perplexus.info :: Numbers : Hyper version of Fermat's Theorem

Hyper version of Fermat's Theorem (Posted on 2006-09-01)

Fermat's Theorem states that aⁿ+bⁿ=cⁿ has no positive integer solutions for n>2. Since this puzzle has been already cracked, let's take the next level:

Prove that ⁿa+ⁿb=ⁿc has no integer solutions for n>1.

The "hyper power" ⁿx is defined recursively by ¹x:=x and ⁿ⁺¹x:=x^(ⁿx), i.e. ²x=x^x, ³x=x^{x^x}, and so on. Note that "hyper powers" are also called "double powers" and may be written as x^^n.

Submitted by JLo

Rating: 4.0000 (2 votes)

Solution: (Hide)

Despite its scary looks, the Hyper-Fermat-Theorem becomes quite obvious as soon as one starts inserting numbers into the equation. As vswitchs observed, c would have to be larger than a and b, but then the right hand side becomes much larger than the left hand side. Just try n=2 (for n > 2 it will be even worse) and you get a^a+b^b=c^c. When trying a=b=2 and c=3 the left hand side gives a mere 8, whereas the right hand side gives a hefty 27. And the larger the numbers get, the larger the difference between left and right hand side becomes.
For the sake of completeness nevertheless a "proper" proof (See also vswitchs post for an alternative). For starters, assume that 0<a<=b<c and note that
(*) ^mx < ^m(x+1) for x, m >= 1
ⁿa+ⁿb
<=
ⁿb+ⁿb
=
2*ⁿb
=
2*b^{(^n-1b)}
<
2*(b+1)^{(^n-1b)}
<=
(b+1)*(b+1)^{(^n-1b)}
<= (*)
(b+1)*(b+1)^{[^n-1(b+1)-1]}
=
(b+1)^{(^n-1(b+1))}
=
ⁿ(b+1)
<=
ⁿc
Reading from top to bottom gives
ⁿa+ⁿb < ⁿc

Comments: ( You must be logged in to post comments.)

	Subject	Author	Date
	Solution - I think	vswitchs	2006-09-01 13:53:54

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