Does there exist $x$, $y \in \mathbb{N}$ such that $x^2 − y^2 = 19$

Question

This is what I got:

$(x-y)(x+y) = 19$ so $x-y \in \mathbb{N}$ and $x+y \in \mathbb{N}$

$ \implies x-y = 19 =x+y$, we know $x+y \geq 8$, not possible

$ \implies x-y = -19 =x+y$, Not possible since $x+y \in \mathbb{N}$

Hence, no $x$, $y \in \mathbb{N}$ exists.

Answer 1

Notice that $19$ is prime. because of this, one of $x-y$ and $x+y$ must be $1$, and the other $19$. Therefore, if we solve the system $x-y = 1, x+y=19$, we can find a solution to this problem. You should be able to find a solution to that system.

Answer 2

You've came to $(x+y)(x-y) = 19$

The only possible natural numbers that multiply to 19 are 1 and 19. And sum/difference of natural numbers are also natural numbers. Since $x-y < x+y$, you have: $$x-y=1,$$ $$x+y=19$$

This gives a solution: $x=10, y=9$.

Answer 3

Consecutive pairs of squares differ by an odd number. In particular, such a difference is in progression: the distance from the $n$-th square and the following is exactly $2n+1$. Thus given an odd integer $k$, there always exist two squares which differs by $k$, namely $$ x=\frac{k-1}{2}, y=\frac{k+1}{2}=x+1 $$