Square-full Problem An integer is square-full if each of its prime factors occurs to at least the second power. Prove that there exists infinitely many pairs of consecutive square-full integers.
 A: Look at the solutions to the Pell equation
$$
x^2 - 2y^2 = 1.
$$
for which $y$ happens to be even. 
They start with the pairs $(x,y)  = (3,2), (17,12), \ldots$ leading to
$(9,8), (289,288), \ldots$
More generally, suppose $d$ is square free and there is a solution to the Pell equation
$$
x^2 - dy^2 = 1 \quad (*)
$$
for which $d$ divides $y$.  Then $(x^2, dy^2)$ is a pair of powerful numbers differing by $1$.
Once you have one such solution you can use it to find infinitely many.
To see why, suppose you have two solutions to $(*)$ for each of which $d$ divides $y$. Write them as  $(p, dq)$ and $(r,ds)$. Then  check that
$$
(pr + d^2q, dqr + dps) \quad (**)
$$
is a third solution with the same property. The two solutions you start with need not be different, so just one will get you infinitely many.
The construction in $(**)$ isn't magic. It comes from expanding the product
$$
(p + q \sqrt{d})(r + s \sqrt{d}) = ? + ?\sqrt{d}
$$
I suspect you can find such a starting pair for every square free $d$. For example, $(26, 15)$ works for $d=3$, yielding the pair $(676,675)$. That comes from the cube of the fundamental unit $2+\sqrt{3}$. Perhaps the $d$th power of the fundamental unit will always work. It does when $p$ is an odd prime since all the binomial coeffic1ents (except the first and last) are divisible by $p$.  I should check $d=6$ next ...
See http://www.zyvra.org/laforth/sqr2.htm
https://www.quora.com/We-know-that-the-square-root-of-2-is-an-irrational-number-but-what-is-the-nearest-fraction-that-can-be-equal-to-the-square-root-of-2-which-we-can-use-for-earthly-calculations
http://mathworld.wolfram.com/PellEquation.html
