When is $a^b-1$ a perfect square?

Question

kainui · Answer

$a^b-1$

ganeshie8 · Answer

Difference between any two perfect squares cannot be less than $2n+1$ 
$$(n+1)^2 - n^2$$

ganeshie8 · Answer

$$a^b -1 = k^2 \implies a^b-k^2=1$$

The earlier observation restricts $b$ to $1$ for nontrivial solutions

kainui · Answer

Oh interesting. But I think that's not completely true, it looks like it restricts $b$ to be an odd number and $a$ to not be a perfect square so there is still some possibilities.

ganeshie8 · Answer

If a^2 - k^2 cannot be less than 2n+1,
then surely a^3-k^2 cannot be less than 2n+1

ganeshie8 · Answer

there must be a more rigorous way to state it...

kainui · Answer

I don't know I kind of see it but the way it's derived is consecutive squares but any square and cube don't seem like they have to be separated in that way hmm.

ganeshie8 · Answer

$a^2 - k^2 \gt 1 \implies a^b - k^2\gt 1$ for all $b\ge 2$ because $a^b\gt a^2$ for all $b\gt 2$

kainui · Answer

Ok I think I found a counter example to your claim:

If we choose n=11 then we have  $2n+1 = 23$

But look the difference here is much less than that:
$$5^3-11^2 = 125-121 = 4$$

ganeshie8 · Answer

Ahh I see the error

kainui · Answer

Here's a little something, 

$$a^b-1 = (a-1)(1+a+a^2+\cdots+a^{b-1})$$
so in order to be a perfect square either $a-1$ is a perfect square or $a-1$ divides $1+a+\cdots+a^{b-1}$.

kainui · Answer

Also I accidentally found a nice way of solving that weird factoring problem from the other day, when we write this in base $n$ does it work in all bases?

$$111111=11*10101=111*1001$$
my proof of this is:
$$\frac{n^{xy}-1}{n-1} = \frac{(n^x)^{y}-1}{n^x-1}\frac{n^{x}-1}{n-1} =\frac{(n^y)^{x}-1}{n^y-1}\frac{n^{y}-1}{n-1}$$

Ok I left out some of the explanation but just a side note there, that kinda makes it clear. Also I realize x=2 and y=3 are things I should have plugged in, to make it match the equations before the proof.

ganeshie8 · Answer

Wow, nice... factoring it in base n and geometric series both look pretty!

ganeshie8 · Answer

$$a^b-1 = (a-1)(1+a+a^2+\cdots+a^{b-1})$$

From above, how to conclude $(a-1)$  perfect square or $a-1$ divides $(1+a+a^2+\cdots+a^{b-1})$ ? If $a-1$ is not a perfect square and not a square free, above conclusion is false ?

bobo-i-bo · Answer

Oh, it's easy to prove that $(a-1)$ can never factorise $(1+a+a^2+...+a^{b-1})$ by using the remainder theorem.

bobo-i-bo · Answer

Also, it cannot be the case that $a-1=1+a+a^2+...+a^{b-1}$.

kainui · Answer

Well ok I'm starting to realize there might be some issues with the reasoning I was thinking of for this:
$$a^b-1 = (a-1)(1+a+a^2+\cdots+a^{b-1})$$

Basically for that to be a perfect square it has a handful of cases to consider and it's kind of ugly. I'll go ahead and rename the parts, the 'subtraction part' $s$ and the 'geometric series part' $g$.

$$s=(a-1)$$$$g=(1+a+a^2+\cdots+a^{b-1})$$

the 3 cases are as follows:

$s$ is a perfect square and $g$ is a perfect square. No divisibility considerations going on here.

$\gcd(s,g)=s$ so that $\frac{g}{s}$ is a perfect square. Apparently @Bobo-i-bo is saying we can prove this case false easily so I'll have to see how that's done that's awesome to know it's easy.

The third and final case, which is mixed up version between the previous two cases:
$s=x^2y$ and $g=yz^2$. So now $s$ does't divide $g$ BUT we have $s*g$ as a perfect square.

bobo-i-bo · Answer

Therefore does it follow that $(1+a+a^2+...+a^{b-1})$ must divide $(a-1)$ which implies that $b=1$

bobo-i-bo · Answer

?

kainui · Answer

How do you use the remainder theorem haha I have heard of it and I think I might have even learned it but I forget how it works. Something to do with polynomial remainder and some possibly questionable step in the derivation of dividing by zero or something not sure that's all I got.

bobo-i-bo · Answer

Ooops, I mean, if $x-c$ divides the polynomial $f(x)$, then $f(c)=0$

kainui · Answer

Ah ok I totally get this it's coming back to me now. Since we always have $f(1)=1+1^2+\cdots +1^{b-1} \ne 0$ then there's always a remainder so it doesn't cleanly divide coool thanks!

kainui · Answer

Actually that's pretty exact, the remainder of $a^b-1$ divided by $a-1$ is $b$. That might be useful in trying to prove that final case there where possibly $a-1=x^2y$ and $1+a+\cdots+a^{b-1}=yz^2$.

kainui · Answer

Whoops I mean the remainder of $1+a+a^2+\cdots+a^{b-1}$ divided by $a-1$ is $b$.

bobo-i-bo · Answer

I think that you then make an argument with Euclid's Algorithm adapted for polynomials to say that, in the third case $y$ is a constant but perhaps the argument can do even better and suggest that $y=1$.