Since $x$ and $y$ are integers and so $v$ is an integer. Then from the above inequality, we get $v \in \Big[ -92, 86 \Big]$.

Step-2: Since from the equation $u^2 + 2u + \left( v- 1994 \right) = 0$. We get, $u= -1 \pm \sqrt{1995-v}$. But since $x>0$ and $y>0$ it follows that $u>0$. So, $u= -1 + \sqrt{1995-v}$.

Now for positive integral value of $u$, $1995-v$ should be a perfect square. Now from the step-1$1995-v \in \Big[ 1909, 2087 \Big]$. There are only two values of $1995 -v$, they are $2025$ and $1936$.

They yield the following pairs of $\left(u,v \right)$. They are $\{ \left(43,59 \right) , \left(44, -30 \right) \}$. But since $u>v$, we get only pair of $\left( u,v\right)$ i.e. $\left( 44, -30 \right)$ which yields $x = 7$ and $y= 37$.

So, therefore only one solution of $\left( x , y \right)$ exists i.e. $\left( 7 , 37 \right)$.

$(x+y)^2 + (x + y) = 1994 - 2x$$\rightarrow (x + y )(x + y + 1) = 1994 - 2x$ Now as 1994 = 44.45 + 14, x = 7 and x + y = 44 is one and only valid solution.

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## Comments

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TopNewestTo make this problem much simpler, let us make a substitution $x = \dfrac{u+v}{2}$ and $y=\dfrac{u-v}{2}$.

So, the given equation converts into

$u^2 + 2u + \left( v- 1994 \right) = 0$

Step-1:Let us first find the lower bound and upper bound of $v$. It is clear that $u = x+y$ and $v = x-y$. So, $u > v$ using this fact we get$\begin{aligned} 1994 = u^2 + 2u +v > v^2 + 3v \\ v^2 + 3v - 1994 < 0 \end{aligned}$

Since $x$ and $y$ are integers and so $v$ is an integer. Then from the above inequality, we get $v \in \Big[ -92, 86 \Big]$.

Step-2:Since from the equation $u^2 + 2u + \left( v- 1994 \right) = 0$. We get, $u= -1 \pm \sqrt{1995-v}$. But since $x>0$ and $y>0$ it follows that $u>0$. So, $u= -1 + \sqrt{1995-v}$.Now for positive integral value of $u$, $1995-v$ should be a perfect square. Now from the

step-1$1995-v \in \Big[ 1909, 2087 \Big]$. There are only two values of $1995 -v$, they are $2025$ and $1936$.They yield the following pairs of $\left(u,v \right)$. They are $\{ \left(43,59 \right) , \left(44, -30 \right) \}$. But since $u>v$, we get only pair of $\left( u,v\right)$ i.e. $\left( 44, -30 \right)$ which yields $x = 7$ and $y= 37$.

So, therefore only one solution of $\left( x , y \right)$ exists i.e. $\left( 7 , 37 \right)$.

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Thanx for the solution!An awesome one!

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@Dev Sharma @Nihar Mahajan

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X=7 and Y=37.

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What have you tried? Where are you stuck?

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I tried a bit of modular arithmetic and tried factorising the expression but nothing worked.A hint sir?

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Well, for one, since $x, y > 0$, we see that $x, y < 2000$ and so we have finitely many possibilities to try.

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$(x+y)^2 + (x + y) = 1994 - 2x$$\rightarrow (x + y )(x + y + 1) = 1994 - 2x$ Now as 1994 = 44.45 + 14, x = 7 and x + y = 44 is one and only valid solution.

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Well,thank you sir,a nice solution!

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I'm getting $x=7,y=37$

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Could you please explain how?

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