Congruence obstructions for three consecutive powerful numbers

Question

A powerful number is an integer $m$ such that if $p$ is prime and $p \mid m$ then $p^2 \mid m$. Powerful numbers can be represented in the form $m=u^2 v^3$.

Erdos conjectured that three consecutive powerful numbers don't exist.

Assume $(a-1,a,a+1)$ is a triple of consecutive powerful numbers.

Choose integers $n,A$ and set $a=A$ and assume the system $$x_1^2 x_2^3=A-1, y_1^2 y_2^3=A+1$$ doesn't have solution modulo $n$.

Then over the integers powerful triple with $a \equiv A \pmod{n}$ doesn't exist.

Some examples with $n$ square:

n | A congruence obstruction
4 [1, 3]
9 [2, 4, 5, 7]
16 [1, 3, 5, 7, 9, 11, 13, 15]
25 [4, 6, 9, 11, 14, 16, 19, 21]
36 [1, 2, 3, 4, 5, 7, 9, 11, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 25, 27, 29, 31, 32, 33, 34, 35]
49 [6, 8, 13, 15, 20, 22, 27, 29, 34, 36, 41, 43]

Q1 Can we find $n$ and congruence obstructions to show non-existence of more naturally defined powerful triples?

Q2 Can we show that for some even $B$ the triple with $a=B^k$ doesn't exist for all $k$? (For odd $B$ this is visible with naked eye)

Answer 1

EDIT: corrected my response to Q2.

Q1: if I understood the question correctly then the answer is negative. Indeed we may assume $(2B,n)=1$ and taking $k=\varphi(n^2)$ we have $n^2|B^k-1,\,n\nmid B^k+1$, so there is no obstruction stemming from $n$.

Q2: I will show that $2^k-1$ can never be powerful, under the assumption that there are no order-3 Wieferich primes, i.e. $2^{p-1}\equiv 1\pmod{p^3}$ (I realize that this is one hell of an assumption, but perhaps the arguments below can be modified into something more useful).

Assume the contrary and let $k$ be the minimal counterexample. If $k$ is even then since $2^{k/2}-1,\,2^{k/2}+1$ are coprime they are both powerful, so $k/2$ is a smaller counterexample - a contradiction.

We now treat the case when $k>2$ is prime. Assume by way of contradiction that $2^k-1$ is powerful. Since $2^k-1$ is not a square, we have $p^3|2^k-1$ for some prime $p>2$. Then $\mathrm{ord}(2\bmod p^3)|k$ is not divisible by $p$ and so must divide $p-1$ and $p$ is an order-3 Wieferich prime, contrary to our assumption.

Hence assume $k$ is odd and composite and let $p>2$ be the smallest prime factor of $k$. Write $k=pm$. We have $2^k-1=ab,\,a=2^m-1,\,b=1+2^m+2^{2m}+\ldots+2^{(p-1)m}$.

First I claim that $p\nmid a$. Indeed if $p|a$ then $2^m\equiv 1\pmod p$ and $(m,p-1)>1$, so $m$ and hence $k$ has a prime factor smaller than $p$ - a contradiction.

Next I claim that $(a,b)=1$. Indeed $b\equiv p\pmod a$ and $(a,p)=1$ so $(a,b)=1$.

Since $ab$ is powerful so is $a=2^m-1$, contradicting the minimality of $k$.