How many pairs of non-negative integers ( m , n ) satisfy the equation below?
2 3 n = 3 2 m − 1
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Damn, I forgot the case with (0,0)
I never thought of using modulo 4 to prove that there are no other solutions. Nice approach, thanks!
Good observation that the more general case 2 x = 3 y − 1 already has a restricted set of solutions, which we can then solve for x = 3 n , y = 2 m . Great setup of the problem :)
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Thank you.
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It is possible for n , m to be negative, and thus x = 3 n need not be an integer.
Can you either make n , m a non-negative integer or add an explanation for why no negative solutions exist?
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@Calvin Lin – If we made either n or m neagtive, then the power will become a decimal, then how an integer solution will exist. Am I right?
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@Hana Wehbi – No, you're missing the point. It's possible that decimal = decimal − 1 .
Calvin is right, you should mention that m , n are both non-negative integers otherwise, your solution is far from complete.
This question is just a small variation of Catalan-Mihailescu theorem .
Nevertheless, nice use of modulo 4. Kudos!!
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We find all solutions of 2 x = 3 y − 1 for positive integers x and y .
If x = 1 , we obtain the solution x = 1 , y = 1 which corresponds to ( n , m ) = ( 0 , 0 ) in the original problem.
If x > 1 , consider the equation modulo 4 .
The left hand side is 0 , and the right hand side is ( − 1 ) y − 1 , so y is even. Thus we can write y = 2 z for some positive integer z , and so 2 x = ( 3 z − 1 ) ( 3 z + 1 ) .
Thus each of ( 3 z − 1 ) and ( 3 z + 1 ) is a power of 2 , but they differ by 2 , so they must equal 2 and 4 respectively.
Therefore, the only other solution is when x = 3 and y = 2 , which corresponds to ( n , m ) = ( 1 , 1 ) in the original problem.