$p = (u + v)^2 + uv$ for primes $p = 5k \pm 1$?

Question

I came up with the following conjecture which I am unable to prove.

For all primes $p$ which can be represented as either $5k + 1$ or $5k - 1$ for some positive integer $k$, we can find positive integers $u$, $v$ such that

$p = (u + v)^2 + u \times v$

For example, $31 = 5 \times 6 + 1 = (2 + 3)^2 + 2 \times 3$

Using a computer, I have verified it for $p < 10000$. Is there a general proof for this?

Answer 1

We'd like to use some algebraic number theory (Kummer-Dedekind) to prove this, and in order to do that, we need to link $(u+v)^2+uv$ to the norm of some quadratic extension of $\mathbb{Q}$. This must be possible, because $(u+v)^2+uv$ is homogeneous of degree $2$. Let $\varphi:=\frac12(1+\sqrt{5})$ be the golden ratio, then: $$ \begin{align*} (u+v)^2+uv&=0\\ (u/v)^2+3(u/v)+1&=0\\ u/v&=\frac12(-3\pm\sqrt{5})\\ u/v&\in\{\varphi-2,-\varphi-1\}. \end{align*} $$ Now we find that $$ \begin{align*} u^2+v^2+3uv &= (u-(\varphi-2)v)(u+(\varphi+1)v)\\ &=\left((u+2v)-\varphi v\right)\left((u+2v)-\bar{\varphi}v\right)\\ &= N_{\mathbb{Q}(\varphi)/\mathbb{Q}}\left((u+2v)-\varphi v\right). \end{align*} $$

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So if $p$ is a prime number, then $p$ can be written as $(u+v)^2+uv$ if and only if $\mathbb{Z}[\varphi]$ contains elements of norm $p$, which is to say that $(p)$ splits in $\mathbb{Z}[\varphi]$ as a product of principal ideals. By Kummer-Dedekind and the fact that the class number of $\mathbb{Z}[\varphi]$ is $1$, this is equivalent to $X^2-X-1$ splitting over $\mathbb{F}_p$.

Note that $X^2-X-1$ is irreducible over $\mathbb{F}_2$. For all odd primes $p$, its roots are given by $\frac12(1\pm\sqrt 5)$, so it splits as the product of two distinct linear factors if and only if $\left(\frac5p\right)=1$. Quadratic reciprocity gives that $$ \left(\frac 5p\right)=\left(\frac p5\right)=\begin{cases}0\quad&\text{if $p=5$}\\1&\text{if $p\equiv \pm 1\pmod {5}$}\\-1&\text{if $p\equiv \pm 2\pmod 5$},\end{cases} $$ so we can write $p=N_{\mathbb{Q}(\varphi)/\mathbb{Q}}(s+\varphi t)$ for integers $s,t$ if and only if $p\equiv \pm 1\pmod 5$ or $p=5$.

Answer 2

Write $p=(x+y)^2+xy = x^2+y^2+3xy=(x-y)^2+5xy$

If $q$ is an integer not divisible by $5$, then $q^2 \equiv 1$ mod $5$. As $p \neq 5$ is prime, it cannot be divided by $5$ and since $x-y$ not divisible by $5$, hence it follows $p \equiv \, 1$ mod $5$. Now, this runs only one way, the converse it not necessarily true from my statement, that for some $p \equiv 1$ mod $5$, then $p=x^2+y^2+2xy$.