Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find all natural numbers $(n,m)$ where $n^{2}-2^{m}=1$.
I have my own answer of that, however I wanted to know if anyone has a better or easier answer or not!

share|improve this question

2 Answers 2

up vote 12 down vote accepted

We have $n^{2}-1=(n-1)(n+1)=2^m$, So let $k=n-1$. Then we know that $k$ and $k+2$ should both be $2^m$'s divisors and also we know that all of divisors of $2^{m}$ are in a geometric progression with the common ratio $2$. Thus, $k+2=2^{i}k$.We take $i=1$ (See the proof at the end), and so $k=2$ and $n=3$. By putting $n=3$ in the equation we may get $m=3$. So the only answers would be $(3,3)$.
Now we must prove that if $k+2=2^{i}k$, Then $i=1$. Suppose $i>1$. Then $2=k(2^{i}-1)$ and so $k=\frac{2}{2^{i}-1}$ which is impossible since $k$ is a natural number and $2<2^{i}-1$ for $i>1$.

share|improve this answer
1  
Technically, that's $k+2=2^i k$, with $i \ge 0$. But the remaining cases are trivial. –  Lord_Farin May 7 '13 at 13:30
1  
Thanks. I added the case at the end. –  CODE May 7 '13 at 14:08

It's similar to your solution.

Obviously, $n \equiv 1 \mod 2$ (since $m>0$). Therefore, there exists $k$ s.t. $2k + 1 = n$.

Now, $4k^2 + 4k + 1 - 2^m = 1$ and $k(k+1) = 2^{m-2}$. Clearly $m \neq 1, 2$. When $m \ge 3$, since one and only one of $k$ and $k+1$ can be even, one of two is $2^{m-2}$ and other is $1$. Clearly, it is possible only when $k = 1$ ($n = 3$) and $m = 3$.

share|improve this answer
1  
+1 Good one. Thanks. –  CODE May 7 '13 at 14:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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