# Integer roots equation

### Integer roots equation

Solve for integers (x,y) : $$4x^2+y^2+5xy+1=3(3x+y)$$
dduclam

Posts: 36
Joined: Sat Dec 29, 2007 10:42 am
Location: HUCE-Vietnam
Reputation: 3

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

icb

Posts: 2
Joined: Wed May 14, 2008 7:55 am
Reputation: 0

### Re: Integer roots equation

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

Let us examine the equation towards y:

$$y^2+(5x-3)y+4x^2-9x+1=0$$

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

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

That shows that there are no such x and y martosss

Posts: 22
Joined: Fri Jun 06, 2008 5:16 am
Reputation: 6

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

Let us examine the equation towards y:

$$y^2+(5x-3)y+4x^2-9x+1=0$$

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

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

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:
$$4x^2+y^2+5xy-9x-3y+2=0$$
<=> $$x(4x+y-1)+y(4x+y-1)-2(4x+y-1)=1$$
<=> $$(x+y-2)(4x+y-1)=1$$
Consider two case $$\left\{\begin{array}{l}x+y-2=1\\4x+y-1=1\end{array}\right.$$ and $$\left\{\begin{array}{l}x+y-2=-1\\4x+y-1=-1\end{array}\right.$$
we've the answer dduclam

Posts: 36
Joined: Sat Dec 29, 2007 10:42 am
Location: HUCE-Vietnam
Reputation: 3

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. MM

Posts: 82
Joined: Tue Jul 22, 2008 7:36 am
Location: Bulgaria
Reputation: 7

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

$$\begin{tabular}{|1}k+3x+1=2\\k-3x-1=2\end{tabular}\cup \begin{tabular}{|1}k+3x+1=\pm 4\\k-3x-1=\mp 1\end{tabular}$$

Now from the first system equasions we have $$x=\frac{1}{3}\notin D.S.$$
We add the two equasions from the second system and we get $$2k=5 \Leftrightarrow k=2.5\notin D.S.$$ Thus it has no solutions.
martosss

Posts: 22
Joined: Fri Jun 06, 2008 5:16 am
Reputation: 6

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

Math Tutor

Posts: 410
Joined: Sun Oct 09, 2005 11:37 am
Reputation: 28

### Re: Integer roots equation

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

### Re: Integer roots equation

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

Put in
Code: Select all perfectmath

Posts: 25
Joined: Fri Jun 29, 2012 12:21 am
Location: Vietnam
Reputation: 1