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\[\begin{align*} 1994 = u^2 + 2u +v > v^2 + 3v \\ v^2 + 3v - 1994 < 0 \end{align*}\]

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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