b^{ad}=8k+3$ for some $k$. If the numbers $a,b$ had the same parity, the number $a^{bc} − b^{ad}$ would be even, but it's not. Therefore, one of them is even, and the other is odd by at least $3$. Then one of the numbers $a^{bc}, b^{ad}$ is divisible by $8$, and the other gives a remainder of $1$ when divided by $8$, because it is the square of an odd number. Thus, their difference cannot give a remainder of $3$ when divided by $8$.

Benjamin Dickman
  • 16,038
Pashtet671
  • 151
  • 7
  • Hm, that's surprising to me that none of the candidates came up with this solution. $\quad$ FYI $8k+5$ also works, by your reasoning. – Calvin Lin Aug 18 '24 at 17:27