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I am looking for the proof that answers the question posed in the title. Wikipedia, and several other sites, list that "Since Pell's equation $x^2 − 8y^2 = 1$ has infinitely many integral solutions, there are infinitely many pairs of consecutive powerful numbers (Golomb, 1970)" (Wikipedia).

However, I cannot find what is being referenced by Golomb's 1970 (supposed) paper/proof. Additionally, I do not understand why $8$ in the place of $n$ for Pell's equation gives a solution.

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    $\begingroup$ Given a solution to the given Pell equation, the integers $x^2$ and $8y^2$ are powerful (obviously) and consecutive. Any cube in place of $8$ would work. $\endgroup$ – lulu May 31 '16 at 0:49
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    $\begingroup$ For that matter, we could also use any $k$th power for $k>2$; as long as the parameter isn't a perfect square, the resulting equation has infinitely many positive integer solutions. $\endgroup$ – Noah Schweber May 31 '16 at 0:52
  • $\begingroup$ I note that it is an open question, as to whether there are any triples of consecutive powerful numbers. $\endgroup$ – Gerry Myerson Jun 8 '16 at 2:43
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Keep in mind that a powerful number is a number $x$ such that, for every prime $p$ dividing $x$, $p^2$ also divides $x$.

Exercise: $x$ is powerful iff $x$ is a product of a square and a cube.

Now suppose $x^2-8y^2=1$. Since $8=2^3$, we have by the exercise that $x^2$ and $8y^2$ are each powerful - and clearly they are consecutive.


The last bit of the equation is: "Why are there infinitely many such solutions?" This is because Pell's Equation $x^2-my^2=1$ has infinitely many solutions as long as the parameter $m$ is not a perfect square; see https://en.wikipedia.org/wiki/Pell%27s_equation.

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