Solve over the integers the equation $2^n+1=m^2$

Question

I tried to discuss two cases when $m$ is even and odd. By discussing both cases, $m$ must be odd, and by using the fact that every odd perfect square is congruent to $1$ modulo $8$.

It is clear that $n=3$ and $m=3$ is a solution, and it seems that it is the only one? I tried to prove that the equation admits one solution, but I failed.

Can I get some help, and thanks in advance.

Where does $p $ comes from ? – The Demonix _ Hermit Dec 20 '19 at 17:33 — The Demonix _ Hermit, Dec 20 '19 at 17:33
Sorry, I meant m not p. I corrected it, thanks. – Kamal Dec 20 '19 at 17:45 — Kamal, Dec 20 '19 at 17:45

score 3 · Answer 1 · answered Dec 20 '19 at 17:44

3

We have $2^n=(m+1)(m-1)$, so, by unique factorization, $m+1$ and $m-1$ must be powers of $2$.

But the only powers of $2$ that differ by $2$ are $4$ and $2$, so $m$ must be $3$.

answered Dec 20 '19 at 17:44

J. W. Tanner

60,406

$2^a-2^b=2\implies2^{b-1}(2^{a-b}-1)=1\implies 2^{b-1}=1, 2^{a-b}=2\implies b=1, a=2$ – J. W. Tanner Dec 20 '19 at 17:54

Alessio K · Answer 2 · 2019-12-20T18:02:45.683

1

Assume we are looking for solutions of the form 2^n=m^2-1=(m-1)(m+1). So this means that both m-1 and m+1 are divisible by 2. By inspection the integer solution is n=3 and m=-+3 since we have 2^3=8=(2)(4)=(-4)(-2). The other solutions are complex. Taking natural logs of both sides we have nln(2)=ln(|m^2-1|) so that n=ln(m^2-1)/ln(2) and ln(2) is irrational. Since if ln(2)=a then e^a=2, but e^a is transcendental. Here: Is the natural log of n rational? So only possible cases for integers solutions is when the numerator can be made of the form nln(2), so the ln(2) cancels this is indeed the case for m=-+3. https://www.wolframalpha.com/input/?i=2%5Ey%3D%28x-1%29%28x%2B1%29

edited Dec 20 '19 at 18:02

answered Dec 20 '19 at 17:51

Alessio K

10,599

1

Hello, sorry if I wrongly rejected your edit. Of course if it is clear that the OP has made a mistake, then I will accept the edit. I usually reject edits when the OP has made a mistake (more than once) when writing the question and needs clarification, this is because you can make a comment and tell the OP this instead. Then the OP will understand their mistake (and not repeat it afterwards) and you can edit the question. Again, sorry if I wrongly rejected your edit. – Alessio K Nov 10 '20 at 08:18

score 0 · Answer 3 · answered Dec 20 '19 at 17:34

Hint:

$$m^2-1=2^n$$

Check for $n=0,1,2$

As $m+1,m-1$ have same parity, both must be even for $n>2$

$$\dfrac{m+1}2\cdot\dfrac{m-1}2=2^{n-2}$$

Now as $\dfrac{m+1}2-\dfrac{m-1}2=1$

$$\left(\dfrac{m+1}2,\dfrac{m-1}2\right)=1$$

So, they must have opposite parities.

Now what is the odd factor of $2^{n-2}$

Solve over the integers the equation $2^n+1=m^2$

3 Answers3