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We have this equation over real numbers: $ \lfloor x^2 \rfloor = x+6$ . How we can solve it without guessing the answers?

My try: I tried to solve it in several intervals and found $-2$ and $3$ as answers but I can't deduct that there isn't any solution in other intervals .

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    Well, since $x+6$ is an integer iff $x$ is, we get that, for a solution, we must actually have $x^2=x+6$. – 2017-01-13
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    In general for problems like this (if, for instance, the $6$ were $6.2$) from $\lfloor x^2 \rfloor = x+6$, you can deduce that $x+5 \le x^2 < x+6$ (for $x + 5 > 0$; for the signs, you have to make up other pairs of inequalities; the messiest is when the floor is zero.). In this case, though, @lulu's answer is clearly a winner. – 2017-01-13
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    Yes , It's impossible sometimes to make up pair of inequalities . Can you provide a good reference to study how we can make up these pairs for solving floor equations? – 2017-01-13
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    @S.H.W learn the definition of the floor function. It will become obvious where the inequality comes from... – 2017-01-13
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    @TheGreatDuck Can you provide some examples ? – 2017-01-13
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    @S.H.W it's not about examples. Go look up the definition of floor. It's literally self-explanatory. – 2017-01-13
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    @S.H.W simply put $\lfloor x \rfloor = m$ if and only if $m \leq x < m+1$ https://en.wikipedia.org/wiki/Floor_and_ceiling_functions#Equivalences – 2017-01-13

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Since LHS is a integer that implies $x+6$ is a integer which then implies that $x$ is a integer.Since $x\in\Bbb{Z}$ we know that $\lfloor x^2\rfloor=x^2$ Hence we have that $$x^2-x-6=0\\x_{1,2}=-2,3$$ Just a note if one of the solutions wasn't a integer then that soulution wouldn't satisfy the equation.

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    Brilliant! Thanks – 2017-01-13