I (who is not a professional mathematician), ended up in the following on which I would like to have your comment, because this is overly simple solution. I know this certainly should not be easy and that is why I'm here.
Powerful numbers are defined as those $n$ which, for all primes, $p \mid n \implies p^2 \mid n$.
Two consecutive powerful numbers may be written as:
$$(A^2+n_1)(B^3+m_1)-A^2B^3=1 \tag{1}$$
Two powerful numbers a difference of which 2 may be written as:
$$(A^2+n_2)(B^3+m_2)-A^2B^3=2 \tag2$$
In both of these, $A, B, m_1, m_2, n_1, n_2$ are integers, and $A,B > 0$.
I suppose $A^2B^3,(A^2+n_1)(B^3+m_1),(A^2+n_2)(B^3+m_2)$ should then represent three consecutive powerful numbers.
Beckon$^1$ shows (as I understood):
Three consecutive numbers are of one of the following forms: $(36k+7, 36k+8, 36k+9), (36k+27, 36k+28, 36k+29), \text{or } (36k-1, 36k, 36k+1)$.
We get from equation $(2): A^2m_2+B^3n_2+m_2n_2=2$, and after division by 2, we get
$$A^2M+B^3N=1-2MN \tag3$$
where $M=m_2/2$ and $N=n_2/2$.
Note that $m_2$ and $n_2$ must be even because $A$, $B$, $(A^2+n_2)$ and $(B^3+m_2)$ are odd. We can now formulate: $M$ and $N$ do not have the same parity.
We can notice that the following equations are constant:
$$\frac{(36k+7)(36k+9)+1}{(36k+8)^2}=1 \tag4$$
$$\frac{(36k+27)(36k+29)+1}{(36k+28)^2}=1 \tag5$$
$$\frac{(36k-1)(36k+1)+1}{(36k)^2}=1 \tag6$$
Examine now the equation $(4)$ of the three alternatives. We can replace the terms in the following manner: $36k+7=A^2B^3$ and $36k+9=(A^2+2N)(B^3+2M)$ in the equation $(4)$ which results in the following:
$$A^2B^2(A^2+2N)(B^3+2M)+1=16(9k+2)^2 \tag7$$
We can solve the equation $(7)$ for M and N:
$$
\begin{align}
M &=\frac{63-A^4B^6-2A^2B^6N+18(72k^2+32k)}{2A^2B^3(A^2+2N)} \implies \\
M &=\frac{\frac{63-A^4B^6}{2}-A^2B^6N+18(36k^2+16k)}{A^2B^3(A^2+2N)}
\end{align}
\tag8
$$
and
$$
\begin{align}
N&=\frac{63-A^4B^6-2A^4B^3M+18(72k^2+32k)}{2A^2B^3(B^3+2M)} \implies \\ N&=\frac{\frac{63-A^4B^6}{2}-A^4B^3M+18(36k^2+16k)}{A^2B^3(B^3+2M)}
\end{align}
\tag9
$$
The equations $(8)$ and $(9)$ should be examined together. Based on the equation $(8)$, $M$ may or may not be divisible by two depending the term $\frac12(63-A^4B^6)$ and $N$. Because the same term is also in the equation $(9)$, $N$ is in the same situation. If the term $\frac12(63-A^4B^6)$ is divisible by two, the divisibility of M depends on N such that both must be divisible by two if one of them is divisible by two. If M is not divisible by two, it also means that N is not divisible by two. Both possibilities thus contradict the assumption that M and N do not have the same parity.
The two other cases $(5)$ and $(6)$ go in the same way (in order to keep this short I don't write them here). Note also that the condition $AB(A^2+2N)=0$ (denominator) is not possible because an even number cannot be an odd number (and $A$ and/or $B$ are/is not $0$).
What is above seems far too simple (Erdös could not solve this problem). OK, the question is, can this super tough problem be solved this easily? Or, can you point out where the calculations go wrong? Thank you for reading.
[1] Beckon, Edward, On consecutive triples of powerful numbers, Undergrad. Math J. 20, No. 2, Paper No. 3, 3 p. (2019). ZBL1435.11010.
Best Answer
Alas, you haven't proven the conjecture. Your calculations go wrong in describing the parity implications of your equation $(8)$:
$$M =\frac{\frac{63-A^4B^6}{2}-A^2B^6N+18(36k^2+16k)}{A^2B^3(A^2+2N)} \tag8$$
Since we're only trying to pry things apart for parity, we can safely say that the denominator is odd (no matter the parity of $N$), the term containing $k$ is even, and $A^2B^6$ is odd. So we rewrite the equation as:
$$M = \frac{Q-ON+E}{O}$$
where $Q$ is the fractional term in the original, and the numbers of known parity are replaced with $O$ or $E$ as appropriate.
You claim that $2 \mid Q \implies (2 \mid M \iff 2 \mid N)$, which contradicts the previously proven $2 \mid M \iff 2 \nmid N$, and this claim is true so far as I can tell.
However, there's a second possibility: $2 \nmid Q$, i.e., $Q$ is odd. And this is where things go off course: you looked at the implications of $Q$ being even, but not of it being odd.
Assume $Q$ is odd. If $N$ is odd, then the numerator is even and thus $M$ is even. If $N$ is even, the numerator is odd, and $M$ must be odd. No contradiction here as long as $Q$ is odd--in fact, using both of those results, $2 \nmid Q \iff (2 \mid M \iff 2 \nmid N)$.
Sadly, what this means is you've proven that if a triple of this sort exists, and $A^2B^3 \equiv 7 \pmod{36}$, then $4 \nmid (63 - A^2B^3)$.
I'm guessing the opposite result (i.e. $Q$ must be even) may come from triples starting with numbers $n_1 \equiv 27 \pmod {36}$, since $27 \equiv 7 \pmod 8$ but $7, 35 \equiv 3 \pmod 8$. But I didn't do the math for the other cases because one counterexample suffices.
I hope this helps! Finding another person's miscalculation often proves an enlightening exercise.