ind all ordered pairs (a,b) of positive integers such that |3^a − 2^b| = 1.

Question

anonymous · Answer

i can eyeball some pairs, but thats about it lol >.<

myininaya · Answer

(1,1) is one I see

anonymous · Answer

(1,2)

anonymous · Answer

(2,3).....i dont see anymore nearby....

anonymous · Answer

(1,1), (1,2), (2,3)

myininaya · Answer

i don't see any more
the numbers are getting too big

anonymous · Answer

if there are anymore pairs, a must be even:$$3^a-2^b=1\iff 2^b=3^a-1=(3-1)(3^{a-1}+3^{a-2}+\cdots+3+1)$$$$2^b=2\cdot(3^{a-1}+3^{a-2}+\cdots+3+1)\iff 2^{b-1}=3^{a-1}+3^{a-2}+\cdots+3+1$$

anonymous · Answer

i dont think there are anymore pairs...

anonymous · Answer

i think i can show that a would have to be even, and a would have to be odd., which is impossible.

anonymous · Answer

We will use mathematical induction on n to prove the slightly stronger result that, for each n greater than or equal to 3,  n! can be represented as the sum of n distinct divisors of itself, with 1 as one of the divisors..

We have the base case, for n = 3:  3! = 1 + 2 + 3, with 1 as one of the divisors.

For the induction step, we assume the inductive hypothesis, that we have a representation for n = k, with 1 as one of the divisors.
That is, k! = 1 + d2 + d3 + ... + dk, with 1 < d2 < ... < dk.
Then (k + 1)!	= (k + 1) + (k + 1)d2 + ... + (k + 1)dk
 	= 1 + k + (k + 1)d2 + ... + (k + 1)dk

That is, the sum of k + 1 distinct divisors of (k + 1)!, with 1 as one of the divisors.

It follows by induction that, for each n greater than or equal to 3,  n! can be represented as the sum of n distinct divisors of itself.

anonymous · Answer

So we have:$$3^a-2^b=1\iff 3\cdot (3^{a-1})-2\cdot (2^{b-1})=1$$using Bezout's Identity, we also have:$$3x-2y=1$$ for some x and y.  All the integer solutions to Bezouts Identity are in the form:$$x=1+2k,y=-1-3k$$ where k is an integer because:$$3(1+2k)+2(-1-3k)=3+6k-2-6k=1$$so now we have:$$3(1+2k)-2(1+3k)=1$$which tells us $$3^{a-1}=1+2k$$ for some k and some a.  Looking at this equation mod 2 gives:$$3^{a-1}\equiv 1 \mod 2$$which says a-1 must be even, but above i showed that a was even. which is a contradiction?