<s>[INST] Find natural number $x,y$ satisfy $x^2+7x+4=2^y$ 
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$

My try: I think $(x;y)=(0;2)$ is only solution. So I try prove $y\ge3$ has no solution, by $(x+1)(x+6)-2=2^y$.
So $2\mid (x+1)(x+6)$, but this is wrong. Done.
This is wrong
Anyone has an idea? Please help, thank you!
 [/INST] If we take mod 3 and $y=2z+1$ then $2^y\equiv -1$ but $$x^2+7x+4 \equiv  0,1$$ a contradiction, so $y=2z$. Let $2^z = t$, then we have

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*If $x>0$ then

$$(x+2)^2\leq \underbrace{x^2+7x+4}_{t^2} \leq (x+4)^2 \implies t = x+3\implies x=5$$

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*If $x<0$ then $x=-n$ with positive $n$. So $$(n-4)^2\leq \underbrace{n^2-7n+4}_{t^2} \leq (n-2)^2 \implies t = n-3\implies n=-5$$so no solution.


*If $x=0$ then $y= 2$.
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