Don't *even* try! This is always odd.
Given the equation 2 n + 1 = 3 k , find all solutions where n , k are positive integers ≤ 1 0 1 0 0 .
Enter the sum of all such n and k . If there are no solutions, enter 0 .
The answer is 7.
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1 solution
You can also explore some symmetry in Case 2:
n = 2 p + 1 is always odd, so when k = 2 q + 1 is odd, you can say 2 ⋅ 2 2 p − 2 = 3 ⋅ 3 2 p − 3 And 2 ( 2 p − 1 ) ( 2 p + 1 ) = 3 ( 3 q − 1 ) ( 3 q + 1 ) The LHS has only one factor of 2 , but the RHS has at least three, except when p = q = 0
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Very nice! Much more elegant than my solution.
You don't need n odd here. LHS is 2 ( 2 n − 1 ) which has only one factor of 2 .
Case 1 : k is even. Let k = 2 k ′ , then 2 n = 3 2 k ′ − 1 = ( 3 k ′ + 1 ) ( 3 k ′ − 1 ) . Note that the factors of the RHS have a difference of 2 . Since each of them is a power of 2 , they can only be 4 and 2 . Indeed, for k ′ = 1 we get ( 3 1 + 1 ) ( 3 1 − 1 ) = ( 4 ) ( 2 ) . Thus, the only solution with k being even is n = 3 , k = 2 .
Case 2 : k is odd. Clearly n = 1 , k = 1 works. Now, let k = 2 k ′ + 1 . For n ≥ 2 we have 4 ⋅ 2 n − 2 ≡ 0 ≡ 3 2 k ′ + 1 − 1 ≡ ( − 1 ) 2 k ′ + 1 − 1 ≡ 2 ( m o d 4 ) , which is a contradiction.
Therefore, the only solutions are n = 1 , k = 1 and n = 3 , k = 2 and thus 1 + 1 + 3 + 2 = 7 .
Note : This is a special cases of Catalan's conjecture .