At least one of $4$ integers is $0$.

Question

I encountered this very interesting problem in my work and it sounds like this:

Let $a, b, c, d\in\mathbb{Z}$ such that $$ca-3bd=5$$ $$ad+bc=2$$ Prove that at least one of them is $0$.

I tried (using the idea that a perfect square is non-negative) to multiply with $d$ the first relation and then with $c$ the second, and then first with $c$ the first relation and then with $d$ the second and by adding/subtracting I obtained $$a(c^2+3d^2)=5c+6d$$ $$b(c^2+3d^2)=2c-5d$$ but it doesn't help much. How should I approach this problem?

Answer 1

If $w=a+b\sqrt{-3}$ and $z=c+d\sqrt{-3}$, this is saying $wz=5+2\sqrt{-3}.$

Then $|w|^2|z|^2=\left|5+2\sqrt{-3}\right|^2=37.$

But $|w|^2=a^2+3b^2, |z|^2=c^2+3d^2.$

So you actually get that either $|w|^2=1$ or $|z^2|=1$, and hence that either $(a,b)=(\pm 1,0)$ or $(c,d)=(\pm 1,0).$

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If you don't want to use complex numbers, then you can prove directly that:

$$(a^2+3b^2)(c^2+3d^2)=(ac-3bd)^2+3(ad+bc)^2$$

Answer 2

Write first equation (you got) like this $$ac^2-5c+(3ad^2-6d)=0$$ since this quadratic equation on $c$ must have a solution we have $D\geq 0$ so

$$ -12a^2d^2 +24ad +25\geq 0$$ write $x=ad\in \mathbb{Z}$, so $$ 12x^2-24x-25\leq 0$$ so $$ 12 (x-1)^2= 12x^2-24x+12\leq 37$$ so $(x-1)^2\leq 3$ so $|x-1|\leq 1$ so $x\in \{0,1,2\}$.

$\bullet$ If $x=0$ so $ad =0$ and we are done.

$\bullet$ If $x=1$ then $a=d=1$ or $a=d=-1$ then...

$\bullet$ If $x=2$ then $bc=0$ and we are done.