Find the image of a two variable function

[Synapse0x]

So on my first course of Algebra in university I have been asked to find the image of the following function:
f : Z x Z -> Z, f(x, y) = x² - y²

In high school I worked only with single variable functions and I could have found the image by drawing its graph, using limits or using some specific shortcuts on well known functions (like the quadratic one).
I really have no idea how should I find the image of this one. Actually I don't even know how to plot it, I think its graph is 3D. Can somebody point me in the right direction?

 

[caffeinemachine]

So on my first course of Algebra in university I have been asked to find the image of the following function:
f : Z x Z -> Z, f(x, y) = x² - y²

In high school I worked only with single variable functions and I could have found the image by drawing its graph, using limits or using some specific shortcuts on well known functions (like the quadratic one).
I really have no idea how should I find the image of this one. Actually I don't even know how to plot it, I think its graph is 3D. Can somebody point me in the right direction?

You are not required to sketch the graph of the function.

The image of a function is a subset of the target space of the function.

You need to say what is $Z$ exactly?

As of now, my answer would be $\text{Im}(f)=\{x^2-y^2: (x,y)\in Z\times Z\}$.

 

[Synapse0x]

Yes, in other words what is Z exactly.

Your current answer makes sense but how should I state it more accurately. I mean what your wrote is not wrong but it would be the same as saying that the image of g, where g : R -> R, g(x) = x² - 4x + 1, is {x² - 4x + 1 : x in R} instead of [-3, +infinity)

 

[caffeinemachine]

Yes, in other words what is Z exactly.

Your current answer makes sense but how should I state it more accurately. I mean what your wrote is not wrong but it would be the same as saying that the image of g, where g : R -> R, g(x) = x² - 4x + 1, is {x² - 4x + 1 : x in R} instead of [-3, +infinity)

I think by $Z$ you mean the set of all the integers.

In that case, the image of $f$ is $\{n\in\mathbb Z: n \text{ is odd }\}\cup \{n\in \mathbb Z: 4 \text{ divides } n\}$.

To see this, first note that $f(y+1,y)=2y+1$.

Thus all the odd integers are in the image.

Now suppose $f(x,y)$ is even.

Then $2$ divides $(x-y)(x+y)$.

But then $4$ divides $(x-y)(x+y)$ (why?)

So if an even integer is in the image, then it have to be divisible by $4$.

Finally note that $f(y+2,y)=4(y+1)$.

Thus all the integers which are divisible by $4$ are in the image.