Is my proof correct? $2^n=x^2+23$ has an infinite number of (integer) solutions.

Question

This is how I tried to prove it. Is it correct? Thanks!!

$2^n = x^2+23$

$x^2$ must be odd, therefore $x^2 = 4k+1$, where $k \in \mathbb{N}$.

$2^n=4k+24$

$k=2(2^{n-3}-3)$

Since $x^2=4k+1$,$ \ \ \ \ $ $k_1 = \frac{x_1^2-1}{4}$

and $k_2=\frac{(x_1+2)^2-1}{4}=\frac{x_1^2+4x_1+3}{4}$

If we substitute $x_1^2=4k_1 + 1$, we end up with:

$k_2=k_1 + \sqrt{4k_1+1} + 1$

Therefore, finding solutions of $k=2(2^{n-3} - 3)$ is comparable to

finding the solutions of $k+\sqrt{4k+1}+1=2(2^{n-3}-3)$

Let $p=2(2^{n-3}-3)$

Therefore, $(\sqrt{4k+1})^2=(p-k-1)^2$, so

$(p-k)^2 = 2(p+k)$

Since $p$ has an infinite number of solutions $(p-k)^2=2(p+k)$ also has an infinite number of solutions, which implies the the original does also.

Answer 1

It can be seen by elementary means that the number of solutions is finite. Then we need only find the possible solutions. I give a heuristic first.
Consider the primefactorization of some expression $$f(b,n) = b^n - 1 = 2^{e_2} \cdot 3^{e_3} \cdot 5 ^{e_5} \cdots =\prod_{q \in \mathbb P} q^{e_q} $$ then we know by Fermat/Euler, that the occurence of the primefactors $q$ cycles when $b$ is fixed and $n$ is increasing. To make the actual value of the exponents explicite I've made myself used to the following notation:
Let's introduce the notations

• $[n:p]$ meaning $[n:p]=1$ if p divides n, and $[n:p]=0$ if not
• $\{n,p\}$ meaning the highest exponent, to which $p$ occurs in $n$

then any primefactor $q$ (where of course $[b:q]=0$ is assumed) occur in $f(b,n)$ at a smallest $n=\kappa_q$ first time and then cyclically with cyclelength $\lambda_q$. Here the "first time" and the cycle-length are identical, we'll see in the next step, where we use this for our actual problem, that they shall differ - but this does not spoil the power of the logic of the arguments.

Also, we denote the exponent, to which the primefactor $q$ occurs in $f(b,\kappa)$ as $\alpha$ such that

• $ \{f(b,\kappa),q\}=\alpha_q$

getting now for instance for the primefactor $q=2$
$$ f(b,n) = b^n - 1 = 2^{e_2} \cdot r $$ $\qquad \qquad $ (where $r$ is some possible remainder)

and $$ e_2 = [n:\lambda_2](\alpha_2 + \{n,2\} ) $$ We can then rewrite the complete expression $$f(b,n) =\prod_{q \in \mathbb P \\\ [b:q]=0} q^{ [n:\lambda_q](\alpha_q + \{n,q\})} $$

The key-observation is here the "valuation-brace" in the exponent, $$e_q = \ldots \{n,q \}$$ varies with n itself; so the multiplicity of some primefactor in $b^n-1$ occurs in the same variation as in $n$ only, so roughly $q^{e_q}$ grows with $n$ and not with $b^n$. The consequence of this is then, that increasing $n$ requires more and more additional primefactors and after a first solution, where $b^n-1$ may have only the single primefactor $q$ (to some power) this shall never happen again when $n$ increases. (This might be also related to the theorem of Szygmondi (correct spelling?).)

Now we turn back to our original question.
We can restate the same function, only fix $n=2$ and let $b$ vary and use another constant, so that we have $$ g(x,2) = x^2 + 23 = 2^{e_2} \cdot 3^{e_3} \cdot 5^{e_5} \cdots = \prod_{q \in \mathbb P} q^{e_q} $$
The arguments about cyclicitiness and level of exponents of the primefactors $q$ are assentially the same, only that we have, that in general $\kappa \ne \lambda$ and also have $2$ disjunct cycles. Your question concerns the primefactor $q=2$ only, so we may look at the level of its exponents heuristically. We get for the first few $x$ the following table, where it is made in $4$ columns each consisting $g=g(x,2)$ and $e_2 =\{ g(x,2),2 \} $ to see the two "cycles". We read the table rowwise from left to right and then downwards for $x=1,2,3,4,5,...$ : $$ \small \small \begin{array} {rr|rr|rr|rr} g & e_2 &g & e_2 &g & e_2 &g & e_2 &\\ \hline 24 & 3 & 27 & 0 & 32 & 5 & 39 & 0 \\ 48 & 4 & 59 & 0 & 72 & 3 & 87 & 0 \\ 104 & 3 & 123 & 0 & 144 & 4 & 167 & 0 \\ 192 & 6 & 219 & 0 & 248 & 3 & 279 & 0 \\ 312 & 3 & 347 & 0 & 384 & 7 & 423 & 0 \\ 464 & 4 & 507 & 0 & 552 & 3 & 599 & 0 \\ 648 & 3 & 699 & 0 & 752 & 4 & 807 & 0 \\ 864 & 5 & 923 & 0 & 984 & 3 & 1047 & 0 \\ 1112 & 3 & 1179 & 0 & 1248 & 5 & 1319 & 0 \\ 1392 & 4 & 1467 & 0 & 1544 & 3 & 1623 & 0 \\ 1704 & 3 & 1787 & 0 & 1872 & 4 & 1959 & 0 \\ 2048 & 11 & 2139 & 0 & 2232 & 3 & 2327 & 0 \\ \ldots \end{array} $$ Although the left part of each column increase dominated by g's quadratic term $ x^2 $, the right part, which contains the exponent of the primefactor $2$, follows only the pattern which is identical to that of increasing $ x^1 $ only. We can also identify the shifts; after we find the high exponent $e_2=11$ and see also, that this consumes the whole number $g=2048$ (which occurs at $x=45$) and constitute thus a solution for your question the previous and the next exponents vary as if $g$ would grow only linearly in steps of $8$ (=$2^3$) and thus $g(x,2)$ needs further primefactors besides $2^{e_2}$ to multiply up to its actual value, except for that finite number of (possible) exceptions (at most 2 because a parabola and a line can meet at most 2 times).

The same is true for the third column; here we have a solution for $g(x,2)=32$ where $e_2=5$ and no other primefactor is needed (which means actually $x=3$) , but again the growth of the g-values in the column is dominated by its quadratic term $x^2$ while the exponents of the primefactor $2$ grow only according to the linear term $x$, besides the offset and the scaling of the stepwith to $8=2^3$.

This is so far heuristic/observation. I cannot yet make the formal proof for the finiteness of the number of solutions; if it is possible I'll append it here - perhaps this is also the line of arguments of Pillai who was linked to in Will @Jagy's answer.

Answer 2

This seems wrong to me.

You went from $k_1 = \frac{x_1^2-1}{4}$ to $x_1^2 = 4 k_1^2+1$. Somehow the $k_1$ got squared.

This is not a correct proof.

Also, having to download that big image is a pain.

Please learn how to enter math in $\LaTeX$.

Answer 3

I get $x=\pm 3$ and $x = \pm 45$ and that's it. It is likely that the set of solutions is finite, see PILLAI'S CONJECTURE There is a fine description of how Pillai's conjecture would be written by @ShreevatsR in comment below:

that "for fixed positive integers A,B,C the equation $Ax^n−By^m=C$ has only finitely many solutions $(x,y,m,n)$ with $(m,n) \neq (2,2).$" Here with $(A,B,C)=(1,1,23)$ it says that $x^n−y^m=23$ has finitely many integer solutions $(x,y,m,n).$ The OP's claim is that it has infinitely many integer solutions of the form $(2,x,2,n).$ This is probably false

Meanwhile, I can describe how to rapidly exhaust possible solutions, by my methods. We know that $n$ must be odd in $x^2 - 2^n = -23.$ So, take $n= 2t+1$ and make a new variable, $y= 2^t.$ The result is $$ x^2 - 2 y^2 = -23. $$ The seed values are $(x,y) = (3,4)$ and $(x,y) = (-3,4).$ We want all solutions such that $y$ turns out to be a power of 2. Now, given a solution $(x,y),$ we get all possible solutions by repeatedly taking the result of applying an element of the automorphism group/isometry group/orthogonal group of $x^2 - 2 y^2,$ namely $$ (3x-4y,-2x+3y). $$

Now, $y=4$ is a power of 2, so that is a start, with $x=\pm 3$

The first string is $$ (3,4), (-7,6), (-45,32),(-263,186),(-1533,1084),(-8935,6318), \ldots $$ So this one gives $$ (x = \pm 3, y = 4), \; \; \; (x = \pm 45, y = 32) $$ as successes

The other string is $$ (-3,4),(-25,18),(-147,104), (-857,606),(-4995,3532),(-29113,20586), \ldots $$

So you can see how I became skeptical about there being any more solutions with $y$ a power of 2.

Note that Erick Wong has pointed out a proof as a simple application of elliptic curves.

Meanwhile, I am now awake, see KATY PERRY.

Answer 4

Here is another heuristic argument along probabilistic lines that there are finitely many solutions.

The distance between perfect squares near $n$ is approximately $2\sqrt{n}$. Thus, the probability of a given integer $n$ to be a perfect square is approximately $\frac1{2\sqrt{n}}$. Summing the probability that $2^n-23$ is a perfect square gives $$ \begin{align} \sum_{n=5}^\infty\frac1{2\sqrt{2^n-23}} &\le\frac16+\sum_{n=6}^\infty\frac1{2\sqrt{2^{n-1}}}\\ &=\frac16+\frac{\sqrt2+1}8 \end{align} $$ According to the Borel-Cantelli Lemma, this indicates that the probability that there are finitely mny solutions is $1$.

Of course, this really proves nothing, because the same argument suggests that there are finitely many perfect squares of the form $2^n$, when in fact $2^n$ is a perfect square whenever $n$ is even. However, in the absence of proof to the contrary, this provides a decent guess.

Answer 5

The substitution $x^2=4k^2+1$ was wrong because $x^2=4k+1$, but nevertheless the next line is correct as if I've substituted the correct thing. The actual problem of the proof comes from when I said "$p$ has infinitely many solutions". I was thinking that $p$ is just an integer, and clearly $2(2^{n-3}-3)$ outputs a positive integer when $n$ is at least $5$. That's why I said that $p$ has an infinite number of solutions. However, I did not realize that since $p=k+\sqrt{4k+a}+1$, I'm already assuming that $2(2^{n-3}-3)=k+\sqrt{4k+a}+1$ has an infinite number of solutions, which would be implied if I was actually correct.Therefore this is a circular argument. But I do have one question about a certain step. Was it correct when I said that finding solutions of $k=2(2^{n-3} - 3)$ is comparable to finding the solutions of $k+\sqrt{4k+1}+1=2(2^{n-3}-3)$?

Answer 6

There are only finitely many solutions. Pillai's conjecture is overkill because it allows for all possible exponents simultaneously, whereas the solutions of $2^n = x^2 + 23$ must lie on one of the three curves $y^3 = x^2+23$, $2y^3 = x^2+23$, $4y^3= x^2+23$, each of which has finitely many integer points.