If $x,y∉ℚ$ and $x^2-y^2∈ℚ_{≠0}$ then $x+y,x-y∉ℚ$?

A rational number times a rational number gives a rational number and a rational number plus a rational number gives a rational number. $\mathbb Q$ is a field.

When we multiply two irrational numbers it is possible that we get a rational number. e.g. $(\sqrt 2)(\sqrt 8) = 4.$ But, a non-zero rational number times an irrational number will always be irrational. The similar rules apply for addition.

$x^2 - y^2 = (x+y)(x-y)$

If $x^2-y^2$ is rational then either $(x+y)$ and $(x-y)$ are both rational or both irrational.

If $(x+y)$ and $(x-y)$ are both rational, then $(x+y) + (x-y) = 2x$ must be rational. But the proposition states that $x,y$ are both irrational. So, it can't be the case that $(x+y)$ and $(x-y)$ are both rational, thus the must be both irrational.