There exist unique $x, y\in \Bbb{N}$ that satisfy $x^2+84x + 2008=y^2$. Find $x+y$.

Question

There exist unique $x, y\in \Bbb{N}$ that satisfy $x^2+84x + 2008=y^2$. Find $x+y$.

I tried this question by completing the square on the left side, but the answer is coming out weird. Please help me.

Answer 1

Hint :Complete the square for $x$ \begin{eqnarray*} (x+42)^2+244=y^2 \\ \color{red}{(y+x+42)}\color{blue}{(y-x-42)} = \color{red}{122} \times \color{blue}{2} \end{eqnarray*}

$y=62,x=18$

Answer 2

The equation over the integers is $$ x^2+84x+2008-y^2=0, $$ which is equivalent to $$ x=\pm \sqrt{y^2-244}-42. $$ Hence $y^2-244=z^2$ for some integer $z$, i.e., $(y-z)(y+z)=244$ For $244=2\cdot 122$ we obtain $y-z=2$ and $y+z=122$, so that $y=62$, and $x=18$.

Answer 3

Perhaps a little less bashing (and I think quicker solution):

We have:

$$ (x+42)^2 <x^2+84x+2008 <(x+45)^2$$ thus $$ (x+42)^2 <y^2 <(x+45)^2$$ so $$ x+42 <y <x+45$$

so $y= x+43$ or $y=x+44$. Plugging in to original equation we get a solution.