Prime-Generating Quadratics of form 2 a squared plus p/5

Theorem

The quadratic form:

$2 a^2 + 5$

yields prime numbers for $a = 0, 1, \ldots, 4$.

Proof

\(\ds 2 \times 0^2 + 5\)

\(=\)

\(\ds 0 + 5\)

\(\ds \)

\(=\)

\(\ds 5\)

which is prime

\(\ds 2 \times 1^2 + 5\)

\(=\)

\(\ds 2 + 5\)

\(\ds \)

\(=\)

\(\ds 7\)

which is prime

\(\ds 2 \times 2^2 + 5\)

\(=\)

\(\ds 2 \times 4 + 5\)

\(\ds \)

\(=\)

\(\ds 8 + 5\)

\(\ds \)

\(=\)

\(\ds 13\)

which is prime

\(\ds 2 \times 3^2 + 5\)

\(=\)

\(\ds 2 \times 9 + 5\)

\(\ds \)

\(=\)

\(\ds 18 + 5\)

\(\ds \)

\(=\)

\(\ds 23\)

which is prime

\(\ds 2 \times 4^2 + 5\)

\(=\)

\(\ds 2 \times 16 + 5\)

\(\ds \)

\(=\)

\(\ds 32 + 5\)

\(\ds \)

\(=\)

\(\ds 37\)

which is prime

$\blacksquare$