Alright so I’ve been a bit busy lately with school and all the homework and such, leaving me with much less time to study and write about mathematics than during the summer. But after a three week break I’m now blogging again. Yay. Anyways.
Catalan’s conjecture is an interesting conjecture in number theory. In this equation
The only solution with is . That is to say, there are no two other powers with a difference of 1.
This was conjectured in 1844, but only recently proven in 2002 by Mihăilescu, so it’s known also as Mihăilescu’s theorem. Mihăilescu’s proof of the theorem uses some very sophisticated techniques, and is beyond the scope of this blog post.
It turns out that Catalan’s conjecture is useful for solving some interesting number theory problems. We can sometimes reduce a problem down to a very specific instance of Catalan’s conjecture, then apply the proof, essentially overkilling the problem. Here are some examples.
Find all triples of positive integers where
We look at powers modulo 4. It can be seen that when is positive, thus and .
Next note that so . As is always odd and is always even, must be odd hence must be odd. Taking the modulo 4, (if or ). Thus , implying that is even.
Then can be written as for some . The equation is then , and with differences of squares,
The product of and is a power of 2, so both must themselves be powers of 2. Then there exists and where and such that
Adding the first two yields
Putting causes a parity error. If then the RHS is divisible by 4 while the LHS clearly isn’t. This makes , giving
This is a case of Catalan’s conjecture. The only solutions to are and . Now we can just substitute the solutions for to get solutions for . They are and .
Here is a very similar problem: find all integer solutions to
The solution is left to the reader.
Hint: rewrite as a difference of squares and subtract
Find all triples of non-negative integers such that
Obviously cannot be 0. We first look at . Then, , and any triple satisfies the equation.
Next we consider when . Then , or as a difference of squares, . The solution then is and , or .
Now we try . Suppose that is even. Then in the equation , and so , thus is even. Then for some , and it is clear that , which is a contridiction.
So must be odd. The RHS, , can be written as , or . This gives us
We claim that for odd , . Let be the GCD of and so that and . Then (the difference). As is odd, and divides it, must be odd so . But so and , thus .
Hence for some integers and where ,
Alternatively, and .
From Catalan’s conjecture, the equation with has no solutions other than . But we have the simultaneous equations and . This is clearly impossible.
Thus the only solutions are and .