Problem 9: Special Pythagorean Triplet

Problem Statement

A Pythagorean triplet is a set of three natural numbers, $a < b < c$, for which, $a^2 + b^2 = c^2$.

For example, $3^2 + 4^2 = 9 + 16 = 25 = 5^2$.

There exists exactly one Pythagorean triplet for which $a + b + c = 1000$.

Find the product $abc$.

Approach

This problem requires finding three natural numbers that satisfy two conditions:

1. They form a Pythagorean triplet: $a^2 + b^2 = c^2$
2. Their sum equals 1000: $a + b + c = 1000$

Brute Force Approach

We can iterate through possible values of $a$ and $b$ and calculate $c$ from the sum constraint:

• For each $a$ from 1 to 1000
• For each $b$ from $a+1$ to 1000
• Calculate $c = 1000 - a - b$
• Check if $a^2 + b^2 = c^2$

Optimized Approach

We can reduce the search space by noting that:

• Since $a < b < c$ and $a + b + c = 1000$, we know $a < 333$ (otherwise $a + a + a > 1000$)
• For each $a$, we can solve for $b$ using the constraint equations

Given:

• $a + b + c = 1000$ → $c = 1000 - a - b$
• $a^2 + b^2 = c^2$ → $a^2 + b^2 = (1000 - a - b)^2$

Complexity

• Time Complexity: O(n²) for brute force approach, where n is the sum constraint
• Space Complexity: O(1)

Solution

View the complete solution on GitHub.

Related Concepts

• Pythagorean theorem
• Number theory
• Algebraic manipulation
• Brute force search optimization