Irrational Diophantine Equation

Suppose there are such integers.

If $k=0$ , then $-\sqrt{3} = m$ is an integer, which is absurd. Therefore, we must have $k\neq 0$ .

Then squaring both sides of $k\sqrt{2} - \sqrt{3} = m$ yields $2k^2 - 2k\sqrt{6} + 3 = m^2 \implies \sqrt{6} = \frac{2k^2 + 3 - m^2}{2k} \text{ is rational}$ Since it's known that $\sqrt{6}$ is irrational, this is a contradiction. Therefore, we can conclude no such integers $k,m$ exist.

We have $k\sqrt 2 - \sqrt 3 = m \implies k\sqrt 2 = m +\sqrt 3\phantom{cc} m ,k\neq 0 \in\mathbb Z$ Squaring on both sides $2k^2 = m^2+2m\sqrt 3 + 3\implies \sqrt 3 = \dfrac{ 2k^2 -m^2 -3 }{2m}$

$\sqrt 3$ is irrational and to the right of it we have rational number which proves that there exist no such $k,m$ .