Question

Let $x,y\in Z$ such that $x^2-2x=y^2-2y+1010$. Then the number of pairs $(x,y)$ satisfying the equation is

• A. only one
• B. infinitely many
• C. more than one but finite
• D. no such pair is possible

Answer 1

The correct option is D no such pair is possible

$x^2-2x=y^2-2y+1010$
$\Rightarrow x^2-2x+1=y^2-2y+1+1010$
$\Rightarrow (x-1{)}^2=(y-1{)}^2+1010$
Let $x-1=X$ and $y-1=Y$
Then, $X^2-Y^2=1010$
$\Rightarrow (X-Y)(X+Y)=1010$

Let both $X$ and $Y$ be even.
Then $(X-Y)(X+Y)$ is a multiple of $4$. But $1010$ is not a multiple of $4$.

If $X$ is even and $Y$ is odd or vice-versa, then $(X-Y)(X+Y)$ is odd.

If both $X$ and $Y$ is odd, then $(X-Y)(X+Y)$ is a multiple of $4$.

So, no value of $x$ and $y$ is possible.