How prove $\sqrt{r^2+c^2}$ is irrational

Question:

Let $a,b,c$ are integer numbers,and $r$ real numbers, and $$ar^2+br+c=0,ac\neq 0$$

show that

$$\sqrt{r^2+c^2}$$ is irrational.

My idea: Note that, $$\sqrt{r^2+c^2}=\sqrt{r^2+(ar^2+br)^2}=r\sqrt{1+(ar+b)^2}$$ Then how to proceed? This is China South East Mathematical Olympiad problem yesterday

Solution 1:

Suppose that $\sqrt{r^2+c^2}\in\mathbb{Q}$, and consider two cases:

Case $b\ne0$.

This implies that $br+c=-ar^2\in\mathbb{Q}$ and consequently $r\in\mathbb{Q}$.

Suppose that $r=\frac{p}{q}$ with $\gcd(p,q)=1$. Since $ap^2+bpq+cq^2=0$ we conclude, as usual that $p|c$ and $q|a$. So let us define integers $a'$ and $b'$ by $$a=qa'\quad\hbox{and} \quad c=pc'.$$

Now $$ \sqrt{r^2+c^2}=\frac{p}{q}\sqrt{1+(c'q)^2}=\frac{p}{q}\sqrt{n}\tag{1} $$ where $n=1+(c'q)^2\in\mathbb{N}$. This implies that $\sqrt{n}\in\mathbb{Q}$, because we assumed that $\sqrt{r^2+c^2}\in\mathbb{Q}$. It follows that $n$ is the square of an integer, which is absurd since $c'q<\sqrt{n}<c'q+1$. This contradiction proves the assertion, in this case.$\qquad\square$

Case $b=0$.

Here we have $r^2=-c/a>0$ so $$\sqrt{r^2+c^2}=\frac{1}{2a}\sqrt{4c^2a^2-4ca}=\frac{1}{2a}\sqrt{n}\tag{2} $$ where $n=4c^2a^2-4ca\in\mathbb{N}$. Again. this implies that $\sqrt{n}\in\mathbb{Q}$, because we assumed that $\sqrt{r^2+c^2}\in\mathbb{Q}$. It follows that $n$ is the square of an integer, which is absurd since $(2ca)^2< n<(2ca-1)^2$. This contradiction proves the assertion, in this case also.$\qquad\square$