Determining for which integers $t\in\mathbb{N}_0$, $\sqrt{t^2 - 4d}$ is an integer when $d = \pm 1$

Question

I am currently stuck on a proof which states that if $d = \pm 1$, then the only non-negative integers $t\in\mathbb{N}_0$ for which $\sqrt{t^2 -4d}$ is an integer are $t \in \{0,2\}$. It is clear that for those values of $t$ the value of $\sqrt{t^2 -4d}$is an integer for sure, but why are these the only two possibilities? For the record, I don't really have a background in number theory, so it is possible that the claim is due to some well-known property of integers. Thanks!

Answer 1

$t^2-4d=s^2 \Leftrightarrow (t-s)(t+s)=4d$ for some $s\in \mathbb N_0$. As 4 divides the RHS, $t$ and $s$ have the same parity: $$\dfrac{t-s}2\cdot \dfrac{t+s}2=d=\pm 1$$ Both factors $\dfrac{t-s}2$ and $\dfrac{t+s}2$ are integers, whence there are 3 cases: i) $t-s=t+s=2$; ii) $t-s=2$, $t+s=-2$; iii) $t-s=-2$, $t+s=2$. The first case yields $t=2$, while the other two return $t=0$.

Answer 2

It means that $t^2+4d$ must be a square, which means that it must be, either $(t±1)^2$ or $(t±2)^2$, which means that, either:

1. $2t±1 = 4$, or:
2. $4t±4 = 4$

The first is impossible because an odd number can't be even, the second yields your current result.

Answer 3

For the case $d=1$, notice that we will have $\sqrt{t^2-4}=n\in\mathbb{N}_0$ when $$\sqrt{t^2-4}=n\iff t^2-4=n^2\iff 4 = t^2-n^2=(t-n)(t+n)$$ and, since $t\geq 0$ and $t-n$ and $t+n$ will be divisors of $4$ with the same parity, it is easy to check that the only possibility is $t-n=t+n=2$ which clearly implies $t=2$. For the other case, i.e. $d=-1$, notice that we will have $\sqrt{t^2+4}=n\in\mathbb{N}_0$ when $$\sqrt{t^2+4}=n\iff t^2+4=n^2\iff 4 = n^2-t^2=(n-t)(n+t)$$ which, by a similar reasoning, gives $n-t=n+t=2\Rightarrow t=0$. Hence, in both cases $t\in\{0,2\}$.