Find the integer solution to when $15 × 2^n + 1$ is a perfect square?

Question

Similar: math.stackexchange.com/questions/2923880

I tried to solve it using the mod but the only reasonable observation made by me was that $n≥3$ except when $n=0$ using $mod 8$, which did not help, unfortunately.
Then I tried to factories it like this: $$15 × 2^n = (k+1)(k-1).$$ I do know that when $2^n = (k+1)(k-1)$ there is only 1 solution which is n=3 and tried to use similar argument somehow but I have not gone anywhere with it yet. From some experimentation I have realised that it has only three solutions (i.e, 0,3 and 6) and after that there are no two consecutive even numbers that multiply to satisfy the left hand side. Similarly, the same case can be observed for other $a × b^n + 1$, where they only have very small number of solutions.
I would really appreciate it if someone can point me in the right direction to solve this and similar questions. Thanks! (If there is a similar question please point me to that because I could not find any.)

Answer 1

The difference-of-squares approach that you have started is often a good approach, and always worth trying out. From there, because you have a factorization of the right-hand side, a prime factorization of the left-hand side is a natural step forward.

Assuming $n>1$, you want $(k-1)\cdot (k+1)$ to have factorization $$ 3\cdot 5\cdot2\cdot2^{n-1} $$ So we merely brute-force distribute those four factors between $k-1$ and $k+1$ and see if it works.

$2$ and $2^{n-1}$ must go to different sides. This yields two main cases:

• $k-1$ gets $2$ and $k+1$ gets $2^{n-1}$. In this case, $k-1$ can be either $2$, $6$, $10$ or $30$. This yields $4$, $8$, $12$ and $32$ as possible $k+1$. We see that only the latter two have the correct factorization, corresponding to $n=3$ and $n=6$.
• $k+1$ gets 2 and $k-1$ gets $2^{n-1}$. In this case, it is $k+1$ that can be $2$, $6$, $10$ or $30$. It is easy to check that none of these yield valid solutions.