# How to determine if $2+x+y$ is a factor of $4-(x+y)^2$?

I know it is a factor but how could have I determined that it was? Feel free to link whatever concept is needed than solve it. Studying for clep and it's one of the practice problems. When I expand it I get nonsense.

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Use difference of squares: $$4 - (x+y)^2 = 2^2 - (x+y)^2 = (2 - (x+y))(2+(x+y)).$$

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Essential hint:

$$a^2-b^2=(a-b)(a+b)$$

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@amWhy: You warmed me up Amy. Thank you for your supportive job. :-) –  Babak S. Jan 17 '14 at 13:31

If one did not know/see the difference of squares, then one could apply the Factor Theorem:

$\ (x\!+\!y\!+\!2)\!\mid\! f(x)\,$ by $\,f\,$ has root $\,x=-(y\!+\!2)\,\ [\,\Rightarrow\, \color{#c00}{x\!+\!y = -2}\,\Rightarrow\, f(x) = (\color{#c00}{x\!+\!y})^2\!-4 =0]$

Or, essentially equivalently, one could employ modular arithmetic

$\ (x\!+\!y\!+\!2)\!\mid\! f(x)\$ since,  modulo $\ x\!+\!y\!+\!2\!:\,\ \color{#c00}{x\!+\!y\equiv -2},\$ therefore $\ f(x) = (\color{#c00}{x\!+\!y})^2\!-4 \equiv 0$

Notice how both methods reduce the divisibility test to a rote arithmetical calculation, eliminating the need for any specialized knowledge about factorization methods such as difference of squares. Furthermore, both methods apply in much more general situations where the dividend is not a difference of squares (or of some other special form whose factorization is known).

The reason that this works so effectively is that replacing the divisibility relation $\ m\mid n\$ by the equivalent equation $\ n\equiv 0\pmod m\,$ allows us to use our well-practiced equational logic of (congruence) arithmetic, replacing arguments of sums and products by congruent numbers, using fractional arithmetic, etc. We have well-honed intuition about such arithmetic because it is very similar to ordinary integer arithmetic. But we have far less intuition about about "arithmetic" of divisibility relations. So divisibility problems often simplify on conversion to arithmetical form.

One ubiquitous example of such simplifcations are certain classes of proofs by induction. When interepreted arithmetically they reduce to trivialities such as $\ 1^n\equiv 1,\$ and $\ A\equiv a\,\Rightarrow A^n\equiv a^n,\$ e.g. see this recent answer.

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@Downvoter If something is not clear, then please feel welcome to ask questions and I will be happy to elaborate. –  Bill Dubuque Jan 16 '14 at 16:32
Tackling the downvoter by this small plus. –  Babak S. Jan 17 '14 at 13:32

The solution's already in the other answers, but in many cases, as in this one, you can try some substitution:

$$t:=x+y\implies\;\text{is}\;\;2+t\;\;\text{a factor of}\;\;4-t^2\;?$$

and now all depends on you remembering the high school algebra's slick formula, namely difference of squares:

$$4-t^2=(2^2)-t^2=(2-t)(2+t)\;\;\text{and etc.}$$

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