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Problem

For the equation:

$12^x=5^y-1$

I want to prove it has no solutions with x,y being positive integers.

Question

Is there a general method for solving this type of equation? (It looks vaguely like a Pell equation, but not close enough that I can see how to solve it with standard methods)

If not, is there an elegant method to prove it for the particular case here (with $A=12$ and $B=5$)?

What I've tried

Thinking modulo 12, the LHS = 0, and the RHS is 0 if and only if y is even. Writing $y=2z$, I can then factorize the RHS into $(5^z+1)(5^z-1)$

Both factors are even and by thinking modulo 3, only one of these factors can be divisible by 3. So I conclude that I need something like $5^z+1=2^?3^x$ and $5^z-1=2^?$ or vice versa.

Subtracting these equations I need $2=2^?3^x-2^?$.

If I now think in binary, these equations look like $10_2 = (11_2)^x100..00_2 - 100...00_2$.

It seems to make sense (but I don't see how to mathematically express this idea) that the only way this equation will work is as $5^1+1=2.3$ and $5^1-1=2.2$ but this solution results in a LHS of 24, which is not a power of 12.

However, I feel there must be a less convoluted proof!

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  • $\begingroup$ what is $\gcd(5^z - 1, 5^z + 1)?$ $\endgroup$ – Will Jagy Oct 11 '16 at 20:15
  • $\begingroup$ Two, so this certainly cuts down the options for the factors a great deal (as one factor must have just a single 2). Is there more I should conclude? $\endgroup$ – Peter de Rivaz Oct 11 '16 at 20:18
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    $\begingroup$ Mihăilescu's theorem kills all of these problems. $\endgroup$ – Fan Zheng Oct 11 '16 at 20:20
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    $\begingroup$ that should be enough. Meanwhile, over the past few days I have been fiddling with questions such as $7^x - 3^y = 100,$ there seems to be a procedure but it is not easy math.stackexchange.com/questions/1946621/… $\endgroup$ – Will Jagy Oct 11 '16 at 20:21
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As an illustration, let us solve $3^A - 2^B = 1,$ which will finish the other answer. We suspect the largest solution is $9-8=1.$ Take $3^A = 2^B + 1$ and subtract $9$ from both sides, for $3^A - 9 = 2^B - 8.$ Divide out both factors and introduce new variables, for $$ 9 (3^x - 1) = 8(2^y - 1).$$ We will show that this is impossible with $x,y \geq 1.$

Little explanation: given $m,n \geq 2$ with $\gcd(m,n) = 1,$ we know that $m^{\varphi(n)} \equiv 1 \pmod n.$ However, there may be a smaller $k$ with $m^{k} \equiv 1 \pmod n.$ If so, we take the smallest such $k$ and call it the order, sometimes multiplicative order, of $m \pmod n.$

We proceed under the assumption that $x \geq 1 $ and $y \geq 1.$

Now, $2^y \equiv 1 \pmod 9.$ This means that $$ 6 | y. $$

jagy@phobeusjunior:~$ ./order 2 9
9     6 = 2 * 3

Furthermore, $2^6 - 1 | 2^y - 1.$ $$ 2^6 - 1 = 63 = 3^3 \cdot 7. $$ Therefore $7 | (3^x - 1),$ $$ 3^x \equiv 1 \pmod 7. $$ Therefore $$ 6 | x, $$ and $3^6 - 1$ divides $3^x - 1.$

jagy@phobeusjunior:~$ ./order 3 7
7     6 = 2 * 3

$$ 3^6 - 1 = 8 \cdot 7 \cdot 13. $$

Therefore $$ 2^y \equiv 1 \pmod {13}, $$ $$ 12 | y. $$

jagy@phobeusjunior:~$ ./order 2 13
13    12 = 2^2 * 3

In particular $$ 4 | y, $$ and $2^y - 1$ is divisible by $15,$ especially divisible by $5.$

$$ 3^x \equiv 1 \pmod 5, $$ so $$ 4 | x. $$

jagy@phobeusjunior:~$ ./order 3 5
5     4 = 2^2

However, $$ 3^4 - 1 = 80 = 5 \cdot 16. $$ This means that $8 (2^y - 1)$ is divisible by $16,$ a contradiction of $$ 9 (3^x - 1) = 8(2^y - 1)$$ with $x,y \geq 1.$

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The powers of $5$ mod $11$ are $5,3,4,9$, and $1$. Thus $5^y-1\in\{4,2,3,8,0\}$ mod $11$. But $12^x\equiv1^x=1$ mod $11$. So $12^x=5^y-1$ can have no solutions in positive integers.

Remark (added later, on reading the question's request for a general method): A key feature of the equation $12^x=5^y-1$ that makes a simple congruence-based approach possible is the fact that we're proving the equation has no solutions. For equations like $2^x=3^y-1$ (see Will Jagy's excellent answer), where you're trying to prove it has just one solution, a simple congruence-based approach doesn't have a chance.

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If $y$ is odd then $5^y-1=1 \pmod 3$ while $12^x=0 \pmod 3$. So there are no solutions whit $y$ odd.

Let $y=2z$. Then the equation is $12^x=25^z-1$. Thinking$\pmod {13}$, the left hand side is $1$ or $-1$ while the right hand side is $0$ or $2$. Then there are no solutions with $y$ even, and therefore no solutions at all.

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That one is easy, clearly $y=1$ has no solutions.

working $\bmod 3$ we get $y$ is even, so $y=2k$ with $k\geq 1$.

From here $12^x=5^{2y}-1=(5^y+1)(5^y-1)$.

Clearly one factor must be $2\times 3^x$ and the other must be $4^{x-1}\times 2$.

Therefore we have $2\times 3^x=4^{x-1}\times 2+2$ or $2\times 3^x=4^{x-1}\times 2 -2$

The first is equivalent to $3^x=4^{x-1}+1$ and the second is equivalent to $3^x=4^{x-1}-1$.

It is easy to see neither have solutions by looking at the following table:

$$\begin{pmatrix} 1 & 1 \\ 3 & 4 \\ 9 & 16\\ 27 & 64\\ 81 & 256\\ 243 & 1024\\ \end{pmatrix} $$

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  • $\begingroup$ I do not understand how the table means there are no solutions, can you explain more? $\endgroup$ – Peter de Rivaz Oct 11 '16 at 20:46
  • $\begingroup$ @PeterdeRivaz I put a proof about $3^A - 2^B = 1$ as an answer. $\endgroup$ – Will Jagy Oct 11 '16 at 22:48

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