Fermat Set is Diophantine Quadruple

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Theorem

The Fermat set $F = \left\{{1, 3, 8, 120}\right\}$ is a Diophantine quadruple:

$\forall a, b \in F: a \ne b: a b + 1 = n^2$

for some $n \in \Z$.


Proof

\(\displaystyle 1 \times 3 + 1\) \(=\) \(\displaystyle 4\)
\(\displaystyle \) \(=\) \(\displaystyle 2^2\)


\(\displaystyle 1 \times 8 + 1\) \(=\) \(\displaystyle 9\)
\(\displaystyle \) \(=\) \(\displaystyle 3^2\)


\(\displaystyle 1 \times 120 + 1\) \(=\) \(\displaystyle 121\)
\(\displaystyle \) \(=\) \(\displaystyle 11^2\)


\(\displaystyle 3 \times 8 + 1\) \(=\) \(\displaystyle 25\)
\(\displaystyle \) \(=\) \(\displaystyle 5^2\)


\(\displaystyle 3 \times 120 + 1\) \(=\) \(\displaystyle 361\)
\(\displaystyle \) \(=\) \(\displaystyle 19^2\)


\(\displaystyle 8 \times 120 + 1\) \(=\) \(\displaystyle 961\)
\(\displaystyle \) \(=\) \(\displaystyle 31^2\)

$\blacksquare$


Sources