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Pythagorean Triples - An Application of Complex Numbers

Solving \(a^2+b^2=c^2\) within the set of natural numbers using complex numbers.

Author

Affiliation

Fei Ye

QCC-CUNY

1 Discovering Pythagorean Triples

Recently, I learned a fascinating trick to find Pythagorean triples, i.e., \((a, b, c)\) such that \(a^2 + b^2 = c^2\), from a YouTube video by Professor Richard E. Borcherds. Here’s how it works:

Consider the equation:

\[a + b\mathbf{i} = (x + y\mathbf{i})^2\]

Expanding this, we get:

\[a + b\mathbf{i} = (x^2 - y^2) + 2xy\mathbf{i}\]

Taking the modulus squared of both sides:

\[ \begin{aligned} |a + b\mathbf{i}|^2 &= |(x + y\mathbf{i})^2|^2 \\ a^2 + b^2 &= (|x + y\mathbf{i}|^2)^2 \\ a^2 + b^2 &= (x^2 + y^2)^2 \end{aligned} \]

By setting \(a = x^2 - y^2\), \(b = 2xy\), and \(c = x^2 + y^2\), we ensure that \(a^2 + b^2 = c^2\).

In other words, we can find Pythagorean triples by choosing integers \(x\) and \(y\) such that \(a = x^2 - y^2\), \(b = 2xy\), and \(c = x^2 + y^2\).

1.1 Examples of Pythagorean Triples

Example 1  

• For \(x = 2\) and \(y = 1\), we get: \[a = 2^2 - 1^2 = 3,\] \[b = 2 \cdot 2 \cdot 1 = 4,\] \[c = 2^2 + 1^2 = 5.\]

• For \(x = 3\) and \(y = 1\), we get: \[a = 3^2 - 1^2 = 8,\] \[b = 2 \cdot 3 \cdot 1 = 6,\] \[c = 3^2 + 1^2 = 10.\]

• For \(x = 3\) and \(y = 2\), we get: \[a = 3^2 - 2^2 = 5,\] \[b = 2 \cdot 3 \cdot 2 = 12,\] \[c = 3^2 + 2^2 = 13.\]