# Fermat Set is Diophantine Quadruple

## 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

 $\displaystyle 1 \times 3 + 1$ $=$ $\displaystyle 4$ $\displaystyle$ $=$ $\displaystyle 2^2$

 $\displaystyle 1 \times 8 + 1$ $=$ $\displaystyle 9$ $\displaystyle$ $=$ $\displaystyle 3^2$

 $\displaystyle 1 \times 120 + 1$ $=$ $\displaystyle 121$ $\displaystyle$ $=$ $\displaystyle 11^2$

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

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

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

$\blacksquare$