# Are there infinite (consecutive) pairs of Achilles numbers?

This was proven true for powerful numbers, but has this been proven for Achilles numbers? I've found a total of 2 sites that claim this to be true but do not provide any sort of reasoning, nor proof as reference, and one site cites the other.

EDIT: I've been asked to explain Achilles numbers and this problem in general so here is a brief explanation:

Q: What is an Achilles number?

A: An Achilles number is a number that is powerful but cannot be expressed in the form of a perfect power. (e.g. $72$)

Q: What is a powerful number?

A: A positive integer $m$ where, if a prime number $p$ divides $m$, then $p^2$ also divides $m$. (e.g. $25$)

Q: What is a perfect power?

A: A positive integer $n$ that can be written as $m^k$, where $m>1$ and $k\geq2$ and both $m$ and $k$ are $\in\mathbb{N}$. (e.g. $8$)

• Well, one of the websites cites Richard P. Stanley, probably his book www-math.mit.edu/~rstan/ec, maybe you should look for it. – Emre May 31 '16 at 10:42
• You should give the definition of an Achilles number here. – Peter May 31 '16 at 11:10

\begin{align}5425069447 &= 7^3 \cdot 41^2\cdot 97^2\\ 5425069448 &= 2^3 \cdot 26041^2\end{align}
By standard reductions to Pell equations, there are infinitely many solutions to the Diophantine equation $5425069448x^2 - 5425069447y^2 = 1$, with the restriction that $x,y \equiv 1 \pmod{2\cdot 7\cdot 41\cdot 97 \cdot 26041}$. Each such solution gives a pair of consecutive powerful numbers which can't possibly be perfect powers.