Question

Suppose a prime can be written as p = x^2 + 5y^2. Show that
p ≡ 1 or 9 (mod 20). Assume that p > 5

Answer 1

Answer on Question #58504 – Math – Combinatorics | Number Theory

Question

Suppose a prime can be written as $p = x^2 + 5y^2$. Show that $p \equiv 1$ or 9 (mod 20).

Assume that $p > 5$.

Solution

Because $p$ is a prime, $p$ cannot be even, hence $p$ is odd.

We may assume that $p$ is an odd prime, greater than 5. Because $p = x^2 + 5y^2$ is odd, one of $x^2$, $y^2$ is odd, and the other is even.

Reducing $p = x^2 + 5y^2 \mod 5$, we see $p = x^2 \pmod{5}$, so $p$ is a quadratic residue mod 5 and thus $p = 1$ or 4 (mod 5).

Reducing $p = x^2 + 5y^2 \mod 4$, we see $p = x^2 + y^2 \pmod{4}$, so $p$ is a sum of two quadratic residues mod 4.

Since the quadratic residues mod 4 are 0 and 1, this rules out the possibility that $p = 3$ (mod 4). Besides, $p = 0$ (mod 4), $p = 2$ (mod 4) are excluded, because $p$ is odd.

So $p = 1$ or 4 (mod 5) and $p = 1$ (mod 4), which means that $p = 1$ or 9 (mod 20).

Check it.

Let

$$p = 1 + 4k \quad \text{and} \quad p = 1 + 5l \quad \text{or} \quad p = 4 + 5m.$$

Suppose that $p = 1 + 4k$ and $p = 1 + 5l$.

Then $1 + 4k = 1 + 5l$, $4k = 5l$, hence $k = 5t$, $l = 4u$.

This means that $p = 1 + 4k = 1 + 4 \cdot 5t = 1 + 20t$, $p = 1 + 5l = 1 + 5 \cdot 4u = 1 + 20u$.

Finally obtain $p = 1$ (mod 20).

Suppose that $p = 1 + 4k$ and $p = 4 + 5m$.

Then $1 + 4k = 4 + 5m$, $4k - 4 - 4m = m - 1$, $4n = m - 1$, hence $m = 4n + 1$.

This means that $p = 4 + 5m = 4 + 5 \cdot (4n + 1) = 4 + 20n + 5 = 9 + 20n$.

Finally obtain $p = 9$ (mod 20).

Given $p = 1$ or 4 (mod 5) and $p = 1$ (mod 4), we came to $p = 1$ or 9 (mod 20), which was to be proved.

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