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Is f(m,n) = m^2 - 4 an onto function for a function that goes from Z x Z -> Z (Where Z means the set of all integers).
I think it is an onto function, but I am not sure how to go about proving it. Can I set an integer y to = m^2 - 4 and then show that if I set n = 0, it will work...?

(Sorry for any formatting errors).

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  • $\begingroup$ What value of $m$ would allow you to have $m^2-4=2$? Also, you write $m^2-4$, there is no $n$ in that... might as well have been $\Bbb Z\to\Bbb Z$ $\endgroup$ – JMoravitz Feb 2 '17 at 2:48
  • $\begingroup$ Oh m would have to be the square root of 6 in this case, which does not fall under the requirement of Integers. So it is not onto... $\endgroup$ – Alex Feb 2 '17 at 2:50
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It is not an onto function.

$$f(m,n) \geq -4$$

Hence there is no $(m,n)$ such that $f(m,n)=-5$ but $-5 \in \mathbb{Z}$

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