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

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14 answers to this question

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Bonanova,

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

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

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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.

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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.

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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.

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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?

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