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\[x,y \in Z^{+}\\ (x+y)^2+3x+y=1994\\ y=?\]

Note by Adarsh Kumar
11 months, 2 weeks ago

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To 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. \(\left( 7 , 37 \right) \). Surya Prakash · 11 months, 2 weeks ago

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@Surya Prakash Thanx for the solution!An awesome one! Adarsh Kumar · 11 months, 2 weeks ago

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I'm getting \(x=7,y=37\) Abdur Rehman Zahid · 11 months, 2 weeks ago

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@Abdur Rehman Zahid Could you please explain how? Adarsh Kumar · 11 months, 2 weeks ago

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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. Rajen Kapur · 11 months, 2 weeks ago

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@Rajen Kapur Well,thank you sir,a nice solution! Adarsh Kumar · 11 months, 2 weeks ago

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What have you tried? Where are you stuck? Calvin Lin Staff · 11 months, 2 weeks ago

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@Calvin Lin I tried a bit of modular arithmetic and tried factorising the expression but nothing worked.A hint sir? Adarsh Kumar · 11 months, 2 weeks ago

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@Adarsh Kumar Well, for one, since \( x, y > 0 \), we see that \( x, y < 2000 \) and so we have finitely many possibilities to try. Calvin Lin Staff · 11 months, 2 weeks ago

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X=7 and Y=37. Siva Meesala · 11 months, 2 weeks ago

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@Dev Sharma @Nihar Mahajan Adarsh Kumar · 11 months, 2 weeks ago

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