Do you have an analytical solution?

We know that $\color{#20A900}{a^2-b^2=(a+b)(a-b)}$ for all real numbers $\color{#EC7300}{a,b}$ . Let this be called Identity 1.

We are given $\color{#69047E}{2^x=x^2-1}$ . Applying Identity 1 to $\color{#EC7300}{x,1}$ , we get:- $\color{#69047E}{(x+1)(x-1)=2^{x}}$

It is east to see that $\color{fuchsia}{x>-1}$ , or we will have a power of $2$ as negative, which is not possible.

Also, we see that $x$ is odd, for if it were even, a power of $2$ would have an odd factor.

So, $\color{#D61F06}{x=2a-1}$ , so that $\color{#69047E}{(a)(a-1)=2^{2a-3}}$

Now, one of $\color{#EC7300}{a}$ and $\color{#EC7300}{a-1}$ must be 1, for it is the only odd number which divides a power of $2$ . $a=1$ does not work, for if $a$ were unity, $a-1$ would've been $0$ , and a power of $2$ would've been $0$ ,

So, $\color{#20A900}{a-1=1}$ , and $\color{#20A900}{a=2}$ , and $x=2a-1=3$ . So the only possible integer is $\color{#3D99F6}{\boxed 3}$ .

Hope everything is explained!

First, we can see ${ 2 }^{ x }={ x }^{ 2 }-1$ is ${ 2 }^{ x }=(x+1)(x-1)$ .

So, here I have two ways to finish the problem:

$1)$ We can see that the power of two ( ${ 2 }^{ x }$ ) is the product of two consecutive even numbers ( $(x+1)(x-1)$ ) one of them divisible by $4$ and the other only by $2$ . So, we know that all the powers of $2$ are the product of other power of two ( as $2$ is a prime number) . Then, we can see that all the numbers ${ 2 }^{ k }$ with $k\ge 2$ are divisible by $4$ so the only solution is ${ 2 }^{ x }=(x+1)(x-1)=4*2=8$ , so $x=\boxed { 3 }$ .

$2)$ Let say $x+1={ 2 }^{ i }$ and $x-1={ 2 }^{ j }$ . See that $(x+1)-(x-1)={ 2 }^{ i }-{ 2 }^{ j }={ 2 }^{ j }({ 2 }^{ i-j }-1)=2$ , so we can see the only solution is $2*1=2$ in which $j=1$ and $i=2$ . And we get the same solution $x=\boxed { 3 }$ .

Another approach is to notice that $2^x$ grows faster than $x^2-1$ . In fact, we can prove by induction that $2^x > x^2-1 ~\forall ~ x\geq 4$ . The base case $x=4$ is true since $16=2^4 > 4^2 -1=15$ . An easy induction shows that $n^2-2>2n$ for all $n\geq 3$ . Assume that $2^n > n^2 - 1$ for some $n\geq 4$ . Then we have: $2^{n+1}=2\cdot 2^n > 2(n^2 - 1) > n^2 + 2n = (n+1)^2 -1,$ so the inductive step is complete. Clearly $x$ can't be negative and even. So we only need to check $x=1,3$ and only $\fbox{3}$ works.

My solution is pretty small.Here goes,we know that 2^x+1=x^2 that implies x=odd.Now write 2^x=(x-1)(x+1).Now let us say that (x-1)=2^a and (x+1)=2^b.Add the two equations and you will get 2x=2^a+2^b.That implies x=2^(a-1)+2^(b-1) but x=odd.That can only happen when one of them =0.Solve it and you get the answer.

2^3=8 3^2-1=9-1=8 So answer is 3.

take log base 2 of both sides, you will have an equation with one variable, that equals to 2, logarithmic function is 1 to 1, consequently only one solution, the rest is easy, 2 doesn't work 3 does.

Exponential will eventually greater than polynomial. Since both sides are equal when x equals to 3, this is the only solution.

2^3=3^2-1 that is 8=8

(x-1)(x+1)=2^x.So,x+1 and x-1 is a factor of 2.So difference of two factores is 2.So the nos. are x-1=2 and x+1=4 is the sol.No other no. satisfies it.Hence,x=3.

I put in 3 directly and got the answer.Lucky me !!

2^3=8 3^2 -1=8

To be frank, initially I just noticed that 'x' has to be odd, because for 'x^2 - 1' to be even, x has to be odd. Then I started plugging in the values, and '3' worked! And then I noticed an obvious pattern, the more you increase 'x' after '3', the bigger 2^x becomes than 'x^2 -1'. So '3' had to be the only answer. But then I tried to prove it analytically by assuming x-1=a, which makes the equation, a(a-2)= 2^(a+1), so 'a' and 'a-2' both have to be some powers of 2, which gives a= 2,4, 8, 16.... and a-2=2,4,8,1....=> a=4. So, x=3 is the answer. It's just like all the other solutions posted here...

we manipulate : 2^{x} = (x+1)(x-1) or (2^{x-1})(2)= (x+1)(x-1). Let x+1=2, then x=1 ( does not hold). Now, let x-1=2, then x=3 ( does hold).

My approach was taking values of x which satisfies the givren equation,with factorisation value of x when 1 and 2 it does not satisfy with x+3 it satisfies so answer is 3. K..K.GARG,India