ROMANIAN MATHEMATICAL OLYMPIAD

For discussing Olympiad Level Number Theory problems
MATHPRITOM
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ROMANIAN MATHEMATICAL OLYMPIAD

Unread post by MATHPRITOM » Tue Feb 07, 2012 12:39 am

Find all the positive integer solution of the equation $ (x+y)^2+3x+y+1= z^2$.

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Nadim Ul Abrar
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Re: ROMANIAN MATHEMATICAL OLYMPIAD

Unread post by Nadim Ul Abrar » Tue Feb 07, 2012 3:43 pm

My Solution
$z^2-(x+y)^2=(z-x-y)(x+y+z)=3x+y+1$
$3x+y+1$ is positive
so that $(z-x-y) \geq 1$
so we can write that
$ z=x+y+k$

SO that $k(2x+2y+k)=3x+y+1$
For $k=2$
we have $4x+4y+4=3x+y+1$
whish is impossible for positive integers . thus for $k=3,4...$
So that only possible value of $k$ is $1$

Now $2x+2y+1=3x+y+1$
Or $x=y$

so that all positive integer triple of $(x,y,z)$ is $(a,a,2a+1)$ ; $\forall a\in \mathbb{Z}^+$
$\frac{1}{0}$

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