Let n be a positive integer greater than 1. Find all values of n such that the equation x^n + y^n = (x + y)^n has infinitely many positive integer solutions (x, y) with x ≠ y.
Let n be a positive integer greater than 1. Find all values of n such that the equation x^n + y^n = (x + y)^n has infinitely many positive integer solutions (x, y) with x ≠ y.
Answer: The equation x^n + y^n = (x + y)^n represents Fermat's Last Theorem for the case when the exponents are equal. According to Fermat's Last Theorem, there are no positive integer solutions (x, y, z) for the equation x^n + y^n = z^n, where n is a positive integer greater than 2.
However, in this case, we are looking for solutions where x ≠ y, so the equation x^n + y^n = (x + y)^n may have infinitely many positive integer solutions for certain values of n.
To find the values of n for which the equation has infinitely many positive integer solutions (x, y) with x ≠ y, we need to consider the equation x^n + y^n = (x + y)^n and see if there are any such values.
Let's analyze the equation for different values of n:
When n = 2:
- In this case, the equation becomes x^2 + y^2 = (x + y)^2, which simplifies to x^2 + y^2 = x^2 + 2xy + y^2.
- Canceling out the common terms, we get 2xy = 0. This implies xy = 0, which means either x = 0 or y = 0.
- Since we are looking for positive integer solutions where x ≠ y, there are no such solutions when n = 2.
When n = 3:
- The equation x^3 + y^3 = (x + y)^3 simplifies to x^3 + y^3 = x^3 + 3x^2y + 3xy^2 + y^3.
- Canceling out the common terms, we get 3x^2y + 3xy^2 = 0. Dividing both sides by 3, we have xy(x + y) = 0.
- This equation is satisfied when x = 0, y = 0, or x = -y.
- Since we are looking for positive integer solutions where x ≠ y, the only valid solution is x = -y.
Therefore, when n = 3, the equation has infinitely many positive integer solutions (x, y) with x ≠ y.
When n > 3:
- According to Fermat's Last Theorem, there are no positive integer solutions (x, y, z) for the equation x^n + y^n = z^n, where n is a positive integer greater than 2.
- Since we are considering the equation x^n + y^n = (x + y)^n, it falls under the case where n > 3.
- Therefore, there are no values of n greater than 3 for which the equation has infinitely many positive integer solutions (x, y) with x ≠ y.
In summary, the equation x^n + y^n = (x + y)^n has infinitely many positive integer solutions (x, y) with x ≠ y when n = 3. For all other values of n greater than 1, there are no such solutions.
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