When is the difference of prime powers 1?

Question

Solving the equation

3^n - 2^m = +/- 1

for integers n and m came up as an aside (that I avoided through some tedious calculations) in this question: http://answers.yahoo.com/question/index?qid=20111016201805AANtKKU

We have the solutions

3^0 - 2^1 = -1
3^1 - 2^1 = 1
3^1 - 2^2 = -1
3^2 - 2^3 = 1

which are in fact the only solutions for 0 <= n, m < 1000. I suspect there are no other solutions, but I'm not sure how to prove such a thing, or look it up, though I feel certain it's been studied (and I even dimly remember reading about something similar).

More generally, of course, one can ask what the solutions of

p^n - q^m = +/- 1

are for p and q prime. However, the left side is even unless precisely one of p, q is 2. There's always a trivial solution p^0 - 2^1 = -1, so we may suppose q=2, n >= 1 and consider

p^n - 2^m = +/- 1

For n=1, the solutions correspond precisely to the Fermat and Mersenne primes, the infinitude of which is an open question in both cases. For n > 1, the only solution I've found is 3^2 - 2^3 = 1, using 2 < p < 1000 (prime), 1 < n < 100, 0 <= m < 1000. I suspect this is the only solution.

So, my question is: (1) solve 3^n - 2^m = +/- 1; (2) give a reference discussing the second form, p^n - 2^m = +/- 1 for n > 1 [I imagine a proof is too much to ask].

2011-10-19T03:39:15Z

Thank you gôhpihán! That resolves my second question, and more (and now I'm almost sure I read the MathWorld article on the Catalan conjecture before). I'm somewhat surprised that it's been proven.

A short proof for the special case (1) still seems at least plausible to me, though maybe I'm hoping in vain.

steiner1745 · Accepted Answer

There is a simple answer to your case (1), which
was discovered by Levi Ben Gershon in the 14th century.
Let's consider 2 cases.
1). 3^n = 2^m -1.
There's no solution for m =1 and if
n = 1, m = 2. So we may assume
n >= 2 and m >= 2. (*)
Next, we have 2^m ≡ 1(mod 3), so
m must be even. So let m = 2k.
This gives 3^n = (2^k + 1)(2^k -1).
Both factors on the right are odd, so they
are relatively prime and we have
2^k - 1 = 3^t
2^k + 1 = 3^(n-t)
Thus 3^t -3^(n-t) = 2,
with 0 <= t < n-t
This yields t = 0, k = 1, m = 2, n = 1,
which contradicts (*).

2). 3^n = 2^m + 1.
Again, if m = 1, n = 1 and there is
no solution for m = 2, so we may assume
n>= 2, m >= 3.
This yields 3^n ≡ 1(mod 8), so n is even.
So let n = 2k.
Then (3^k-1)(3^k + 1) = 2^m.
The 2 factors on the left are relatively prime and we
conclude
3^k - 1 = 2^t
3^k + 1= 2^(m-t)
with 0 <= t < m-t.
So 2^(m-t) - 2^t = 2.
or 2^t(2^(m-2t) -1) = 2.
This yields t = 1, k = 1, m -1 = 2
so m = 3 and n = 2.

Anonymous · Answer

You could try proving (3^n - 2^m = +/- 1) in two cases, 3^n - 2^m = 1 and 3^n - 2^m = -1, both requiring two dimensional induction for n>2 and m>3.

eatherly · Answer

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gôhpihán · Answer

I don't know if this helps or not but your question resembles Catalan's Conjecture

http://en.wikipedia.org/wiki/Catalan%27s_conjecture

When is the difference of prime powers 1?

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