If $x^2 - y^2 = 1995$, prove that there are no integers x and y divisible by 3

Question

If $x^2 - y^2 = 1995$, prove that there are no integers $x$ and $y$ divisible by $3$.

I'm quite new to this. As in the title I would like to ask if my reasoning here is correct and if my answer would be considered acceptable.

1. If $x$ or $y$ is divisible by $3$, but not both, then: $\begin{align}x \equiv \pm 1 \pmod{3} \land y \equiv 0 \pmod{3} &\iff x^2 \equiv 1 \pmod{3} \land y^2 \equiv 0 \pmod{3} \\ &\iff x^2 - y^2 \equiv 1 \pmod{3}\end{align}$

So if $x$ or $y$ is divisible by $3$ but not both then no values for $x$ and $y$ will ever make $x^2 - y^2$ divisible by $3$.

2. If $x$ and $y$ are divisible by $3$:

Since both $x$ and $y$ are divisible by $3$ they can be expressed $x = 3m$ and $y = 3n$ for some integers $m$, $n$.

$(3m)^2 - (3n)^2 = 1995 \iff 9m^2 - 9n^2 = 1995 \iff 3(m^2-n^2) = 665$

And it's easy to see that there are no whole number solutions for m and n, and thus there are no whole number solutions for x and y such that $x^2 - y^2 = 1995$ where x or y (or both) is divisible by 3.

Is this a reasonable proof? Are there more elegant ways to do it?

Answer 1

Your argument is fine and well expressed. More succinctly you could paraphrase parts of your argument as follows.

If precisely one of $x,y$ is divisible by $3$ then $x^2-y^2$ is not divisible by $3$.

If both of $x,y$ are divisible by $3$ then $x^2-y^2$ is divisible by $9$.

Answer 2

The question is a bit ambiguous. Anyway, you can indeed answer both interpretations.

If both $x$ and $y$ are divisible by $3$, then the left-hand side is divisible by $9$, but the right-hand side isn't (which is your proof).

Suppose $x$ is divisible by $3$. Then $-y^2\equiv0\pmod{3}$, because $1995\equiv0\pmod{3}$. Thus implies $y\equiv0\pmod3$ and we already have excluded the case.

Similarly if $y$ is divisible by $3$.

Answer 3

$1995$ I'd divisible by $3$ but not by $9$ hence $x , y $ both simultaneously are not multiple of $3$. $$ $$ let either of $x,y$ is multiple of $3$ and other is not multiple of $3$ then $x^2-y^2$ is not multiple of $3$ where as $ 1995$. Is multiple of $3$ hence not possible.

Answer 4

Another method is $x^2-y^2=1995$ where $x,y \in I$ can be written as $$(x+y)(x-y)=3×5×133$$ Hence $x+y=1995,665,399,133$ and $x-y=1,3,5,15$ Hence $(x,y)=(\pm 998,\pm 997),(\pm 334,\pm 331),(\pm 202,\pm 197),(\pm 74,\pm 59)$ where none is multiple of $3$.