Suggested Proof of Fermat’s Theorem

G.R.Dixon, 1/19/2006

Author Note: This "proof" is not exhaustive, but its ideas are provided for information purposes, and for possible use for others who also enjoy trying to figure out what Fermat might have been thinking.

Let a be a positive integer (a=1,2,…) and let b and c be such that

(1a)

and

(1b)

Prove that, given integer a, there are integer solutions for b and c only if n=1,2.

Case 1. n=1,3,…

For n=1, Eq. 1b becomes

. (2)

For every integer a there are an infinite number of integer couplets (b,c) satisfying Eq. 2. And for any instance of Eq. 2 it is also true that

. (3)

Or, defining

, (4a)

, (4b)

, (4c)

Eq. 3 can be written as

. (5)

When a and b are integers, b’ cannot be an integer. For every integer a can be factored into primes:

(6)

where ai is a prime and ai is an integer. Therefore

(7)

The only way b’ can be an integer is if, for every aiai, there is an equal bibi and vice versa. Since it is given that a<b this condition cannot be satisfied. Similarly for c’. If a’ in Eq. 5 is an integer, and if b and c are integers, then c’ and b’ are necessarily non-integer.

The same holds for n=5,7,…For if b’5=a4b, then b’ must be non-integer, etc. It can thus be concluded that, for odd n, only Eq. 2 (n=1) has integer solutions for (b,c).

n=2,4,…

For n=2, Eq. 1 becomes

. (8)

As previously demonstrated, this equation has integer solutions (a,b,c) only for a=3,5,6,… And for any of these values of a, there are only a finite number of integer couplets (b,c). For example, if a=3 then there is only one couplet: (b,c)=(4,5). If a=9 then there are two: (b,c)=(12,15), (40,41).

For any instance of Eq. 8, it is also true that

. (9)

Or, defining

(10a)

, (10b)

, (10c)

Eq. 9 can be written as

. (11)

Here again it is readily shown that, when a, b and c are integers, then b’ and c’ cannot be integers.

Similar remarks apply to n=6,8,… Thus it can be concluded that, for even values of n, only Eq. 8 (n=2) has integer solutions for (b,c).

QED.