Finding the positive integer solutions $(x,y)$ of the equation $x^2+3=y(x+2)$

Finding the positive integer solutions $(x,y)$ of the equation $x^2+3=y(x+2)$

Source: Art of Problem Solving Vol. 2

Any help would be appreciated.

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Considering what $gcd(x+2,x^2 + 3)$ could be would be a help. –  Geoff Robinson Dec 31 '11 at 18:02
What have you tried? –  lhf Dec 31 '11 at 18:53

Divide the polynomial $x^2+3$ by the polynomial $x+2$. We get $$\frac{x^2+3}{x+2}=x-2+\frac{7}{x+2}.$$ For an integer $x$, the value of $x-2+\frac{7}{x+2}$ is an integer if and only if $x+2$ divides $7$. The only positive integer $x$ for which this is true is given by $x=5$.
Comment: If we are interested in integer solutions, not necessarily positive, note that $x-2+\frac{7}{x+2}$ is an integer also at $x=-1$, $x=-3$, and $x=-9$.
HINT $\$ Factor Theorem $\rm\:\Rightarrow\ x+2\ |\ f(x)\ \iff\ x+2\ |\ f(-2)\:,\$ for all polynomials $\rm\ f(x)\in \mathbb Z[x]$
So, in particular, for $\rm\:x\in \mathbb N\:,\ \ x+2\ |\ x^2+3\ \iff\ x+2\ |\ 7\ \iff\ x\ =\ \ldots$