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

Note by Adarsh Kumar
4 years, 1 month ago

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To make this problem much simpler, let us make a substitution x=u+v2x = \dfrac{u+v}{2} and y=uv2y=\dfrac{u-v}{2}.

So, the given equation converts into

u2+2u+(v1994)=0u^2 + 2u + \left( v- 1994 \right) = 0

Step-1: Let us first find the lower bound and upper bound of vv. It is clear that u=x+yu = x+y and v=xyv = x-y. So, u>vu > v using this fact we get

1994=u2+2u+v>v2+3vv2+3v1994<0\begin{aligned} 1994 = u^2 + 2u +v > v^2 + 3v \\ v^2 + 3v - 1994 < 0 \end{aligned}

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

Step-2: Since from the equation u2+2u+(v1994)=0u^2 + 2u + \left( v- 1994 \right) = 0. We get, u=1±1995vu= -1 \pm \sqrt{1995-v}. But since x>0x>0 and y>0y>0 it follows that u>0u>0. So, u=1+1995vu= -1 + \sqrt{1995-v}.

Now for positive integral value of uu, 1995v1995-v should be a perfect square. Now from the step-1 1995v[1909,2087]1995-v \in \Big[ 1909, 2087 \Big] . There are only two values of 1995v1995 -v, they are 20252025 and 19361936.

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

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

Surya Prakash - 4 years ago

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Thanx for the solution!An awesome one!

Adarsh Kumar - 4 years ago

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@Dev Sharma @Nihar Mahajan

Adarsh Kumar - 4 years, 1 month ago

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X=7 and Y=37.

siva meesala - 4 years, 1 month ago

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What have you tried? Where are you stuck?

Calvin Lin Staff - 4 years ago

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

Adarsh Kumar - 4 years ago

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

Calvin Lin Staff - 4 years ago

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(x+y)2+(x+y)=19942x(x+y)^2 + (x + y) = 1994 - 2x(x+y)(x+y+1)=19942x\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 - 4 years ago

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Well,thank you sir,a nice solution!

Adarsh Kumar - 4 years ago

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

Abdur Rehman Zahid - 4 years ago

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Could you please explain how?

Adarsh Kumar - 4 years ago

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