13
$\begingroup$

The problem is: Find all natural numbers $n$ for which $2^n + 1$ is a perfect square?

I am having a bit of trouble finding a generic way of finding these numbers. Of course the first obvious solution is $n = 3.$ For which we have $8 + 1 = 3^2.$ Anyone has any smart ideas?

$\endgroup$
35
$\begingroup$

$$ 2^n+1=m^2\implies 2^n=(m-1)(m+1). $$ This implies that $(m-1)$ and $(m+1)$ are both powers of $2$. In particular, $m-1$ divides $m+1$ so that $m-1$ divides $(m+1)-(m-1)=2$. This means that $m-1$ is either $2$ or $1$. Only $m-1=2$ works, so $m=3$ and $n=3$ constitute the only solution.

$\endgroup$
1
  • 7
    $\begingroup$ Glorious leader has done it again! He is truly all-knowing! $\endgroup$ – PyRulez Apr 13 '15 at 23:58
4
$\begingroup$

Method $\#1:$

If $2^n+1=m^2\iff2^n=(m+1)(m-1)$

We can easily test for $n\le2$

For $n>2,$ clearly $m$ is odd

and we have $$2^{n-2}=\frac{m-1}2\cdot\frac{m+1}2$$

But as $\displaystyle\frac{m+1}2-\frac{m-1}2=1,\left(\frac{m-1}2,\frac{m+1}2\right)=1,$ at least one of them is odd

But each divides $2^{n-2},$ the odd must be $\pm1$

If $\displaystyle\frac{m-1}2=1, m=3$ which is a legitimate solution

If $\displaystyle\frac{m+1}2=-1, m=-3$ which is also a valid solution

But, $\displaystyle\frac{m-1}2=-1$ and $\displaystyle\frac{m+1}2=1$ makes $2^n=0$

Method $\#2:$

If $n\ge1,2^n+1$ is odd, for the existence of a square we need $2^n+1=(2a+1)^2\iff2^n=4a(a+1),4$ must divide $2^n\implies n\ge2$

So, we have $2^{n-2}=a(a+1)$ and again, $(a+1,a)=(a+1-a,a)=1$

As exactly, one of $a,a+1$ is odd

Now any odd divisor of $2^{n-2}$ must be $\pm1$

Case $\#1:$ If $a$ is odd

Case $\#2:$ If $a+1$ is odd

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.