Equal Quadratics?

Relevant wiki: Using the Quadratic Formula

By rearranging the equation, and rewrite it as a quadratic equation of $y$ , we have $y^2 (2) + y(-8) + (3-x^2) = 0$ . Using the quadratic formula , we have $a = 2, b = -8, c = 3-x^2$ , then $y = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4(2)(3-x^2)}}{2(2)} = 2 \pm \dfrac{\sqrt{10+8x^2}}2.$ Since we're solving for integers $x$ and $y$ , we must have $\dfrac{\sqrt{10+8x^2}}2 = m$ , where $m$ is some integer. Rearranging this equation gives $10 + 8x^2 = 4m^2\Rightarrow 5 + 4x^2 = 2m^2$ which shows that an odd number plus an even number yields an even number, which is absurd. Hence, there is no solution.

Since every square is $0$ or $1 \mod 4$ , we have if $y$ is even, say $y=2k$ then $2(2k)^2-8(2k)+3\equiv 3\mod 4$ and this possibility can be ruled out, so $y$ must be odd. This time reducing modulo $8$ we obtain; $2(2k+1)^2-8(2k+1)+3\equiv 5\mod 8$ . But the only squares $\mod 8$ are $0,1,4$

$y^2-8 y+(3-x^2)=0$

Using Sridharachariya's Rule and after simplifying

$y=\dfrac{4\pm \sqrt{2 x^2-10}}{2}$

So the discriminant part $\sqrt{2 x^2-10}$ will be a perfect square

Now consider $2 x^2-10=m^2$ where $m$ is an integer.

Left side is even so right will also be even

Then putting $m=2 p$ where $p \in \mathbb{I}$

Then equation becomes
$2 x^2-10=4 p^2$

$x^2-5=2 p^2$

Now right side is even so left side will also be even.So $x$ will be odd integer.

So $x=2 l+1$ .putting this in the equation it becomes

$4 l^2+4 l+1-5=2 k^2$

$k^2=2 (l^2+l-1)$

$k^2=2 (l (l+1) -1)$

We know that $l (l+1)$ is always even .So even -odd is always odd.But left hand side is a perfect square and there is already a factor 2 is present .To become a perfect square there should be another factor 2 is needed but the expression $(l (l+1)-1)$ is always odd.

So L.H.S. cannot be a perfect square.So there is no integral solution.

We know that LHS $x^{2}$ is a non-negative integer.

But the RHS $2y^2-8y+3$ :If we check the Discriminant it turns out to be positive ,implying that the result of the quadratic is always negative.

But a non-negative number cannot equal a positive number.Hence there are no solutions.

Without doing any major calculations, it is actually quite obvious why there is no integral solutions to the following equation:

$x^{2} = 2y^{2} - 8y + 3$

By Completing the Square on the right hand side, the equation above becomes:

$x^{2} = 2(y-2)^{2} - 5$

From here, one can avoid any calculations that the previous solutions may have given and conclude that no solutions would exist because Difference of Perfect Squares performed on the right hand side would contain a $\sqrt{2}$ component, which immediately should raise any red flags.