To solve this problem using the given polynomial identity, we start with the equation \(2xy = 96\). From this, we can find \(xy\):
\[ xy = \frac{96}{2} = 48 \]
According to the polynomial identity \((x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2\), we will substitute \(2xy\) into it. So, we need to calculate \( (2xy)^2 = 96^2 = 9216 \).
Now, we can denote \(x^2 + y^2\) and \(x^2 - y^2\) in terms of \(m\) and \(n\):
\[ (x^2 + y^2)^2 = m^2 + 9216 \]
With \(xy = 48\) and \(x\) and \(y\) being the lengths of the two sides of the triangle, we can rewrite \(x\) and \(y\) in terms of \(s\) and \(d\) (where \(s\) is the shorter side and \(d\) is the longer side):
\[ x = s, \quad y = d \]
From \(s \cdot d = 48\) and \(x = s\) and \(y = d\), we can find some possible pairs \((s,d)\) such that \(s\) and \(d\) are coprime.
Let’s find pairs \((s, d)\) such that \(s \cdot d = 48\).
The factor pairs of 48 are:
- (1, 48)
- (2, 24)
- (3, 16)
- (4, 12)
- (6, 8)
Now, we can find the lengths corresponding to those pairs (keeping in mind that we need integer lengths).
The values of \(s\) and \(d\) must be different lengths, and we also need \(s + d\) to comply with the constraints of a right triangle given \(a^2 + b^2 = c^2\).
Given the side lengths suggested (24 inches, 20 inches, 80 inches, 100 inches), we check which one meets our conditions.
Checking Possible Combinations
- If we take 24 inches and 48/24 = 2 => (2, 24)
- If we take 20 inches, then \(d = 48/20\) => not an integer.
- If we take 80 inches, then \(d = 48/80\) => not an integer.
- If we take 100 inches, then \(d = 48/100\) => not an integer.
Conclusion
The only pairs of integer side lengths we could achieve here, keeping the existing constraints, point to 24 inches satisfying our conditions.
So, the length of the two undetermined sides that would work is indeed:
24 inches.
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