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
\(\ds 1 \times 3 + 1\) | \(=\) | \(\ds 4\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^2\) |
\(\ds 1 \times 8 + 1\) | \(=\) | \(\ds 9\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3^2\) |
\(\ds 1 \times 120 + 1\) | \(=\) | \(\ds 121\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 11^2\) |
\(\ds 3 \times 8 + 1\) | \(=\) | \(\ds 25\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 5^2\) |
\(\ds 3 \times 120 + 1\) | \(=\) | \(\ds 361\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 19^2\) |
\(\ds 8 \times 120 + 1\) | \(=\) | \(\ds 961\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 31^2\) |
$\blacksquare$
Sources
- 1969: A. Baker and H. Davenport: The equations $3x^2 − 2 = y^2$ and $8 x^2 − 7 = z^2$ (Quart. J. Math. Vol. 20: pp. 129 – 137)
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $120$
- 2015: Yifan Zhang and G. Grossman: On Diophantine triples and quadruples (Notes on Number Theory and Discrete Mathematics Vol. 21, no. 4: pp. 6 – 16)