Prime-Generating Quadratics of form 2 a squared plus p/3

Theorem

The quadratic form:

$2 a^2 + 3$

yields prime numbers for $a = 0, 1, 2$.

Proof

\(\ds 2 \times 0^2 + 3\)

\(=\)

\(\ds 0 + 3\)

\(\ds \)

\(=\)

\(\ds 3\)

which is prime

\(\ds 2 \times 1^2 + 3\)

\(=\)

\(\ds 2 + 3\)

\(\ds \)

\(=\)

\(\ds 5\)

which is prime

\(\ds 2 \times 2^2 + 3\)

\(=\)

\(\ds 2 \times 4 + 3\)

\(\ds \)

\(=\)

\(\ds 8 + 3\)

\(\ds \)

\(=\)

\(\ds 11\)

which is prime

$\blacksquare$