Show that there exists no integer coordinates on curve

Question

Problem: Show that there does not exist any integer coordinates to the curve $$y = \frac{x^2-3}{4}, x\in \mathbb{R}.$$

My attempt: The problem is equivalent of saying that there does not exist any $x\in \mathbb{R}$ such that $$x^2-3 \equiv 0 \text{ mod } 4.$$

I've tried to factor $x^2-3 \equiv (x+\sqrt{3})(x-\sqrt{3}) \text{ mod } 4$ and reason that this implies that $$ x+\sqrt{3} \text{ mod } 4$$ or $$ x-\sqrt{3} \text{ mod } 4.$$ This would then mean that $x = \pm \sqrt{3} + 4l, l \in \mathbb{Z}$, which is not an integer. I don't think such a solution would be valid though for example for $x = \sqrt{3}$ on of the factors $x \pm \sqrt{3}$ wouldn't be an integer. Would appreciate any insight on this problem.

Answer 1

This problem gives you a chance to review basic modulo arithmetic with the squares. So any integer can be either odd or even. Hence you can write: For any $x \in \mathbb{Z}$, $x = 2k$ or $x = 2k + 1$.

Thus: $x^2 = (2k)^2 = 4k^2 \equiv 0 \pmod 4$ or $x^2 = (2k + 1)^2 = 4(k^2 + k) + 1 \equiv 1\pmod 4$.

But your equation gives: $x^2 \equiv 3 \pmod 4$ and this can't happen. So the answer follows.

Answer 2

LAcarguy's answer is very succinct and correct, so rather than repeat it, let me focus on where your own solution is flawed.

You have:

I've tried to factor $x^2-3 \equiv (x+\sqrt{3})(x-\sqrt{3}) \text{ mod } 4$ and reason that this implies that $$ x+\sqrt{3} \text{ mod } 4$$ or $$ x-\sqrt{3} \text{ mod } 4.$$

Nope! First of all, you seem to be trying to generalize the "zero product property" (if $ab=0$ then either $a=0$ or $b=0$) to the case of modular arithmetic. But it isn't the case that if $ab \equiv 0 \mod{4}$ then $a \equiv 0$ mod 4 or $b \equiv 0$ mod 4. (Consider what happens if both $a$ and $b$ are $\equiv 2$ mod 4.) In fact the zero-product property only holds in modular arithmetic when you are working mod $p$ for some prime number -- if you are working modulo a composite number it fails.

The second issue has to do with how you factored that polynomial. Factorization in modular arithmetic can behave in unexpected ways. This is a bit harder to explain, so let me give you a different example. Suppose you are working modulo 11, and you want to factor $x^2-5$. Your instincts are probably to factor it as $(x-\sqrt{5})(x+\sqrt{5})$, which would be correct if we were factoring over $\mathbb{R}$. But over the integers mod 11, we have $x^2-5 \equiv (x-4)(x+4)$ (don't take my word for it, multiply it out and see!). So the equation $x^2 \equiv 5 \mod{11}$ has two solutions: $x \equiv 4$ and $x \equiv -4 \equiv 7$. No need to mess around with irrational numbers at all.

As a general rule, if you are working over the integers, and looking for integer solutions, don't try to factor over the reals -- it will just lead you astray.