# An Ingenious Exponent Puzzle

1. Nov 8, 2006

### K Sengupta

(A) Determine all possible values of R that yields multiple solutions to the equation:

2^P – 3^Q = R; where P, Q and R are all positive integers.

(B) Determine all possible values of R that yields multiple solutions to the equation:

2^P – 3^Q = R; where P, Q are positive integers, but R is a negative integer.

2. Nov 8, 2006

### Sane

I'm confused. Is this homework and you're asking for help? If so, what have you tried doing?

If not, this looks very interesting. Either way, I'll have to try it some time soon! If you're posting the answer, please but a "spoiler ahead" sign to notify other readers!

3. Nov 8, 2006

### Werg22

~~ Possible spoiler ~~

Any integer value for P and Q will give an integer R. Now if R is positive, we at least now that P > Q. However since there's now bound on both P and Q, R dosen't have any bound either. This means that there's infinite number of solutions for R. Same concept with negative R.

Edit: Rereading the question, I figured out why it wasen't as innocent as I firstly thought :p. This is indeed a interesting problem, I doubt you can solve it without knowledge in number theory.

Last edited: Nov 8, 2006
4. Nov 9, 2006

### K Sengupta

Re: Ingenious Exponent Puzzle

On Today 09:38 AM; Sane wrote:

I apologise for any instance of ambiguity arising out of the choice of the texts comprising the original post.

The problem under reference, at the outset requires one to determine a given R which will yield more than one pair (P,Q). Then, the problem requires one to find all possible values of R with this property.

For, example in (A); 2^3 - 3 = 2^5-3^3 = 5; so R =5 generates more than one pair (P,Q) satisfying conditions of the problem. Problem (A) then requires one to find all possible values R, each of which generates more than one pair (P,Q).

Accordingly, I am looking for an analytic method to determine all possible value of R for each of (A) and (B).

Last edited: Nov 9, 2006
5. Nov 9, 2006

### AlbertEinstein

I think that this is a problem related to elliptic curves. there are problems in which u have to find rational points on the curves. However i do not know the technique, :(

6. Nov 9, 2006

### AlbertEinstein

Ok,one thing is certain R is an odd number.

7. Nov 12, 2006

### NanoTec

~~ spoiler ahead ~~
R=13= 2^8-3^5 = 2^4-3^1 (Which I wish Sengupta had posted as an example too…)

For 2^P-3^Q > 0 there exist: (I.) for a value of P a maximum value of Q, and/or for a value of Q a minimum value of P. [as already stated by Werg22] P>Q/log3(2)

For a value of R with multiple solutions, one of those solutions must satisfy (I.)
The other solutions will have lower values of P and Q.

Something of a proof: R(P,Q) for the values defined in (I.) increases exponentially; from a point R(P,Q) (a.) an increase in P will increase R, (b.) Q can not be increased alone, (c.) an increase in Q necessitates an increase in P in proportion to (I.) which would increase R ~ For solutions satisfying (I.) to have another solution for R it must have lower P, and Q.

R2(P,Q)-R1(P,Q) = {6,18,54,162,486,1458,4374,…} for ‘higher’ values of P and Q (ie the differences in R are quantized 6*3^N)

Doing the math in base2… R=1P0s – 11^Q since the Qterm is odd it must have one fewer binary digits than P to satisfy (I.) therefore for a solution satisfying (I.) R is 1 plus the inverse of the bits of 3^Q. The second P is the number of digits in R, and 3^secondQ is the bits after the first zero in 3^Q.

It may be that there are only two solutions to this problem. I do not believe I have proved many if any of the statements above yet.

Reference:
[1/Log3(2)=~1.585]
[3^1=11]
[3^2=1001]
[3^3=11011]
[3^4=1010001]
[3^5=11110011]