Expert: Sherry Wallin - 12/1/2010

Question
QUESTION: Expert Sherry Wallin:
Please confirm the following statements are true in Real numbers when x,y,z are not zero:

A.) Any square can be subdivided into an infinite number of two squares such that x^2 + y^2 = z^2.

B.) Given: n>2. If x^2 + y^2 = z^2, and we multiply both sides of the equation thus, z^n-2(x^2)+ z^n-2(y^2) = z^n-2(z^2), then z^n-2(x^2) + z^n-2(y^2) = z^n.
For example, when n=3 and 2^2 + 2.236067...^2 = 3^2.
Then:         3^1(2^2) + 3^1(2.236067...^2) = 3^1(3^2)
And:          12 + 15 = 27
And:          27 = 9^3

C.) Whenever x^2 + y^2 = z^2, and n>2, then x^n + y^n =/= z^n.
For example, when n=3 and 2^2 + 2.236067...^2 = 3^2.
Then:     2^1(2^2) + 2.23...^1(2.23...^2) =/= 3^2
And:          8 + 11.180339... =/= 27
And:          19.180339... =/= 9^3


Lee




ANSWER: Lee~
Part a is true. Consider Pythagoreans Theorem. Given any hypotenuse, c, square it, c^2, there has to be two sides a and b to a right triangle that will be such that when you square each side a^2 and b^2 and you take their sum that that sum is equal to the hypotenuse squared. a^ + b^2 = c^2. Another way to look at this is to select any length for c, square it, c^2, and choose some other length for a, just so that a < c, and square it, a^2, then all you need to do is solve for b^2. This can be done only because you are in the real numbers and you are not restricting your domain to the natural numbers or rational numbers p/q with p and q natural numbers.

Part b is true. All you are doing is multiplying each term by z^(n-2). Please note the use of parentheses, these are necessary and you didn't use them. When you said z^n-2(x^2) you were really saying:
z^n - 2x^2, so be careful. Notice that z^(n-2)*z^2 is z^(n-2+2) = z^n

Part c is true also because you are not multiplying each term by the same amount. You multiplied the 1st term by x^(n-2) and the 2nd term by y^(n-2) and finally you multiplied z by z^(n-2). This could only be true if x = y = z. To show part c it is sufficient to find one example that does not work and you were provided with one.

Math Prof


---------- FOLLOW-UP ----------

QUESTION: Expert Sherry Wallin:
Given that the foregoing is true, can it be said that for the singular case were the 'sum of two squares', x^2 + y^2 = z^2, the following is also true?:

Assume: x^n + y^n = z^n, n>2, positive Real numbers, x,y,z =/= 0.
Then: x^n/z^(n-2) + y^n/z^(n-2)= z^n/z^(n-2), which means that no matter what values x^n/z^(n-2) and y^n/z^(n-2) are, they are one of the infinite number of the 'sum of two squares' equal to z^2. And, since all 'sum of two squares' such that x^2 + y^2 = z^2, when each term is raised to a power n>2, have been shown to produce an inequality, we can say that the assumption x^n + y^n = z^n is false, and x^n + y^n =/= z^n.

Lee  

Answer
Hi Lee~

I'm not sure what your question is. Did you want me to confirm or deny your proof? I will make comments on how I would prove the statement:

Assume x^n + y^n = z^n then x^n/z^(n-2) + y^n/z^(n-2) = z^n/z^(n-2).

Proof:
Let x^n + y^n = z^n  for positive real numbers such that x,y,z ~= 0
x^n/z^n + y^n/z^n = z^n/z^n =>
x^n/z^n + y^n/z^n = 1 =>
(x/z)^2 + (y/z)^2 = 1 =>
this is the equation of a circle centered at the origin with radius 1, therefore it is true.

Math Prof