If $x,y∉ℚ$ and $x^2-y^2∈ℚ_{≠0}$ then $x+y,x-y∉ℚ$? Show that if $x$ and $y$ are irrational numbers such that $x^2-y^2$ is a non-zero rational, then $x+y$ and $x-y$ are both irrational numbers.
I know that $\mathbb{Q}$ is closed under addition and product, however I have no clue how to solve this problem. I would really appreciate help. I tried assuming that the sums are rational to reach a contradiction, but I failed.
 A: 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.
A: With
$x, y \notin \Bbb Q \tag 1$
but
$0 \ne x^2 - y^2 \in \Bbb Q, \tag 2$
first note that (2) implies that
$x - y, x + y \ne 0, \tag 3$
lest
$x^2 - y^2 = (x - y)(x + y) = 0; \tag 4$
then in light of (3), we have either
$x - y, x + y \in \Bbb Q \tag 5$
or
$x - y, x + y \notin \Bbb Q, \tag 6$
for
$x + y = \dfrac{x^2 - y^2}{x - y} \tag 7$
and
$x - y = \dfrac{x^2 - y^2}{x + y}. \tag 8$
Now if (5) binds, then
$x = \dfrac{(x + y) + (x - y)}{2} \in \Bbb Q \tag 9$
and
$y = \dfrac{(x + y) - (x - y)}{2} \in \Bbb Q \tag{10}$
in contradiction to (1); therefore (6) must hold.
$OE\Delta.$
