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.

  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.