Prime-Generating Quadratics of form 2 a squared plus p/5
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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 |
$\blacksquare$