Equation - what first? I have this equation:
$$
  (x+y)(x^x + y^y) = 2009.
$$
I must designate all pairs of integers satisfying the equation.
What first? I tried multiply brackets , but to no avail
 A: There are no solutions with $x,y\in\Bbb Z$.
If $x\ge -1$ and $y\ge -1$, then $x^x,y^y$ are also integers and we need only check the few possible factorizations of $2009=7^2\cdot 41$. 
The values of $z^z$ for $z=-1,0,1,2,\ldots$ are
$$ -1,1,1,4,27,256$$
and from then on  all values are $\gg 2009$.
No two of these numbers sum up to one of the factors $\pm1,\pm7,\pm41,\pm49,\pm287,\pm2009$ of $2009$, hence no solution with $x,y\ge -1$.
If $x\ge-1$ and $y\le -2$ let $p$ be a prime dividing $y$, say $p^k\mid y$ with $k\ge 1$. Then $x^x+y^y$ is a rational number with smallest denominator a multiple of $p^{kp^k}$. Thus $x+y$ must be a multiple of $p^{kp^k}$.
This gives us a direct contradiction if $x=0$ and $k$ is choosen maximal, so we conclude  $x\ne0$. This also implies $p^k\mid x$, and so $$p^{kp^k}\le |x^x|\le |x^x+y^y|+1\le\frac{2009}{p^{kp^k}}+1.$$ One easily handles all possible cases from these constraints. In fact, $p\ge 5$ is ruled out immediately, and likewise $9\nmid y$ and $8\nmid y$.  The remaining cases $y=-2,-3,-4,-6$ are left as an exercise.
Remains the case $x\le -2$ and $y\le -2$.
As above, $p^k\mid y$ implies $p^k\mid x$ and vice versa.
This implies $|x|=|y|$ and hence also $x=y$. This leads to
$$4 = 2009\cdot |x|^{-(x+1)} \gg 2009$$
which is impossible
A: hint: $2009=7^2\times 41$ so keeping in mind that $x+y\lt x^x+y^y$ we're looking for integers $x$ and $y$ such that
$$(x+y)(x^x+y^y)=7^2\times 41$$
This means finding integers such that $x+y=41$ and $x^x+y^y=49$ or $x+y=7$ and $x^x+y^y=7\times 41$
A: The first step is to factor 2009.
Then, for every possibility of 2009=ab, try to satisfy $x+y=a and $x^y+y^x=b$.
A: Hint, 
you want to find all integers $x,y$ that solves
$$(x+y)(x^x+y^y)=2009$$
Since $x,y$ are integers so also $x+y$ and $x^x+y^y$ (I believe you know why).
I guess you can figure out that $2009=7^2\times 41$.
And hope that you know what to do from here.
