<s>[INST] Find all $x,y \in \mathbb{N}$ such that $x^2+y^2-8x=9$ This is a math Olympiad problem.
my attempt :
By solving the quadratic equation in $x$, i’ve got this:
$$x=\frac{8 \pm \sqrt{100-4y^2}}{2}$$
And from this it’s easy to see that $100-4y^2$ it’s a perfect square and it’s divisible by $4$.
$$\cases{100-4y^2=m^2 \\ 100-4y^2=4n }$$
Dividing the first equation by the second we get:
$$m^2=4n \iff m=2\sqrt{n} \implies n=a^2$$
By plugging this value to $100-4y^2=4n$:
$$100-4y^2=4a^2 \iff 25=a^2+y^2$$
so :$y \in \{0,5,3,4\}$
And by plugging these values to $x=\frac{8 \pm \sqrt{100-4y^2}}{2}$, we get:(we ignore the cases where $x \notin \mathbb N$)
$$(x,y) \in \{(9,0),(4,5),(8,3),(0,3),(1,4),(7,4)\}$$
I think that my solution is a bit long, so if is there a short solution please post it.
 [/INST] From the equation, we have
$$(x-4)^2+y^2 = 25$$
Hence, we can deduce that  $(|x-4|,y) = (0,5),(3,4),(4,3)$ or $(5,0)$.
The solution is then
$$(x,y) = (4,5),(7,4),(1,4),(0,3),(8,3),(9,0)$$
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