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Question. $p$ is a prime, $k$ is a given natural number. If $a, b, c, d$ exists s.t. $p=a^2+kb^2$, $pn=c^2+kd^2$, proof that integer $x, y$ such that $n=x^2+ky^2$ exists.

My approach. Let $n=x^2+ky^2$.

$pn=(a^2+kb^2)(x^2+ky^2)=(ax \pm kby)^2+k(ay \mp bx)^2$.

Let $(x_1, y_1)$ the root of $ax+kby=c, -bx+ay=d$.

Let $(x_2, y_2)$ the root of $ax-kby=c, bx+ay=d$.

If one of $(x_1, y_1)$ or $(x_2, y_2)$ is integer pair, the proof will be done.

$$x_1=\frac{ac-kbd}{p}, y_1=\frac{ad+bc}{p}$$ $$x_2=\frac{ac+kbd}{p}, y_2=\frac{ad-bc}{p}$$

So when we proof these three,

  1. $p \mid (ad+bc)(ad-bc)$
  2. $p \mid ad+bc \Rightarrow p \mid ac-kbd$
  3. $p \mid ad-bc \Rightarrow p \mid ac+kbd$

The proof will be done.

How should I prove this?

Bill Dubuque
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coding1101
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    Here's a hint to move you forward: Try reducing the equations for $p$ and $pn$ modulo $p$, and solve them both for $k$. This will relate $a$,$b$,$c$ and $d$. – FredH Feb 21 '19 at 08:21

2 Answers2

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$(1)\ \ \ \ p^2n = (ac\mp kbd)^2+ k(\color{#0a0}{ad\pm bc})^2\ $ by composing $\,p\times pn\,$ (or by norm multiplicativity)

Note $\ \color{#c00}p\mid d^2\color{#c00}p - b^2\color{#c00}pn = d^2(a^2\!+\!kb^2)\!-\!b^2(c^2\!+\!kd^2) = (ad)^2\!-\!(bc)^2 = \color{#0a0}{(ad\!+\!bc)(ad\!-\!bc)}$

So $\,p\mid \color{#0a0}{ad\!+\!bc}\,$ or $\,p\mid \color{#0a0}{ad\!-\!bc}\,$ so $(1)\Rightarrow$ $\,p\mid ac\!-\!kbd\,$ or $\,p\mid ac\!+\!kbd\,$ so we can cancel $\,p^2$ from $(1)$ to get the sought representation of $\,n$.

Bill Dubuque
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feb. 21 2019

LEMMA: with integers, if $v^2 | w^2,$ then $v | w.$

With $b \neq 0,$ since $p$ is a prime rather than a square, $$ p = a^2 + k b^2 $$ $$ np = c^2 + k d^2 $$ $$ kb^2 = p - a^2 $$ $$ b^2 np = b^2 c^2 + (k b^2) d^2 = b^2 c^2 + (p-a^2)d^2 = (b^2 c^2 - a^2 d^2) + p d^2 $$

Thus $$ p | (bc-ad)(bc+ad) $$ and $p$ divides at least one of them. Define some $$ \phi = \pm 1$$ so that $$ p | bc - \phi ad. $$ We then have an integer $\tau$ with $$ bc - \phi ad = p\tau, $$ $$ bc + \phi ad = p\tau + 2 \phi a d , $$ $$ (b^2 c^2 - a^2 d^2) = p^2 \tau^2 + 2 \phi adp\tau . $$ Recall $$ b^2 np = (b^2 c^2 - a^2 d^2) + p d^2 $$ $$ b^2 np = p^2 \tau^2 + 2 \phi adp\tau + p d^2 $$ Divide by $p$ and switch order $$ b^2 n = d^2 + 2 \phi da\tau + p \tau^2 $$ Add and subtract $a^2 \tau^2, \;$ $$ b^2 n = (d^2 + 2 \phi da\tau + a^2 \tau^2) + (p \tau^2 - a^2 \tau^2) $$ $$ b^2 n = (d + \phi a \tau)^2 + \tau^2 (p-a^2) $$ Recall $p - a^2 = k b^2$ $$ b^2 n = (d + \phi a \tau)^2 + k b^2 \tau^2 $$ Now, $b^2 | (d + \phi a \tau)^2,$ so the LEMMA says $b | d + \phi a \tau.$ We may introduce an integer $\psi$ with $$ d + \phi a \tau = b \psi, $$ whence $$ (d + \phi a \tau)^2 = b^2 \psi^2. $$ Then $$ b^2 n = b^2 \psi^2 + k b^2 \tau^2, $$ finally $$ n = \psi^2 + k \tau^2. $$

Will Jagy
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