Expert: Sherry Wallin - 2/27/2011

Question
Ms. Sherry Wallin, MathProf:
Please review and critique:
Given: in Positive Real numbers;  x,y,z=/=0; x,y<z; exponents are whole numbers; all numbers congruent in "size"

To Prove: where x^2 and y^2 are any values such that x^2 + y^2 = z^2, then x^n + y^n =/= z^n, n>2
                       Proof #1:
If x^2 + y^2 = z^2, then, raising power to n>2, and substituting (x^2 + y^2) for z^2, we would get x^2(x^n-2)+ y^2(y^n-2)=/= z^n-2(x^2 + y^2), or x^n + y^n =/= z^n, n>2.
                          QED



To Prove: If x^n + y^n = z^n, n>2, is assumed true, the assumption is false, and x^n + y^n =/= z^n, n>2.
                      Proof #2:
If, by assumption, x^n + y^n = z^n, then x^n/z^n-2 + y^n/z^n-2 = z^n/z^n-2.
Let x^n/z^n-2 = x^2, and y^n/z^n-2 = y^2, where x^2 and y^2 are any values such that x^2 + y^2 = z^2.
However, by Proof #1, there is no x^2 + y^2 = z^2 such that x^n + y^n = z^n, n>2.
Therefore, the assumption x^n + y^n = z^n, n>2, is false.
                          QED

To Prove: If x^2 + y^2 < > z^2, then x^n + y^n =/= z^n, n>2.
                        Proof #3:
If x^2 + y^2 < > z^2, then x^2 + y^2 = z^2 +/-(m^2),where -(m^2) = z^2 - (x^2 + y^2); and +(m^2) = (x^2 + y^2) - z^2.
Raise exponents to n>2 and get x^n + y^n < = > z^n +/-(m^n).
If x^n + y^n = z^n +/-(m^n), then x^n + y^n =/= z^n, n>2.
If x^n + y^n < > z^n +/-(m^n), let x^n + y^n < = > z^n, n>2.
If x^n + y^n < > z^n, n>2, then x^n + y^n =/= z^n, n>2.
If we say x^n + y^n = z^n, n>2, then by Proof #2 it cannot be equal, and x^n + y^n =/= z^n, n>2.
                             QED

Therefore, in Real numbers, for any x,y,z values congruent in square,
                   x^n + y^n =/= z^n, n>2

Sincerely,

Lee

Answer
Hi Lee~

I don't follow your first proof. How can you multiply each term by a different amount? If you want to raise each of x^2, y^2, and z^2 so that each of x^2, y^2, and z^2 are to the nth power you are not going to be able to do so without introducing other terms that are nonzero. For example:

(x^2+y^2)^2 = x^4 +2x^2y^2+y^4 => here you have extra terms 2x^y^2
(x^2+y^2)^3 = x^6 + 3x^4y^2 + 3x^2y^4 + y^6 => here you have extra terms 3x^2y^2 + 3x^2y^4
Hopefully you can see a pattern here...
let n/2 = m, then (x^2+y^2)^(n/2)
= (x^2+y^2)^m = x^2m + ax^(m-2)y^2 + bx^(m-4)y^2 +...+ kx^2y^(m-2)+ y^m
where ax^(m-2)y^2 + bx^(m-4)y^2 +...+ kx^2y^(m-2) is your bunch of other stuff and a,b,k are coeffcients, thus
(x^2 + y^2)^(n/2) = (z^2)^(n/2)=> x^n + (a bunch of other stuff in x and y) + y^n = z^n and the only way x^n + y^n = z^n is if a bunch of other stuff in x and y = 0, which can't be zero since both x,y =/= 0 and you are adding all terms.

In your 2nd proof you are dividing by z^(n-2). If you have x^n/z^(n-2)= x^2 and y^n/z^(n-2) = y^2 this implies that x^(n-2) = z^(-2n) and y^(n-2) = z^(-2n) => x^(n-2) = y^(n-2) => x = y which certainly limits your cases. I don't think there is any need to continue to critique this version until you have figured out a way to fix 'this'.

Since you use proof 2 in your proof 3 this makes this method invalid as well. Likewise your proof two is suppose to follow from your proof one and I don't think your proof one is valid either for the reasons given above.

I am sorry if this has been an eye opener and not what you were hoping to hear. Good luck.

Math Prof