Prime factorizations of consecutive numbers, can they both contains powers of greater than or equal to $1000$?

Question

Say that a positive integer n is highly exponential if there is an exponent greater than or equal to 1000 in its prime factorization. Can there exist two such numbers consecutively?

Answer 1

You're essentially asking for positive integer solutions to the Diophantine equation $$ax^{1000}-by^{1000}=1,$$ for $x,y\neq 1$. The answer is that there is certainly a solution. To see why, recall Bezout's identity:

Bezout's identity. Given any coprime integers $a,b$, there exist integers $x,y$ so that $ax+by=1$.

So if we were to choose $y=x+1$, then certainly $x^{1000}$ and $y^{1000}$ would be coprime, so there must be integers $a,b$ that satisfy the equation. Of course, don't expect to be able to find an explicit example, since the order of magnitude of the quantities involved are so large.