Prove there are no integer solutions for $x^2 + 3y^2 = 8$

Question

I know there are analogous questions to mine, but I'm seeking a different approach. All the solutions I saw use some sort of modular arithmetic trick, but I'd like to prove this without resorting to that. This question is from an exercise that asks to prove things using proof by cases. Here is what I came up with:

Proof. Suppose, for the sake of contradiction, that there are $x, y \in \mathbb{Z}$ such that $$x^2 + 3y^2 = 8.$$ Then $x^2 = 8 - 3y^2 \geq 0$, since $x^2$ is always positive. Thus $$3y^2 \leq 8 \implies y^2 \leq \frac{8}{3}.$$ Since $y$ is an integer, it follows that $$0 \leq y^2 \leq \left \lfloor{8/3}\right \rfloor \implies 0 \leq y^2 \leq 2.$$ Thus, there are three cases to consider:

(1) $y^2 = 0 \implies x^2 = 8 \implies x = \pm\sqrt{8}$.

(2) $y^2 = 1 \implies x^2 + 3 = 8 \implies x = \pm\sqrt{5}$.

(3) $y^2 = 2 \implies x^2 + 6 = 8 \implies x = \pm\sqrt{2}$.

Since all cases contradict the assumption that $x$ is an integer, there are no integer solutions to $x^2 + 3y^2 = 8$. $\quad\blacksquare$

Probably this is unnecessarily verbose... What do you think? If there are no errors, is there a better way to do it? (without using modular arithmetic xD)

Answer 1

In case (2), you cannot conclude $y=1$, it may also be $y=-1$. Likewise, case (3) rather leads to $y=\pm\sqrt 2$. In the end this doesn't change much, but formally, you left out those cases.

For the sake of the arguments, I might prefer to do cases from the opposite point of view:

1. $y=0$. Then $x^2=8-3y^2=8$, but $8$ is not a perfect square.
2. $y=\pm1$. Then $x^2=8-3y^2=5$, but $5$ is not a perfect square.
3. Anything else, i.e, $y\ge 2$ or $y\le -2$. Then $y^2\ge 4$, $0\le x^2=8-3y^2<0$, contradiction

Answer 2

Suppose (x, 3)=1 and also (y, 3)=1 then we ca write:

$x^2\equiv 1 \bmod 3$

$y^2\equiv 1\bmod 3$

$\Rightarrow x^2+3y^2\equiv 1 \bmod 3$

where $8\equiv 2 \bmod 3$

So there can not be integer solution.

Answer 3

We have the equation for an ellipse $ \ x^2 + 3y^2 = 8 \ $ , which is centered on the origin with semi-major axis $ \ \sqrt{8} < 3 \ $ and semi-minor axis $ \ \sqrt{\frac83} < 2 \ . $ Since plainly $ \ x = 0 \ $ and $ \ y = 0 \ $ don't produce integer solutions for the equation, the only other values to be tested are the pairings of $ \ x^2 = 1 \ $ or $ \ x^2 = 4 \ $ with $ \ y^2 = 1 \ $ ; none of these solve the equation either. (The graph shows that there are no points of the integer-grid that lie on the ellipse.)

This sort of approach won't be as helpful for large coefficients. (Using hyperbolas to "solve" Pell's equations can become rather misleading...)

Answer 4

Perhaps it has been all said already.

Note:

0)We can restrict our search to non--negative integers.

1)$x \not=0$, since $x=0$ would imply $3$ divides $8$

$y\not =0$, since $y=0$ would imply $8$ is a square.

Hence. $(x,y) \not =(0, 0);$

2)If $x$ is even, $y$ is even; if $x$ is odd $y$ is odd.

3)Left to do: Check

$x=1$; and $y=1;$

No integer solutions to this equation.