BMAD 63 Report post Posted April 8, 2013 Find pairs of positive integers where 2^{a }= 3^{b} + 5. Prove that you have found them all. 1 1 Share this post Link to post Share on other sites

0 bonanova 78 Report post Posted April 9, 2013 2, 2 1 Share this post Link to post Share on other sites

0 BMAD 63 Report post Posted April 9, 2013 Bonanova, Do you mean 2^2 = 3^2 + 5 ? Share this post Link to post Share on other sites

0 TimeSpaceLightForce 11 Report post Posted April 9, 2013 (a,b): (3,1) (5,3) .. Share this post Link to post Share on other sites

0 BMAD 63 Report post Posted April 9, 2013 This is a good start. (a,b): (3,1) (5,3) .. Share this post Link to post Share on other sites

0 bonanova 78 Report post Posted April 9, 2013 Bonanova, Do you mean 2^2 = 3^2 + 5 ? I plead drowsiness. I read the puzzle incorrectly. Minus 5 obviously. Share this post Link to post Share on other sites

0 BMAD 63 Report post Posted May 5, 2013 Since (2,4,3,1) and (3,4,2,1) are cyclic residues (mod 5) for integer powers of 2 and 3 respectively, the combinations that give a residue zero (mod 5) for 2^a - 3^b are 1) a and b both even and congruent mod 4; 2) both odd with a=4n+3, b=4m+1; 3) both odd with a=4n+1, b=4m+3. Share this post Link to post Share on other sites

0 TimeSpaceLightForce 11 Report post Posted May 9, 2013 I cant find the those elusive evens ..using excel i formulate the 2 expressions with 1-300 exponents , add both on a column , sort ascending and check for equal consecutive.. Share this post Link to post Share on other sites

0 BMAD 63 Report post Posted May 9, 2013 Case 1: Suppose a=2u and b=2v, then (2^u - 3^v) (2^u + 3^v) = 5, and since 5 is prime it must be (2^u - 3^v) = 1 and (2^u + 3^v) = 5. The second equation implies u=v=1, which doesn't satisfy the first one, so no solution of type 1 exists. Share this post Link to post Share on other sites

0 BMAD 63 Report post Posted May 11, 2013 The solution a=3, b=1 (n=0, m=0) can be checked directly. Suppose another solution exists, with n>0, m>0. Then 2^(4n+3) - 3^(4m+1) = 2^3 - 3, i.e. 2^3 (2^4n - 1) = 3 (3^4m - 1) = 3 (3^2m - 1) (3^2m + 1) = 3 (3^m - 1) (3^m + 1) (3^2m + 1). There are three even factors at second member, but one of (3^m - 1) and (3^m + 1) must be divisible by 4, so second member is divisible by 2^4, whilst the first one is only divisible by 2^3. Therefore no other solution of this type exists. Share this post Link to post Share on other sites

0 mastermath20 0 Report post Posted May 14, 2013 2^3=3^1+5 2^5=3^3+5 1 1 Share this post Link to post Share on other sites

0 BMAD 63 Report post Posted May 14, 2013 2^3=3^1+5 2^5=3^3+5 Are these the only cases that work? Share this post Link to post Share on other sites

0 genius124 0 Report post Posted May 26, 2013 3,1 as 8=3+5 and 5,3 as32=27+5 so the answer is 3,1 and 5,3 Share this post Link to post Share on other sites

0 BMAD 63 Report post Posted May 26, 2013 3,1 as 8=3+5 and 5,3 as32=27+5 so the answer is 3,1 and 5,3 but are these the only ones that work? Share this post Link to post Share on other sites

0 TimeSpaceLightForce 11 Report post Posted June 1, 2013 maybe a complex no Share this post Link to post Share on other sites

Find pairs of positive integers where 2

^{a }= 3^{b}+ 5. Prove that you have found them all.## Share this post

## Link to post

## Share on other sites