Powers separated by 5

[BMAD]

Find pairs of positive integers where 2^a = 3^b + 5. Prove that you have found them all.

[bonanova]

2, 2

[BMAD]

Bonanova,

Do you mean 2^2 = 3^2 + 5 ?

[TimeSpaceLightForce]

(a,b): (3,1) (5,3) ..

[BMAD]

This is a good start.

(a,b): (3,1) (5,3) ..

[bonanova]

Bonanova,

Do you mean 2^2 = 3^2 + 5 ?

I plead drowsiness. I read the puzzle incorrectly. Minus 5 obviously.

[BMAD]

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.

[TimeSpaceLightForce]

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..

[BMAD]

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.

[BMAD]

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.

[mastermath20]

2^3=3^1+5

2^5=3^3+5

[BMAD]

2^3=3^1+5

2^5=3^3+5

Are these the only cases that work?

[genius124]

3,1 as 8=3+5 and 5,3 as32=27+5 so the answer is 3,1 and 5,3

[BMAD]

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?

[TimeSpaceLightForce]

maybe a complex no