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The theorem is from Niven's number theory "Binary quadratic forms" 3.4 Theorem 3.10 page 163-164

Theorem $3.10 $: Let $f(x, y) = ax^2 + bxy + cy^2$ be a binary quadratic form with integral coefficients and discriminant $d$. If $d \neq 0$ and $d$ is not a perfect square , then $a \neq 0, c \neq 0$, and the only solution of the equation $f(x, y) = 0$ in integers is given by $x = y = 0$.

Proof : We may presume that $a \neq 0$ and $c \neq 0$, for if $a = 0 $or $c = 0$ then $ac = 0$ and $d = b^2 - 4ac = b^2$, a perfect square. Suppose that $x_0$ and $y_0$ are integers such that $f(x_0, y_0) = 0$. If $y_o = 0$ then $ax_0^2 = 0$, and hence $x_0 = 0$ because $a \neq 0$. If $x_0 = 0$, a parallel argument gives $y_0 = 0$. Consequently we take $x_0 \neq 0$ and $y_0 \neq 0$. By completing the square we see that $$4af( x, y) = (2ax + by)^2-dy^2$$ and hence $(2ax_0 + by_0)^2 = dy_0^2$ since $f(x_0, y_0) = 0$. But $dy^2 \neq 0$, and it follows by unique factorization that $d$ is a perfect square. The proof is now complete

I am quite confused, why unique factorization would lead d be a perfect square?
It seems for me to prove d is indeed the perfect square we have to prove that $\frac{x_0}{y_0}$ is an integer, since $$d=(\frac{(2ax_0+by_0)}{y_0})^2$$ and such fact doesn't have much to do with Unique factorization. Thank you for your advices!

Arturo Magidin
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Tong Su
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  • This is a matter of style in proof by contradiction. If we say the assumption(s) are $x_0, y_0 \neq 0$ and $d\neq0$ also not a square. The contradiction may come in more than one way. The lemma they are using is: If $u,v,w$ are nonzero integers and $u^2 = v w^2,$ then $v$ is a square – Will Jagy Oct 11 '22 at 23:16
  • @ Will Jagy, my understanding is that we want to prove this theorem by contradiction, and we want to prove this theorem by showing that $x_0 \neq 0$ $y_0 \neq 0$ would leads to the contradiction that the discriminant d will be a square, Is this understanding correct? If what I just said is indeed correct, then the only thing I don't understand is why the last statement shows d is indeed a square, or why the lemma which you showed is true. – Tong Su Oct 11 '22 at 23:35
  • As proved in the linked dupes (& many others) on irrationality of sqrts: if $\sqrt d$ is rational then it is integral. – Bill Dubuque Oct 11 '22 at 23:42
  • Proofs typically use properties intimately connected to unique prime factorization (e.g. properties of gcds). This result may fail in domains without unique factorization, e.g. in quadratic number rings, e.g. see here and here. – Bill Dubuque Oct 11 '22 at 23:54
  • @Bill Dubuque， Then for similar theorem in such quadrtic rings which unique factorization property fails , Is such similar theorem still holds even without the property of unique factorization? If it is, which proof strategy is commonly use? – Tong Su Oct 12 '22 at 00:01
  • See here for citations on root closed domains. – Bill Dubuque Oct 12 '22 at 00:11

1 Answers1

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First you prove the following result about greatest common divisor:

Fact 1:

If $p, q \in \mathbb{N}$ and if $(p,q) = 1$, then $(p^2,q^2) = 1$.

Proof: We use Bezout's theorem for this. That is there exists integers $s,t \in \mathbb{Z}$ such that: $sp+tq = 1$. Square both sides: $s^2p^2+2stpq + t^2q^2 = 1 \implies s^2p^2+2stpq(sp+tq) + t^2q^2 = 1$. Rewrite to obtain: $s^2p^2+ 2s^2tqp^2+2st^2pq^2+t^2q^2 = 1$, and this gives: $(s^2+2s^2tq)p^2+(2st^2p+t^2)q^2 = 1$. By Bezout's theorem, $(p^2,q^2) = 1$. Now use this result to prove one more result, namely:

Fact 2:

If $a \in \mathbb{Q}$, and $a^2 \in \mathbb{N}$ then $a \in \mathbb{N}$.

Proof: For if $a \notin \mathbb{N}$, then since $a \in \mathbb{Q}$, we can write $a = \dfrac{p}{q}$ where $(p,q) = 1$ and $p, q \in \mathbb{N}$. First we show: $q = 1$. Suppose $q \neq 1$, then $a^2 = \dfrac{p^2}{q^2}\implies p^2 = a^2q^2\implies (p^2,q^2) \ge q^2 > 1$, contradiction to fact 1. Thus $q = 1$. Hence $a = p \in \mathbb{N}$.

You have: $d = \left(\dfrac{2ax_0+by_0}{y_0}\right)^2 = a^2 \in \mathbb{N}\implies a = \dfrac{2ax_0+by_0}{y_0} \in \mathbb{N}$ by fact 2. Thus $d = a^2$ which is a perfect square.

Wang YeFei
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