I (non-mathematician) asked a similar kind of question 5 days ago. Now I revisit the case in a different manner. The powerful numbers may be written in the form $A^2B^3$, where $A$ and $B$ are positive integers. Erdös conjectured in 1975 that there do not exist three consecutive powerful integers. The problem is notoriously tough. My simple calculations seem to prove the conjecture, and hence I would like someone to try to find the mistake (I failed in my search), which is probably more probable than not. And please point out where you found the mistake. Still, I hope my calculations are entertaining (at least they should be easy to follow).
Two powerful numbers, a difference of which is $2$, may be written as follows:
$$(A + N_2)^2 (B + M_2)^3 – A^2 B^3 = 2 \tag1$$
where $N_2$ and $M_2$ are integers, and they must be even because the largest powerful number is odd [Beckon]. We can now perform the following replacements $N_2=2N$ and $M_2=2M$.
$$(A + 2N)^2 (B + 2M)^3 – A^2 B^3 = 2 \tag2$$
We may expand the terms and divide both sides of equation $(2)$ by $2$.
$$3MA^2 B^2 + 2A^2M(3 B M + 2 M^2) + 2 A N (B + 2 M)^3 + 2 N^2 (B + 2 M)^3=1 \tag3$$
In equation $(3)$, the only possible term that could be odd is $3MA^2B^2$. Hence, $M$ must be odd. Two powerful numbers, a difference between which is $2$, may also be written as follows:
$$(A^2+n_2)(B^3+m_2)-A^2B^3=2 \tag4$$
Expanding the terms of equation $(4)$ and dividing both sides by $2$, we get
$$A^2m+B^3n=1-2mn \tag5$$
It is apparent that $m$ and $n$ do not have the same parity.
We have the following dependence:
$$(A^2+n_2)=(A+N_2)^2 \implies n_2 = N_2 (2 A + N_2) \implies n=2N(N+A) \tag6$$
It can be seen that $n$ must be even. As a result, $m$ must be odd. Equation $(6)$ may also be solved for $N$:
$$N=n-2A \tag7$$
We may now solve equation $(2)$ for $M$:
$$ \frac{\sqrt[3]{(A^2 B^3 + 2) (A + 2 N)}}{2 (A + 2 N)}-\frac{B}{2} \tag8$$
We may insert the result of equation $(7)$ into equation $(8)$, and the result is
$$M = \frac{\sqrt[3]{(n – A)(A^2B^3+2)}}{2 (n – A)}-\frac{B}{2} \tag9$$
As a result of a comparison of equations $(8)$ and $(9)$, we may write the following equations:
$$n=2(A+N) \tag{10a}$$
$$n=-2N \tag{10b}$$
Because of equation $(6)$, we may write the following equation
$$2(A+N)=2N(A+N) \implies N=1 \tag{11a}$$
$$N=-(A+1) \tag{11b}$$
The result of equation $(11$b) is effectively the same as $N=1$. We also know now that $n=2A+2$ (this result comes from both $(11$a) and $(11$b)). We may also write the first term of the largest powerful number: $(A+2)^2=A^2+4A+4$. By inserting into equation $(2)$, we get
$$M = \frac{1 – B^3}{A^2 + 2} \tag{12}$$
We may now solve equation $(12)$ for $A^2$
$$A^2 = -\frac{B^3 + 2 M – 1}{M} \tag{13}$$
In equation $(13)$, the nominator is even and the denominator is odd. In such a case, the result is even. However, $A$ is odd. This leads to a contradiction with the parity of $A$. The parity of $M$ is determined to be odd in conjunction with the equation $(3)$. When $B^3=B=1 \implies M=0$ and $A=0$.
As a consequence, three consecutive powerful numbers do not exist. There is a risk that I made a mistake somewhere, because I am blind to my own mistakes (like people tend to be).
[Beckon]: Beckon, Edward (2019), ”On Consecutive Triples of Powerful Numbers”, Rose-Hulman Undergraduate Mathematic Journal, Vol. 20: Iss. 2, Article 3.
Best Answer
Equation (7) is incorrect; it should read $2N=\frac nN-2A$, not $2N=n-2A$. Dropping this factor of $N$ leads to the deduction that $N=1$ -- without this deduction, the rest of the proof falls apart.
(One reason this proof must fail is that it seems to predict that no two consecutive odd numbers, i.e. odd numbers which differ by 2, may be powerful. There are quite a few examples of such numbers; the smallest pairs are $(25,27)$ and $(70225, 70227)$, where $70225=5^253^2$ and $70227=3^517^2$.)