Tetration Equation

How many pairs of non-negative integers ( m , n ) (m,n) satisfy the equation below?

2 3 n = 3 2 m 1 \large 2^{3^{n}} = 3^{2^{m}} -1

0 1 2 3 4

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1 solution

Hana Wehbi
Dec 6, 2016

We find all solutions of 2 x = 3 y 1 2^x= 3^y- 1 for positive integers x x and y y .

If x = 1 x = 1 , we obtain the solution x = 1 , y = 1 x = 1, y = 1 which corresponds to ( n , m ) = ( 0 , 0 ) (n, m) = (0, 0) in the original problem.

If x > 1 x > 1 , consider the equation modulo 4 4 .

The left hand side is 0 0 , and the right hand side is ( 1 ) y 1 (-1)^y - 1 , so y y is even. Thus we can write y = 2 z y = 2z for some positive integer z z , and so 2 x = ( 3 z 1 ) ( 3 z + 1 ) 2^x= (3^z-1)(3^z+1) .

Thus each of ( 3 z 1 3^z-1 ) and ( 3 z + 1 (3^z+ 1 ) is a power of 2 2 , but they differ by 2 2 , so they must equal 2 2 and 4 4 respectively.

Therefore, the only other solution is when x = 3 x = 3 and y = 2 y = 2 , which corresponds to ( n , m ) = ( 1 , 1 ) (n, m) = (1, 1) in the original problem.

Damn, I forgot the case with (0,0)

Konstantin Zeis - 4 years, 5 months ago

I never thought of using modulo 4 to prove that there are no other solutions. Nice approach, thanks!

Christopher Boo - 4 years, 6 months ago

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Thank you.

Hana Wehbi - 4 years, 6 months ago

Good observation that the more general case 2 x = 3 y 1 2^x = 3^y - 1 already has a restricted set of solutions, which we can then solve for x = 3 n , y = 2 m x=3^n, y = 2^m . Great setup of the problem :)

Calvin Lin Staff - 4 years, 6 months ago

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Thank you.

Hana Wehbi - 4 years, 6 months ago

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It is possible for n , m n, m to be negative, and thus x = 3 n x = 3^n need not be an integer.

Can you either make n , m n,m a non-negative integer or add an explanation for why no negative solutions exist?

Calvin Lin Staff - 4 years, 6 months ago

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@Calvin Lin If we made either n n or m m neagtive, then the power will become a decimal, then how an integer solution will exist. Am I right?

Hana Wehbi - 4 years, 6 months ago

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@Hana Wehbi No, you're missing the point. It's possible that decimal = decimal 1 \text{decimal } = \text{ decimal } - 1 .

Calvin is right, you should mention that m , n m,n are both non-negative integers otherwise, your solution is far from complete.

Pi Han Goh - 4 years, 6 months ago

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@Pi Han Goh Ok, I will. Thank you.

Hana Wehbi - 4 years, 6 months ago

This question is just a small variation of Catalan-Mihailescu theorem .

Nevertheless, nice use of modulo 4. Kudos!!

Pi Han Goh - 4 years, 6 months ago

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