# Another equation

1. Jan 30, 2007

### murshid_islam

how to find the integer solutions of the following equation:
$$x^3 + x^{2}y + xy^2 + y^3 = 8(x^2 + xy + y^2 + 1)$$

after simplifying, i got,
$$(x^2 + y^2)(x + y - 8) = 8(xy + 1)$$

from this, i can see that x and y are both odd or both even. but what should i do next. i tried putting x = 2p and y = 2q but that's not going anywhere.

2. Jan 31, 2007

3. Feb 1, 2007

### tehno

Murshid ,my man,here it is

Obviously $x\equiv y (mod2)$.
WLOG,put x=u+v,y=u-v and it is sufficient to consider:
$$(u^2+v^2)(u-4)-2(u^2-v^2+1)=0$$
• if v=0,then u3-6u2-2=0
This equation has no solutions in $\mathbb{Z}$
• if u=0,then v2+1=0 has no solutions in $\mathbb{Z}$
• if $|u|\geq 1,|v|\geq 1$ ,then, $u^2+v^2>|u^2-v^2+1|$ with consequence that $|u-4|=1$ must hold.
• if u=3 ,than v2-29=0 has no solutions in $\mathbb{Z}$
• if u=5 ,than v2-9=0 has solutions v1=3,v2=-3

Thus, the diophantine equation we started with, has only two solutions in x,y :
S={(2,8),(8,2)}

Last edited: Feb 1, 2007
4. Feb 2, 2007

### murshid_islam

i didn't understand a few things. for example,

where is this used?

how did you get this?

5. Feb 3, 2007

### murshid_islam

ok, i understood this much:
if x = u+v and y = u-v, then x - y = 2v. therefore,
$$x - y \equiv 0 (\bmod \\ 2)$$
$$x \equiv y (\bmod \\ 2)$$

but what about the second one i mentioned in my last post? what i want to know is how did tehno get:

Last edited: Feb 3, 2007
6. Feb 3, 2007

### tehno

by splitting it in two cases:

a) $|u|\geq|v|\geq 1\rightarrow u^2-v^2+1\leq u^2<u^2+v^2$

b) $|v|>|u|\geq 1\rightarrow |u^2-v^2+1|=v^2-u^2-1<v^2<u^2+v^2$

Joining these two in other words means that:

$$|u-4|=|2(u^2-v^2+1):(u^2+v^2)|<2\longrightarrow|u-4|=0\ or \ |u-4|=1$$
which I specified in the post .

Last edited: Feb 3, 2007
7. Feb 4, 2007

### murshid_islam

is this how you got a) and b) ?:
$$|u| \geq |v| \geq 1$$

$$u^2 \geq v^2 \geq 1$$

$$-u^2 \leq -v^2 \leq -1$$

$$0 \leq u^2 - v^2 \leq u^2 -1$$

$$1 \leq u^2 - v^2 + 1 \leq u^2$$

now $$u^2 - v^2 + 1 \leq u^2 < u^2 + v^2$$

again,
$$|v| > |u| \geq 1$$

$$v^2 > u^2 \geq 1$$

$$-v^2 < -u^2 \leq -1$$

$$0 < v^2 - u^2 \leq v^2 - 1$$

$$1 < v^2 - u^2 + 1 \leq v^2$$

now,
$$v^2 - u^2 - 1 < v^2 - u^2 + 1 \leq v^2 < u^2+ v^2$$
$$v^2 - u^2 - 1 < v^2 < u^2 + v^2$$

but how did you get the following: