Squares of the form $2^j\cdot 3^k+1$ Squares of the form $2^j\cdot 3^k+1$, for j,k nonnegative.
Is it known if there are infinitely many? And if  $2^j\cdot 3^k+1=N$ is a square, then it must be necessarly a semi-prime?
Do you think that 3-smooth neighbour squares is a good name for these squares?
 A: There are finitely many pairs $(j,k)$, and this follows from results on $S$-unit equations. Moreover, the solutions can be effectively determined. Here is a quick treatment going back to the classical work of Thue (1909).
Assume that $2^j3^k+1=n^2$ is a square. Then $n-1=2^a3^b$ and $n+1=2^c3^d$ for some $a,b,c,d\in\mathbb{N}$. Classifying the quadruple $(a,b,c,d)$ according to its residue modulo $3$, we are led to $81$ equations of the form $Ax^3-By^3=2$, where the coefficients $A$ and $B$ lie in $\{2^p3^q:p,q\in\{0,1,2\}\}$. Each of them has finitely many solutions by a theorem of Thue (1909), so we are done. See also Thue equations.
P.S. This method generalizes to arbitrary many primes instead of just $2$ and $3$. For the original problem at hand, an elementary treatment fits better. See Fedor Petrov's answer.
A: There's a relevant paper by D.H. Lehmer proving finiteness and giving an elementary algorithm for finding all $n$ with $n^2-1$ smooth up to any prescribed smoothness bound.
On a problem of Størmer
Illinois J. Math.
Volume 8, Issue 1 (1964), 57-79
https://projecteuclid.org/euclid.ijm/1256067456
Luca and Najman used a related method to find all $n$ where $n^2-1$ is $100$-smooth:
On the largest prime factor of $x^2-1$
Math. Comp. 80 (2011), 429-435
https://doi.org/10.1090/S0025-5718-2010-02381-6
A: It is also possible to reduce to 9 Mordell equations. Denoting $x := 2^{[i/3]}3^{[k/3]}$, we get 9 equations:
$$d x^3 + 1 = y^2$$
indexed by $d\mid 2^23^2$.
Equivalently,
$$(dy)^2 = (dx)^3 + d^2,$$
which are Mordell equations with known solutions.
A: We may find them all by elementary methods. Assume that $2^j3^k=(n-1)(n+1)$. Since $\gcd(n-1,n+1)\leqslant 2$, we get either $j=0$, $n-1=1$, $n+1=3$, or $j\geqslant 1$, $(n-1)/2$ and $(n+1)/2$ are consecutive 3-smooth numbers. They must be a power of 2 and a power of 3, so we should solve the equations $2^a=3^b\pm 1$. This is not hard.

*

*$2^a=3^b+1$. Either $b=0,a=1$ or $b\geqslant 1$, 3 divides $2^a-1$, so $a$ is even and $3^b=2^a-1=(2^{a/2}-1)(2^{a/2}+1)$, the multiples are powers of 3 which differ by 2, so $a=2$, $b=1$.


*$2^a=3^b-1$. Either $a=1$, $b=1$ or $a\geqslant 2$, 4 divides $3^b-1$, $b$ is even, $2^a=(3^{b/2}-1)(3^{b/2}+1)$, the multiples are powers of 2 which differ by 2, $b=2$, $a=3$.
