# $(x+1)^2 + (y+1)^2 + xy(x+y+3)=2$

I've came across this problem some hours ago and, although it looks (and possibly is) just some algebra calculus, I can't get on the right track.

Find $x$, $y$ integers such that $$(x+1)^2 + (y+1)^2 + xy(x+y+3)=2$$

Some basic algebra will lead to:

$$(x+y)^2 + x(xy+2) + y(xy+2) + xy = 0$$ $$(x+y)(xy+x+y+2)=-xy$$ Now, some divisibility properties should end the problem. Well, maybe my approach isn't the right one, but I'm just stuck at this point.

A piece of advice or a hint would be apreciated.

• There is no $z$ anywhere! – ÍgjøgnumMeg Mar 17 '16 at 18:45
• @Ed_4434 sorry, I've fixed it. – scummy Mar 17 '16 at 18:45
• @daniel your solution does not satisfy the equation. – SchrodingersCat Mar 17 '16 at 19:09
• $(-2+1)^2+(0+1)^2+(-2)0(-2+0+3)=2$? – daniel Mar 17 '16 at 19:13
• Hint: $(x+1)(y+1)(x+y+1) = 1$. – achille hui Mar 17 '16 at 19:16

HINT:

$$(x+y)(x+y+xy+2)=-xy$$

There exist no solution for positive integers $x,y$, since L.H.S. is +ve and R.H.S. is -ve.

What if both -ve or one +ve, one -ve?

Of course $(0,0)$, $(0,-2)$, $(-2,0)$ are solutions, and these are the only ones with $x=0$ or $-2$. There are no solutions for $x = -1$. If $x \ge 1$ or $x \le -3$, the left side is a quadratic in $y$ whose values at $y=-1$ and $y=0$ are $1$ and $(x+1)^2 + 1 > 2$, so there is a solution (non-integer of course) between $y=-1$ and $y=0$. Similarly, the values at $y=-x-1$ and $y=-x-2$ are $1$ and $(x+1)^2 + 1$, so there is a different non-integer solution between $-x-1$ and $-x-2$. But a quadratic has only two roots, so we conclude there are no integer solutions with $x \ge 1$ or $x \le -3$.

Here's a hint: Try factoring the LHS of the equation. To see how you can do that, you need to pick up your pen and expand the powers and write down all the nine terms.

You might have to carry a constant to the other side of the equation.

The answers are: $$(0,0),(0,-2) \text{ and } (-2,0)$$

Replace $x+1$ by $a$ and $y+1$ by $b$ and you get

$$2=a^2+b^2 +(a-1)(b-1)(a+b+1)+1=a^2b+b^2a-ab =ab(a+b-1)+1.$$

Since $x$ and $y$ are integers, $a,b$ are integers too. Therefore $a,b,a+b-1 \in \{-1,1\}$ and so $a,b \in \{-1,1\}$ and $a+b \in \{0,2\}$. If $a=b=-1$ we get a contradiction. Since the starting equation was symmetric in $a$ and $b$, lets assume $a=1$. Then $b=1$ and $b=-1$ are still valid.

So the only solutions for $(x,y)$ are $(0,0), (2,0), (0,2)$.

The origin $(0,0)$, $(0,-2)$ and $(-2,0)$ are the only diophantine (integer) solutions. Simplifying the equation and taking note of the symmetry about $y=x$ ($x$ & $y$ are interchangeable, so we will eventually only see symmetric terms such as xy and x+y), we see that \begin{align} 0 &= (x+1)^2 + (y+1)^2 + xy(x+y+3) - 2 \\&= x^2+2x + y^2+2y + xy(x+y+3) \\&= x^2+y^2 + 2(x+y) + xy(x+y+3) \\&= (x+y)^2 + 2(x+y) + xy(x+y+1) \\&= (x+y+1)^2 + xy(x+y+1) - 1 \\&= u^2 + 2uv - 1 \\&= \left(u+v\right)^2 - \left(1+v^2\right) \end{align} for $u=x+y+1$ & $v=xy/2$, a quadratic equation with solution $u=-v\pm\sqrt{1+v^2}$, or $2(x+y+1)=-xy\pm\sqrt{4+(xy)^2}$. Next, note that $x$ & $y$ can't both be odd, because if they were, then we would have $u$ odd $\implies 2v=1-u^2$ even $\implies v\in\mathbb{Z}\implies xy$ even, leading to a contradiction. But then $2|xy$ so that $v\in\mathbb{Z}$. However, then the radical $\sqrt{1+v^2}$ can only be an integer if $v^2$ is both a perfect square and one less than a perfect square. This can only happen when $v=xy=0$, i.e., when at least one of $x,y$ is zero as noted above.

Another approach: \begin{align}0 &=(x+y)^2+x(xy+2)+y(xy+2)+xy\\ &=(x+y)^2+(x+y)(xy+2)+xy\\ &=u^2+ut+t-2 \end{align} for (integers) \begin{align}t&=xy+2\\u&=x+y\end{align} so that $$t(1+u)=2-u^2$$ or $$t=\frac{2-u^2}{1+u}=\frac{1-u}{1+u}+\frac1{1+u}.$$ For $t$ to be an integer, we must have $u+1=\pm1$, or $u=x+y\in\{0,-2\}$. In both cases, $t=2\implies xy=0\implies x$ or $y$ is zero. This only admits the three solutions $(x,y)=(0,0),(0,-2),(-2,0)$.