Integer roots equation

Integer roots equation

Postby dduclam » Tue Mar 25, 2008 2:18 am

Solve for integers (x,y) : [tex]4x^2+y^2+5xy+1=3(3x+y)[/tex]
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Postby icb » Wed May 14, 2008 8:07 am

The problem is very interesting.
Could you post any direction, please?

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Re: Integer roots equation

Postby martosss » Fri Jun 06, 2008 5:46 am

dduclam wrote: Solve for integers (x,y) : [tex]4x^2+y^2+5xy+1=3(3x+y)[/tex]


Let us examine the equation towards y:

[tex]y^2+(5x-3)y+4x^2-9x+1=0[/tex]

[tex]D=25x^2-30x+9-16x^2+36x-4=9x^2+6x+5=(3x+1)^2+4[/tex]
Now [tex]D[/tex] should be an integer, because [tex]y[/tex] is an integer, so [tex](3x+1)^2+4=k^2,\: x,k\in Z[/tex] and I think this has no integer solutions, because the distance between the integer exact squares grows constantly with 2

[tex]0^2=0,[/tex]
[tex]1^2=1,[/tex]
[tex]1-0=1, 1-0=1[/tex]
[tex]2^2=4, 4-1=3 = 1+2[/tex]
[tex]3^2=9, 9-4=5 = 3+2 e.t.c.[/tex],
so the only solution is when [tex](3x+1)^2=0,\: that\: is\: x=-\frac{1}{3}\notin Z[/tex]

That shows that there are no such x and y ;)

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Postby dduclam » Sun Jun 08, 2008 11:59 pm

martosss wrote:
dduclam wrote: Solve for integers (x,y) : [tex]4x^2+y^2+5xy+1=3(3x+y)[/tex]


Let us examine the equation towards y:

[tex]y^2+(5x-3)y+4x^2-9x+1=0[/tex]

[tex]D=25x^2-30x+9-16x^2+36x-4=9x^2+6x+5=(3x+1)^2+4[/tex]
Now [tex]D[/tex] should be an integer, because [tex]y[/tex] is an integer, so [tex](3x+1)^2+4=k^2,\: x,k\in Z[/tex] and I think this has no integer solutions, because the distance between the integer exact squares grows constantly with 2

[tex]0^2=0,[/tex]
[tex]1^2=1,[/tex]
[tex]1-0=1, 1-0=1[/tex]
[tex]2^2=4, 4-1=3 = 1+2[/tex]
[tex]3^2=9, 9-4=5 = 3+2 e.t.c.[/tex],
so the only solution is when [tex](3x+1)^2=0,\: that\: is\: x=-\frac{1}{3}\notin Z[/tex]

That shows that there are no such x and y ;)


Your solution's not close,martosss,although the answer's true ;)

We can solve as follows:
[tex]4x^2+y^2+5xy-9x-3y+2=0[/tex]
<=> [tex]x(4x+y-1)+y(4x+y-1)-2(4x+y-1)=1[/tex]
<=> [tex](x+y-2)(4x+y-1)=1[/tex]
Consider two case [tex]\left\{\begin{array}{l}x+y-2=1\\4x+y-1=1\end{array}\right.[/tex] and [tex]\left\{\begin{array}{l}x+y-2=-1\\4x+y-1=-1\end{array}\right.[/tex]
we've the answer :)

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Postby MM » Tue Jul 22, 2008 12:25 pm

Actually martosss hadn't finished his solution. After (3x+1)2+4=k2, we have to solve some system equations. (k+3x+1)(k-3x-1)=4.

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Postby martosss » Mon Sep 15, 2008 2:05 pm

That's right, MM, I just can't finish them the right way, so the system is:

[tex]\begin{tabular}{|1}k+3x+1=2\\k-3x-1=2\end{tabular}\cup \begin{tabular}{|1}k+3x+1=\pm 4\\k-3x-1=\mp 1\end{tabular}[/tex]

Now from the first system equasions we have [tex]x=\frac{1}{3}\notin D.S.[/tex]
We add the two equasions from the second system and we get [tex]2k=5 \Leftrightarrow k=2.5\notin D.S.[/tex] Thus it has no solutions.
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Postby Math Tutor » Thu Jul 29, 2010 9:41 am

The form use LaTeX.
You can use Latex help buttons down.

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Re: Integer roots equation

Postby Guest » Thu Jan 05, 2012 6:56 am

hi ,
how we will use latex in this forum ................Please post some solutions .
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Re: Integer roots equation

Postby perfectmath » Sat Jun 30, 2012 3:36 am

Guest wrote:hi ,
how we will use latex in this forum ................Please post some solutions .

Put in
Code: Select all
[tex][/tex]

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