^{n}+ b

^{n}= c

^{n}for any integer value of n greater than two.

All variables denote positive integers.

Where p is a prime number, n = 2, 1 < a < b < p,

(a

^{n}+ b

^{n}) can be evenly divided by p

^{n-1}.

Examples:

(2

^{2}+ 4

^{2}) / 5

^{1}= 4

(3

^{2}+ 4

^{2}) / 5

^{1}= 5

(2

^{2}+ 10

^{2}) / 13

^{1}= 8

Therefore, since (a

^{2}+ b

^{2}) can be evenly divided by p

^{1}, a

^{2}+ b

^{2}= c

^{2}has solutions.

Where p is a prime number, n > 2, 1 < a < b < p,

(a

^{n}+ b

^{n}) is not evenly divisible by p

^{n-1}. Therefore, it is reasonable to assume a

^{n}+ b

^{n}= c

^{n}cannot be satisfied when n > 2.

Adding terms to the left of the equation:

Let the variables in equation a

^{n}+ b

^{n}= c

^{n}be renamed to a

_{1}

^{n}+ a

_{2}

^{n}= b

^{n}.

Adding two additional terms to the left of the equation would be:

a

_{2}

^{n}+ a

_{2}

^{n}+ a

_{3}

^{n}+ a

_{4}

^{n}= b

^{n}.

Conjecture:

Where p is prime, 1 < a

_{1}< a

_{2}+...+ < a

_{m}< p, if there exists

{n, a

_{1}, a

_{2}, ..., a

_{m}, p, x} that can satisfy (a

_{1}

^{n}+ a

_{2}

^{n}+...+ a

_{m}

^{n}) / p

^{n-1}= x, then there exists, with the same number of terms (a) and exponent (n), {n, a

_{1}, a

_{2}, ... , a

_{m}, b} that can satisfy a

_{1}

^{n}+ a

_{2}

^{n}+...+ a

_{m}

^{n}= b

^{n}.

An example of a {n, a

_{1}, a

_{2}, a

_{3}, a

_{4}, p, x} that satisfies

(a

_{1}

^{5}+ a

_{2}

^{5}+ a

_{3}

^{5}+ a

_{4}

^{5}) / p

^{4}= x, is {5, 31, 40, 43, 51, 61, 45}.

(31

^{5}+ 40

^{5}+ 43

^{5}+ 51

^{5}) / 61

^{4}= 45

Therefore, there exists {n, a

_{1}, a

_{2}, a

_{3}, a

_{4}, b} that can satisfy

a

_{1}

^{5}+ a

_{2}

^{5}+ a

_{3}

^{5}+ a

_{4}

^{5}= b

^{5}.

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Last update: August 9, 2010

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