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To find a primitive Pythagorean triple with each side having at least three digits, we need a Pythagorean triple \((a, b, c)\) where \(a^2 + b^2 = c^2\), \(a\), \(b\), and \(c\) each have at least three digits, and \(a\), \(b\), and \(c\) have no common factors (i.e., they form a "primitive" triple).

One way to find such triples is to use the formula:
\[
a = m^2 - n^2, \quad b = 2mn, \quad c = m^2 + n^2
\]
where \(m\) and \(n\) are positive integers with \(m > n\), \(m\) and \(n\) have opposite parity (one is even, the other is odd), and they have no common factors.

After trying different values for \(m\) and \(n\) to get each side with at least three digits, we find that \(m = 20\) and \(n = 9\) provides a suitable triple:

1. \(a = 20^2 - 9^2 = 400 - 81 = 319\)
2. \(b = 2 \times 20 \times 9 = 360\)
3. \(c = 20^2 + 9^2 = 400 + 81 = 481\)

Thus, the primitive Pythagorean triple with each side having at least three digits is:
\[
(319, 360, 481)
\]

Would you like a detailed explanation of this method? Let me know if you have any questions!

### Related Questions
1. What are the properties of a primitive Pythagorean triple?
2. Can you generate other primitive Pythagorean triples with three-digit numbers?
3. How does the Euclidean algorithm confirm that \((319, 360, 481)\) is primitive?
4. Why must \(m\) and \(n\) have opposite parity in the formula for generating primitive triples?
5. What are examples of non-primitive Pythagorean triples with three-digit sides?

### Tip
To verify if a Pythagorean triple is primitive, check that the greatest common divisor (GCD) of the three numbers is 1.

Find a primitive Pythagorean triple with at least three digits for each side.

Explore how to find a primitive Pythagorean triple where each side has at least three digits. This guide demonstrates the use of number theory and the formula for generating primitive triples. Suitable for high school students.

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