For which $n,p$ does $\cfrac {5n^2-3n}{2}=2^p-1$

Question

For which $n,p$ does $\cfrac {5n^2-3n}{2}=2^p-1$, where $n,p$ are positive integers

Answer 1

Not an elementary solution. But you can solve it in integers.

$(10n-3)^2=5\cdot 2^{p+3}-31$. Clearly $p+3\ge 0$.

• If $p+3=3k, k\ge 0$, then $(5(10n-3))^2=\left(5\cdot 2^k\right)^3-775$. But $a^2=b^3-775$ has $14$ integral solutions (see http://oeis.org/A081120 and http://oeis.org/A081120/b081120.txt), which can be found with a program or these tables.

  $$(a,b)=(\pm 15,10), (\pm 78,19), (\pm 85, 20), (\pm 585,70),$$

  $$(\pm 715,80), (\pm 2167815, 16750), (\pm 4321215,26530)$$ This gives the solutions $(n,p)=(0,0),(2,3),(-14,9)$.

• If $p+3=3k+1, k\ge 0$, then $(10(10n-3))^2=\left(5\cdot 2^{k+1}\right)^3-3100$. But $a^2=b^3-3100$ has $2$ integral solutions (http://oeis.org/A081120/b081120.txt), which are $(a,b)=(\pm 70, 20)$. This gives $(n,p)=(1,1)$.

• If $p+3=3k+2, k\ge 0$, then $2^{p+3}\equiv 4\pmod{7}$, so $5\cdot 2^{p+3}-31\equiv 3\pmod{7}$, which is not a quadratic residue.

This uses results from $1998$ in Mordell Equations (from this paper). In particular, $y^2=x^3+k$ is solved in integers for all $0<|k|\le 10^4$, and OEIS gives lists of the numbers of solutions for each $k$.