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# Help needed friends!

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

1 year, 10 months 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. $$\left( 7 , 37 \right)$$. · 1 year, 10 months ago

Thanx for the solution!An awesome one! · 1 year, 10 months ago

I'm getting $$x=7,y=37$$ · 1 year, 10 months ago

Could you please explain how? · 1 year, 10 months ago

$(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. · 1 year, 10 months ago

Well,thank you sir,a nice solution! · 1 year, 10 months ago

What have you tried? Where are you stuck? Staff · 1 year, 10 months ago

I tried a bit of modular arithmetic and tried factorising the expression but nothing worked.A hint sir? · 1 year, 10 months ago

Well, for one, since $$x, y > 0$$, we see that $$x, y < 2000$$ and so we have finitely many possibilities to try. Staff · 1 year, 10 months ago

X=7 and Y=37. · 1 year, 10 months ago