$ x^2=y(2009-2y)\to 0\le y\le 1004$. $ 2009=7^2*41$. Because $ (\frac{-2}{7})=-1$ for $ y\not =0\mod 7$ $ x^2\not =-2y^2\mod 7$. Therefore $ y=7z,x=7t$ and $ t^2=z(287-2z)$, by analogy $ t=7x_1,z=7y_1$ or $ x=49x_1,y=49y_1$. We get $ x_1^2=y_1(41-2y_1).$ $ (\frac{-2}{41})=-1$, therefore $ 41|x_1,41|y_1$. Therefore unique solution $ x=y=0$.

Rust wrote: $ x^2 = y(2009 - 2y)\to 0\le y\le 1004$. $ 2009 = 7^2*41$. Because $ (\frac { - 2}{7}) = - 1$ for $ y\not = 0\mod 7$ $ x^2\not = - 2y^2\mod 7$. Therefore $ y = 7z,x = 7t$ and $ t^2 = z(287 - 2z)$, by analogy $ t = 7x_1,z = 7y_1$ or $ x = 49x_1,y = 49y_1$. We get $ x_1^2 = y_1(41 - 2y_1).$ $ (\frac { - 2}{41}) = - 1$, therefore $ 41|x_1,41|y_1$. Therefore unique solution $ x = y = 0$. $ x = 588,y = 784$

I have solved the problem in an effective but long way: Just write $ d=gcd(x,y)$ (exclude the case when $ x=y=0$) Then you write $ x=ad$ $ y=bd$ yielding to: $ da^2=b(2009-2bd)$ and since $ gcd(a,b)=1$ then $ bk=d$ Replacing we get the following expression: $ k(a^2+2b^2)=2009$ and then it follows that $ k$ divides $ 2009$ so we have only six possible values of $ k$ Then it is just casework ( some might be eliminated easily, working with divisibility properties ) You obtain the solution $ x=588$ $ y=784$ (Plus the trivial solution) Daniel

I am sorry $ (\frac{-2}{41})=1$, because $ 41=1\mod 8$ (I think $ 41=5\mod 8$). $ x_1^2=y_1(41-2y_1)$. Because $ (y_1,41-2y_1)=1$, ($ 0\le y_1\le 24$) we get for($ y_1>0$) $ x_1=x_2x_3,y_1=x_2^2,41-2y_1=x_3^2$ or $ x_3^2+2x_2^2=41.$ It had unique (non trivial positive) solution $ x_3=3,x_2=4$, therefore $ (x,y)=(0,0),(49*3*4,49*4^2)$.

Also $ (x,y)=(-588,784)$ satisfies the equation.

Note that $ 2009 = 41\cdot 49$, therefore, $ x^2 + 2y^2$ is divisible by $ 7$, and it is easy to ascertain that it is possible iff both $ x$ and $ y$ are divisible by $ 7$, so we may assume $ x = 7a, y = 7b$. Thus the problems reduces to finding all integer solutions of the following equation: \[ a^2 - 41b + 2b^2 = 0 \] Note that $ b$ is positive, otherwise we get a contradiction. On the other hand, due to the equation $ b(41 - 2b)$ must be positive. Furthermore, if $ b$ is divisible by $ 41$, then so is $ a$, but then $ a = b = 0$. So we may assume that $ b(41 - b)$ is a perfect square, but ever since $ (b,41 - 2b) = (b,41) = 1$, we conclude that $ 41 - 2b$ and $ b$ are both perfect squares, but it is possible only if $ b = 0$ ,or $ b = 16$. So all possible solutions are: $ (x,y): (0,0); ( - 588,784); (588,784)$.

@Rust

here is my solution: Since x and y are integers then the discriminant of the quadratic equation $ 2y^2-2009y+x^2=0$ which is $ 2009^2-8x^2$ should be a perfect square, so $ 2009^2-8x^2=a^2$ for some integer $ a$. Now grom here twice modulo 7 and after that modulo 3 we conclude that $ 3.7.7|x$ i.e. $ 147|x$ so $ x=174k$ for k-integer and $ 49|a$ so $ a=49b$ for $ b$-integer and then we reach: $ 41^2=72k^2+b^2$ and then $ k^2<25$. Of course since $ x$ i solution so is $ -x$ and that's the reason why we can look just for nonnegative $ x$. then $ 0\le k \le 4$ and 5 easy chechs give $ k=0$ and $ k=4$ for solutions. then $ x=0$ and so $ y=0$ or $ x=4.147=588$ and $ y=784$. Now adding and $ -x$ as solution all the solutions become (0,0);(588,784);(-588,784)

Erken wrote: Find all pairs of integers $ (x,y)$, such that \[ x^2 - 2009y + 2y^2 = 0 \] We rewrite it as $ 2x^2 + (2y - 2009)^2 = 7^4 \cdot 41^2$. Since $ - 2$ is not a quadratic residue $ \bmod 7$ we get $ 7^2 \mid x,2y - 2009 \iff 7^2 \mid x,y$. Then let $ x = 7^2 \alpha, y = 7^2 \beta$. So we just have to solve: $ \alpha^2 - 41\beta + 2\beta^2 = 0$. Which we rewrite as $ \alpha^2 = \beta(41 - 2\beta)$. It's easy to see that $ 41 \mid \alpha \Rightarrow \alpha = \beta = 0$, which is a solution. Now assume that $ 41 \nmid \alpha \Rightarrow 41 \nmid \beta \Rightarrow \gcd(\beta,2\beta - 41) = 1$. Thus there exists numbers $ n,m$ such that $ \beta = n^2$ and $ 41 - 2\beta = m^2$. (It's obvious from the first equation, that $ y \ge 0 \Rightarrow \beta \ge 0$). Then $ m^2 + 2n^2 = 41 \Rightarrow (|m|,|n|) = (3,4) \Rightarrow \beta = 16$. Then $ \alpha^2 = 16 \cdot 9 = 12^2$ so $ \alpha = \pm 12$. And thus $ (x,y) = (\pm 2^2 \cdot 3 \cdot 7^2, 2^4 \cdot 7^2)$ or $ (x,y) = (0,0)$ are the only solutions.

let x,y>0 x^2 give 0,1,2,4 mod 7 and 2*y^2 give 0,1,2,4 mod 7. 7|2009=> it's easy to see 7|x,y. since 49|2009=> 49|x,y let x=49*a,y=49*b. a^2-2*b^2=41b <=> a^2=b*(41-2*b) <=>let gcd(b,41-2*b)=d if d>1 then d|b =>d|41=>d=41 since d|b then b>=d=41 ,it means 41-2*b<0 contr. d=1 =>b is square.let b =p^2 and 41-2*p^2=q^2>=0 41>2*p^2 <=>20>p^2 <=> 4>=p p=1 41-2=39 not square p=2 41-8=33 not square p=3 41-18=23 not square p=4 41-32=9 square b=16,q^2=9 =>a^2=144 =>a=+12,-12 x=588,-588 y=784 if x=o then y=o if y=0 then x=o solution (x,y)=(588,784),(-588,784),(0,0)

$x^2 - 2009y + 2y^2 =0$ $\implies$ $x^2 + 2y^2=2009y$ and $x^2=7k$ $x^2=7k+1$ $x^2=7k+2$ $x^2=7k+4$ and $2y^2=7k_1$ $2y^2=7k_1+1$ $2y^2=7k_1+2$ $2y^2=7k_1+4$ $\implies$ $x^2=49k$ and $2y^2=49k_1$ $\implies$ $7$/$x$ and $7$/$y$ (2009=41 x 7 x 7 ) We get $x=49m$ and $49=7n$ $\implies$ $m^2 - 41n + 2n^2 = 0$ $\implies$ $m^2=(41-2n)n$ $(41-2n,n)=(41,n)=1$ ($m^2$>$0$ $\implies$ $n<41$) $\implies$ $n=p^2$ and $41-2n=q^2$ $\implies$ [$n=0$ and $m=0$] and [$n=16$ and $m=12$ $\implies$ $x=0;y=0$ and $x=588$ and $y=784$ $O.Y.SH.$

Wonderful

Reducing the equation modulo $7$ implies that $7$ divides both $x$ and $y$. Letting $x = 7x_{1}$ and $y = 7y_{1}$ for integers $x_{1}$ and $y_{1}$ yields the equation $x_{1}^2 - 287y_{1} + 2y_{1}^2 = 0$, which similarly to the first one implies that $x_{1} = 7x_{2}$ and $y_{1} = 7y_{2}$ for integers $x_{2}$ and $y_{2}$. Finally, we get to the equation $x_{2}^2 - 41y_{2} + 2y_{2}^2 = 0$. This is simple enough to solve directly by rewriting it as $x_{2}^2 = y_{2} (2y_{2} - 41)$. As the left-hand side is non-negative, we get $0 \leq y_{2} \leq 20$ and also $\gcd(y_{2}, 2y_{2} - 41) = 1$ for $y_{2} > 0$, so either $(x,y) = (0,0)$, or $y_{2} = z^2$ and $2y_{2} - 41 = t^2$ for integers $z,t \geq 0$. Here $t$ must be odd, so $t \in \{1,3,5\}$ and a direct check leads to the solutions $(x,y) = (\pm 588, 784)$.