I am trying to show that $\sqrt{p^2 - 4q}$ is always irrational with $p, q$ as described above. Any hints?
For distinct odd primes $p, q$ we have $\sqrt{p^2 - 4q}$ is always irrational?
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On
$p^2$ leaves a remainder $1$ when divided by $8$ and $4q$ leaves a remainder $4$ when divided by $8$ as $q$ is odd, hence $p^2-4q$ leaves a remainder $5$ when divided by $8$, hence cannot be a perfect square as odd perfect squares leave a remainder $1$ when divided by $8$.
And $\sqrt n$ is an irrational if $n$ is not a perfect square, hence the fact $\sqrt{p^2-4q}$ is irrational is not only true for odd primes $p,q$ but also for odd numbers $p,q$.
On
$\sqrt {p^2-4q}\;\in \mathbb Q$ iff $p^2-4q=m^2$ with $m\in \mathbb Z^+\cup \{0\}. $
If $m\in \mathbb Z\cup \{0\}$ and $p^2-4q=m^2$ then $(p-m)(p+m)=4q.$ Now $p$ is odd so $m$ is odd. (If $m$ were even then $(p-m)$ and $(p+m)$ would both be odd, causing $4q$ to be odd.). So $(p-m)$ and $(p+m)$ are even.
So $\frac {p-m}{2}\cdot \frac {p+m}{2}=q\;$ with non-negative integers $\frac {p-m}{2}$ and $\frac {p+m}{2}.$
But $q$ is prime and $p>0$ and $m\geq 0 ,$ so this requires $1=\frac {p-m}{2}$ and $q=\frac {p+m}{2}.$
Adding, we have $1+q=\frac {p-m}{2}+\frac {p+m}{2}=p.$ But $1+q=p$ contradicts the hypothesis that $p$ and $q$ are both odd.
Note that $q$ is required to be prime but $p$ is only required to be an odd positive integer.
First idea that came to mind: If $n$ is odd, $n^2\equiv 1\pmod 8$. However, if $q$ is odd, $$p^2-4q\equiv -3\pmod 8$$