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^xx, and so on. Note that "hyper powers" are also called "double powers" and may be written as x^^n.

Did you mean "no positive integer" solution? If we choose b=0, a=c is a solution for all n.  I am not sure about whether negative a, b, c might be a solution - then the equation becomes complex-valued for n>2.

Here is my proof for a, b, c>0:
The idea is that for a given n the hyper powers are "too far apart" for two smaller ones to add up to a larger one.

As a, b>0, c^^n has to be > a^^n, b^^n, so c > a, b.  So all we have to prove is that

(a+1)^^n > 2 a^^n

(As c must be at least a+1, and b can be assumed to be <=a WOLOG, this proves the hypothesis.)

Here goes:

(a+1)^^n - a^^n
= (a+1)^((a+1)^^(n-1)) - 2 a^(a^^(n-1))
= a^((a+1)^^(n-1)) + ((a+1)^^(n-1)) *
        a^((a+1)^^(n-1)-1) + ... - 2 a^(a^^(n-1))
>= a^((a+1)^^(n-1)) + ((a+1)^^(n-1)-2) *
        a^((a+1)^^(n-1)-1) + ...
> 0

"..." indicates the omitted terms from the binomial formula for the relevant power, all of which is >0.  The last but one step is true because for n>=2 the exponent of the second term is larger than or equal to the exponent of the last:

(a+1)^^(n-1)-1 >= a^^(n-1)
or equivalently:
(a+1)^^(n-1) > a^^(n-1)
or:
(a+1)^((a+1)^^(n-2)) > a^(a^^(n-2))

For n=2, the second formulation is obviously true.  For n>2, the third formulation follows by induction, because both the base and the exponent on the left-hand side are larger than on the right.


(q.e.d.)^^n

Posted by vswitchs on 2006-09-01 13:53:54