Vitanuova for 2003 June 2 (entry 6)

A toy theorem

Fermat was the most comical character in Partition, and his presence and incessant discussion of his Last Theorem reminded me that Hofstadter made a joke where you invert Fermat's Last Theorem to say that

n^a = n^b + n^c

has no integer solutions for n > 2. (It appears on pp. 334-5 of Gödel, Escher, Bach, where it is attributed to "Lierre de Fourmi" and captioned "Johant Sebastiant [Fermant]'s Well-Tested Conjecture".)

This is easy to show (how remarkably much easier than the trivially typographically different "aⁿ = bⁿ + cⁿ"!).

Suppose that n > 2 and n^a = n^b + n^c. Then, if b = c, we have n^a = 2n^b, quod fieri nequit.

So if we continue, assuming without loss of generality that b > c, we can divide by n^c and get

n^a-c = n^b-c + 1

or

n^a-c - n^b-c = 1

or

n^b-c(n^a-b-1) = 1

or consequently

n^b-c = 1

and taking the logarithm of both sides

b - c = 0

contrary to our original assumption that b > c.

I sent this to Sumana as an "in Soviet Russia" joke, but afterward I wanted to see if it generalizes so that we can say that a power of n can be decomposed into n powers of n, but not into any other number of powers of n. This is true for n=2, as above. In fact the same technique works to prove it in general.

Suppose that

n^a = n^b + n^c + n^d + n^e + ... n^z

Now (without even making an assumption about whether b, c, d, etc.,are distinct), we simply assume that a>=b, b>=c, etc. This is no loss of generality because any number of integers can be arranged into descending order. Now divide by n^z:

n^a-z = n^b-z + n^c-z + n^d-z + ... + n^y-z + 1

and again

n^a-z - n^b-z - n^c-z - ... - n^y-z = 1

Factor out n^y-z:

n^y-z(n^a-y - n^b-y - n^c-y - ... - n^x-y - 1) = 1

So we conclude that y=z, and we also have

n^a-y - n^b-y - n^c-y - ... - n^x-y = 1

and we can follow the same procedure by induction showing that x=y, w=x, v=w, u=v, t=u, etc. Thus, whenever a power of n is decomposed into any number of other powers of n, they are all equal powers. Thus, there must be exactly n of them.

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