What Values of R Yield Multiple Solutions to 2^P-3^Q = R?

In summary, the problem requires finding all possible values of R that yield multiple solutions to the equation 2^P - 3^Q = R, where P, Q, and R are positive integers. It is a complex problem related to elliptic curves and finding rational points on the curves. It is known that R is an odd number and for a solution to have multiple solutions, it must satisfy certain conditions. Further investigations have led to the discovery that there may be only two solutions to this problem. Further research and analysis is needed to fully solve this problem.
  • #1
K Sengupta
113
0
(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.
 
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  • #2
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
~~ 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.
 
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  • #4


On Today 09:38 AM; Sane wrote:

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

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).
 
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  • #5
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
Ok,one thing is certain R is an odd number.
 
  • #7
~~ 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]
 

Related to What Values of R Yield Multiple Solutions to 2^P-3^Q = R?

1. What is an Ingenious Exponent Puzzle?

An Ingenious Exponent Puzzle is a mathematical puzzle that involves using exponent rules to solve a given equation or sequence of numbers.

2. How do you solve an Ingenious Exponent Puzzle?

To solve an Ingenious Exponent Puzzle, you need to understand the basic rules of exponents, such as the product rule, quotient rule, and power rule. Then, you can apply these rules to the given equation or sequence to simplify it and find the solution.

3. What are some tips for solving an Ingenious Exponent Puzzle?

Some tips for solving an Ingenious Exponent Puzzle include breaking the equation or sequence into smaller parts, using substitution to simplify expressions, and checking your work for errors. It also helps to have a good understanding of basic algebra and exponent rules.

4. Can an Ingenious Exponent Puzzle have multiple solutions?

Yes, an Ingenious Exponent Puzzle can have multiple solutions. This is especially true for more complex puzzles that involve multiple variables and operations. It is important to carefully check your work to ensure you have found all possible solutions.

5. Why are Ingenious Exponent Puzzles important?

Ingenious Exponent Puzzles are important because they help develop critical thinking and problem-solving skills, especially in the field of mathematics. They also help reinforce the rules of exponents and their applications in real-life situations.

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