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Find $x \in N$ such that $12+13^x$ be a perfect square

I am going to limit $k < 12 + 13^x < k+i$ so that I can have $t<x<t+u$, I don't know how to do it, if $x=2k$, it pretty easy but x can also equal $2k +1$ too. So... Stuck here

Update 2: I can prove that $x$ can't be $2k$, if so, x = 2k $(k \in \mathbb{N})$ then $13^{2k}<12+13^x = 12 + 13^{2k}<(13^k+1)^2$ => $12+13^x$ can't be a perfect square.

~# if $x=2k+1$

=> $12+13^x = 12+13^{2k+1}$. Now we need prove that $k$ can not greater than $1$ (how to do that ?, stuck again)

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$x=1$ works, but you probably knew that. – Gerry Myerson Feb 7 '13 at 5:03
Where is this question from? – Will Jagy Feb 7 '13 at 5:40
Ramanujan and Nagell are associated with the similar equation $-7+2^x=y^2$, which has several solutions. The methods used for finding all the solutions would probably be a good starting point for the current problem. – Gerry Myerson Feb 7 '13 at 6:00
How were you led to this question? – Will Jagy Feb 7 '13 at 6:01
A couple of papers that might be relevant: MR0856715 (87m:11027a) Pethö, A.; de Weger, B. M. M., Products of prime powers in binary recurrence sequences, I, The hyperbolic case, with an application to the generalized Ramanujan-Nagell equation, Math. Comp. 47 (1986), no. 176, 713–727 and MR0856716 (87m:11027b) de Weger, B. M. M., Products of prime powers in binary recurrence sequences, II, The elliptic case, with an application to a mixed quadratic-exponential equation, Math. Comp. 47 (1986), no. 176, 729–739. – Gerry Myerson Feb 7 '13 at 6:09

There are many ways to solve such problems -- I'm not sure that any of them are particularly easy. One way, since you've observed that your exponent $x$ is necessarily odd, would be to find all the integral points on the elliptic curves given by the equations $$ y^2 = 13^\delta u^4+12 \; \mbox{ for } \; \delta \in \{ 1, 3 \}. $$ One can do this in, for example, magma by typing : IntegralQuarticPoints([13,0,0,0,12]); and IntegralQuarticPoints([13^3,0,0,0,12]); which lead to the two known solutions (with $|u|=1$ and $|y| =5$ and $47$). These routines are using lower bounds for linear forms in logarithms (elliptic, I believe).

Another approach (which has some similarities) would be to use an argument of de Weger (from his thesis, again based on linear forms in logarithms). This would enable you, for example, to tackle the more general equation $$ 13^x + 2^y 3^z = w^2. $$ I haven't worked out the details, but one should be able to show that the only solutions are with $$ \begin{array}{r} (x,y,z) = (0,0,1), (0,3,0), (0,3,1), (0,4,1), (0,5,2), (1,0,1), (1,0,5), \\ (1,2,1), (1,2,2), (1,2,3), (2,0,3), (2,6,1), (2,10,5), (3,2,1). \\ \end{array} $$

Yet another way to solve such problems is to use the hypergeometric method of Thue and Siegel. In this context, it enables one to prove an inequality of the shape $$ \left| y^2 - 13^x \right| > |y|^{0.4}, $$ valid for all integers $y$ and odd $x$. Such an approach is also useful for bounding the number of solutions to equations like the one under consideration here. One can, for example, show that given any odd prime $p$ and integer $D$, there are at most $3$ positive integers $x$ such that $$ p^x+D = y^2 $$ for integer $y$. This is, of course, not quite sharp when $p=13$ and $D=12$, but it's close.

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I may be two weeks late, but welcome to Math Stack Exchange professor Bennett! – Eric Naslund Feb 7 '13 at 18:59
Thank you, Eric. – Mike Bennett Feb 8 '13 at 1:57

$x=3 \implies 13^x+12=2209=47^2$.

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I don't think there's a nice way to do this. You can, however, use a calculator or a computer to find solutions. I used the following line in Mathematica:

      Intersection[12+13^(2#-1) & /@Range[100000],#^2&/@Range[300000]]

And it returned

{25, 2209}

Implying the only answers less than $100000$ are $1$ and $3$.

EDIT: changed code and results

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We want: $$13^x + 12 = a^2$$ Not an answer just compiling results:
$$ x=1,3 $$ Are the first two solutions.
There are no solutions for $$ 3<x<100000 $$ Note that $$ 12+13^x \equiv 1 \mod 8, \;\; \forall x>1, \mbox{ such that $x$ is odd}\\ 12+13^x \equiv 5 \mod 8, \;\; \forall x>1, \mbox{ such that $x$ is even}\\ $$ Hence $$ a^2 \equiv 1 \ $$ So we have that: $$ a \equiv 1,3,5 \text{ or } 7 \mod 8 $$
and $x$ is odd since $5$ is a non-residue $\mod 8$

A similar result yields that: $$ a \equiv 2 \text{ or } 5 \mod 7 $$

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I think algebraic approach is appropriate to this problem. With some calculation, we get $$(y+2\sqrt{3})(y-2\sqrt{3})=(4+\sqrt{3})^x (4-\sqrt{3})^x.$$ and $4\pm\sqrt{3}$ is prime on $\mathbb{Z}[\sqrt{3}]$. And $$\frac{5-2\sqrt{3}}{4+\sqrt{3}}=2-\sqrt{3}$$ $$\frac{47+2\sqrt{3}}{(4+\sqrt{3})^3}=2-\sqrt{3}$$

So I conjectured following propositions:

  1. If $(x,y)$ is solution of this equation, then $y+2\sqrt{3}$ associates $(4+\sqrt{3})^x$ or $(4-\sqrt{3})^x$.

  2. And each case ($(4+\sqrt{3})^x$ associates $y+2\sqrt{3}$ or $(4-\sqrt{3})^x$ associates $y+2\sqrt{3}$) gives only one solution.

But I can't get more.

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By working out the unit group, you have $$y \pm 2 \sqrt{3} = (4 + \sqrt{3})^x (2 - \sqrt{3})^a$$ for some integer $a$ and positive integer $y$. A quick exhaust shows $a > 0$ too. Subtracting the conjugates on both sides, you get what is essentially a quadratic equation in $(2 - \sqrt{3})^a$, so there is effectively one solution for each choice of $(4 \pm \sqrt{3})^x$. But there isn't always a solution where $a$ is an integer. – Hurkyl Feb 7 '13 at 8:54

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