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A function $f:N->N$ is defined by $f(x) = x^2 + x + 1$. Is the function $f(x)$ One - One Onto (Bijection) ?

My Try :

For proving injection, I need to show that $f(x) = f(y)=> x=y$

So, $f(x) = f(y)=> (x-y)(x+y+1)=> x=y$ but one more value comes $x = -y-1$

Hence, I am getting $f(x)$ as many- one function.

Am I right here ?

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    Notice that $f(-1) = f(0) = 1$. – 2017-01-11
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    @AlexisOlson So, Am I right ? – 2017-01-11
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    @AlexisOlson I think the domain is only natural numbers so your counter example will not work. In fact it will be injective because $x+y+1 \neq 0$ when in $\mathbb{N}$. It is not onto because you cannot map to even numbers (because $(x(x+1)$ is even.) – 2017-01-11
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    @AnuragA What is the difference between $N$ and $Z$ ? – 2017-01-11
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    $\mathbb N$ is the set of positive (or sometimes non-negative) integers, $\mathbb Z$. For non-negative values, the function $f$ is injective, but not surjective. – 2017-01-11
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    @Garrick $\mathbb{N}$ is the set of natural numbers $\{1,2,3, \ldots\}$ (note some include $0$ as well). $\mathbb{Z}$ is the set of integers $\{0, \pm 1, \pm 2, \ldots \}$. – 2017-01-11
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    @AnuragA Thanks !! I did not know exactly this much difference . – 2017-01-11

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$$f(x)=x^2+x+1\\ x=1 \to f(1)=3 \\x=2 \to f(2)=7 \\x=3 \to f(3)=13 \\x=4 \to f(4)=21 \\,...$$ so it is one by one ,but not onto function from $N \to N$

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    Your argument for one-one is not a mathematical proof. Also why is not onto? – 2017-01-11