Number Theory or Algebra? Part 1

Spoilers: By guessing, 1 is one of the answer.

You can see that $x$ must be odd because $L.H.S.$ is always even for any positive integer $x$ .

$x$ can be written as $2k+1$ for some positive integer $k$ .

We get $2^{2k+1} = (2k+1)^{2} + 1$

$2^{2k+1} = 4k^{2} + 4k + 2$

Divide by 2 we get $2^{2k} = 2k^{2} + 2k + 1 = 2(k^{2} + k) + 1$ .

Since L.H.S. is even, but R.H.S. is odd ( $k^{2} + k$ is positive integer).

Therefore, no such $k$ exists that makes the equation true.

The only answer is $\boxed{1}$ . ~!~!~

By guessing we will get x = $1$ as an answer.

But, for $x \geq 2:$

Because $2^{2} = 4, L. H. S = 2^{x} \equiv 0 (mod 4)$

For the R. H. S.,

From $x^{2} \equiv 0, 1 (mod 4)$ ----> Please refer to my solution from my problem "Are these squares perfect?"

We will get $R. H. S. = x^{2} + 1 \equiv 1, 2 (mod 4)$

Thus $L. H. S. \not\equiv R. H. S.$

Thus $L. H. S. \neq R. H. S., for x \geq 2$

So we only have $\boxed{1}$ as our only positive INTEGER $x$ .

Remarks: There is also a second value of $x$ , but it is not an integer.

Notice that if we have x > 1, then $x^2+1$ is not a power of $2^x$ . Thus, the only positive integer that can fit x is 1. Only $\boxed{\text{one}}$ integer fits this description. $\textbf{Ans.}$

The graphs of 2^x and x^2 +1 reveals that the system has only one solution. By trial that is 1.

Derivative of x^2 +1 = 2x and derivative of 2^x = 2^xln2 , now 2^x and x^2 +1 are equal at x =1 , now plot y=2x and y= 2^x ln2 , since after x=1, 2^xln2 is greater than 2x , it implies x^2+1 will never intersect 2^x after x=1

First, you can see that $2^x$ is even("positive integers" don't include $0$ ) while $x^2+1$ is odd at an even $x^2$ ( $\text{even} +1=\text{odd}$ ) and even at an odd $x^2$ ( $\text{odd} +1=\text{even}$ ) and for $n^2$ to be even, $n$ must be even(or irrational), same goes for odd numbers, so $x$ must be odd.

Next, $2$ power an odd number has an alternating pattern of ones digits made of $2$ s and $8$ s( meaning $2^{\text{odd number} }$ always ends in either 2 or 8. example: $2^1={\color{#D61F06}2}, 2^3={\color{#D61F06}8}, 2^5=3{\color{#D61F06}2}, 2^7=12{\color{#D61F06}8}, …$ ) so the $2^x$ part must have a ones digit of $2$ or $8$ , so does $x^2+1$ , so $x^2$ must end in either $1$ or $7$ . Because perfect square numbers don't have $7$ as their ones digit, only $1$ can be the ones digit for $x^2$ so $x$ must end in either $1$ or $9$ .

Testing some numbers that work : $2^1=1^2+1$ which is true, $2^9=512\neq82=9^2+1$ , $2^{11}=2048\neq122=11^2+1$ . As you can see, $2^x$ grows much faster than $x^2$ because the difference between $2^x$ and $2^{x-1}$ is $2^{x-1}$ while the difference between $x^2$ and $(x-1)^2$ is $2x-1$ or consecutive odd integers and, clearly, $2^{x-1}>2x-1$ . So $2^x$ would grow much larger than $x^2+1$ after $x=1$ , making sure they won't meet again. So only ${\color{#20A900}1}$ solution works.

If 2^x = x^2 + 1, and x > 0, then x is odd. Let x = 2k + 1, so 2^(2k + 1) = (2k + 1)^2 + 1 = 4k^2 + 4k + 2. Dividing by 2, 2^(2k) = 2k^2 + 2k + 1, whose only solution can be k = 0, otherwise, the left side is even, the right side is odd. If k = 0, x = 1, and can be the only solution. Ed Gray.

$2^1$ = $1^2 +1$ .

P.S. 0 is an answer but they asked for POSITIVE integers.

1,is a obvious solution.....but can anyone help me out....

2^x =x^2+1

Take log(e base)/ln on both sides

xln(2)=2ln(x)+0 (lna^x=xlna...... And .... ln(1)=0)

x =2 satisfies the eq. ....so 2 is also an ans.

But how come 2 be the solution of the original equation????????

(Is it an extraneous root????)

We are pretty much familiar with the exponential and parabolic curves

by just seeing we can say that they intersect at two pts 0,1

But it is asked for positive integers so answer is 1

Why could it not be 0 aswell?

we can graphically also see that 0 and 1, are only 2 points where $2^x$ and $x^2+1$ will intersect.

Using Mathematical Induction, we will prove that for $x \ge 5$ , we have $2^x > x^2+1$ , so it is impossible to have $2^x = x^2+1$ .

When $x=5$ , $2^5 = 32 > 26 = 5^2 + 1$ .

Assume that $2^k > k^2+1$ for $k \ge 5$ .

For $k+1$ :

$\begin{array}{cl}&2^{k+1}\\=& 2 \times 2^k\\>& 2k^2 + 2\\=& k^2 + k^2 + 2\\>& k^2 + 2k +2\\=& (k+1)^2 + 1\\\end{array}$

Therefore, for $x \ge 5$ , we have $2^x > x^2+1$ by Mathematical Induction.

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So, we have reduced our problem to $x \in \{1,2,3,4\}$ , which can be easily checked.

Clearly it can be seen by plugging in some +ve values, That 0&1 are the solutions. Since ques. Asks for +ve integers hence 1 is the only sol.!

I do know that all potency of base 2 is pair. Meantime, all number pair plus 1 it is odd. Then, only 2¹ = 1¹+1, being 1 positive integer. Thus, 2^0 = 1^0 +1, but zero is not integer positive.