# Discrete math. Finding a perfect square.

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?

## 2 Answers

$$2^n+1=m^2\implies 2^n=(m-1)(m+1).$$ 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.

• Glorious leader has done it again! He is truly all-knowing! – PyRulez Apr 13 '15 at 23:58

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