Sum of 2 Squares in 2 Distinct Ways/Examples/65

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Theorem

$65$ can be expressed as the sum of two square numbers in two distinct ways:

\(\ds 65\) \(=\) \(\ds 8^2 + 1^2\)
\(\ds \) \(=\) \(\ds 7^2 + 4^2\)


Proof

We have that:

$65 = 5 \times 13$


Both $5$ and $13$ can be expressed as the sum of two distinct square numbers:

\(\ds 5\) \(=\) \(\ds 1^2 + 2^2\)
\(\ds 13\) \(=\) \(\ds 2^2 + 3^2\)


Thus:

\(\ds \) \(=\) \(\ds \paren {1^2 + 2^2} \paren {2^2 + 3^2}\)
\(\ds \) \(=\) \(\ds \paren {1 \times 2 + 2 \times 3}^2 + \paren {1 \times 3 - 2 \times 2}^2\) Brahmagupta-Fibonacci Identity
\(\ds \) \(=\) \(\ds \paren {2 + 6}^2 + \paren {3 - 4}^2\)
\(\ds \) \(=\) \(\ds \paren {2 + 6}^2 + \paren {4 - 3}^2\)
\(\ds \) \(=\) \(\ds 8^2 + 1^2\)
\(\ds \) \(=\) \(\ds 64 + 1\)
\(\ds \) \(=\) \(\ds 65\)


and:

\(\ds \) \(=\) \(\ds \paren {1^2 + 2^2} \paren {2^2 + 3^2}\)
\(\ds \) \(=\) \(\ds \paren {1 \times 2 - 2 \times 3}^2 + \paren {1 \times 3 + 2 \times 2}^2\) Brahmagupta-Fibonacci Identity: Corollary
\(\ds \) \(=\) \(\ds \paren {2 - 6}^2 + \paren {3 + 4}^2\)
\(\ds \) \(=\) \(\ds \paren {6 - 2}^2 + \paren {3 + 4}^2\)
\(\ds \) \(=\) \(\ds 4^2 + 7^2\)
\(\ds \) \(=\) \(\ds 16 + 49\)
\(\ds \) \(=\) \(\ds 65\)

$\blacksquare$


Sources

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