Answer:
a. (x² + y²)² = (x² - y²)² + (2xy)²
b. The conjecture works in all cases.
c. Sides of 119, 120, and 169
Explanation:
a. Equation that models this conjecture
x² + y² = sum of squares
x² - y² = difference of square
2xy = twice the product of the integers
If these are the sides of a right triangle then
(x² + y²)² = (x² - y²)² + (2xy)²
b. Test the conjecture
(i) Try x = 2, y = 1
(2² + 1²)² = (2² - 1²)² + (2×2×1)²
5² = 3² + 4²
25 = 9 + 16
(ii) Try x = 3, y = 1
(3² + 1²)² = (3² - 1²)² + (2×3×1)²
10² = 8² + 6²
100 = 64 + 36
(iii) Try x = 3, y = 2
(3² + 2²)² = (3² - 2²)² + (2×3×2)²
13² = 5² + 12²
169 = 25 + 144
The conjecture appears to work in all cases.
c. A possible triangle
We must have one side greater than 100. That means,
x² > 100 or x >1 0.
Let x = 12
One side = 12² + y²
The second side = 12² - y²
The third side must have 2xy > 100
24y > 100
y > 4.2
Try y = 5
(12² + 5²)² = (12² - 5²)² + (2 × 12 × 5)²
169² = 119² + 120²
So, one right triangle could have sides of 119, 120, and 169.
Furthermore, these sides have no common factors.
Check:
169² = 119² + 120²
28561 = 14161 + 14400
28561 = 28561