# How do I solve a question like this?

If $p$ and $n$ are integers such that $p>n>0$ and $p^2-n^2=12$, which of the following can be the value of $p-n$?

I.  1
II.  2
III.  4

(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III

The first thing I did was to simplify $p^2-n^2=12$:

$$\displaystyle\large(p+n)(p-n)$$

Then I listed the factors of 12, one of which could be the value of $p+n$.

$$\displaystyle\large\{(1, 12), (2, 6), (3, 4)\}$$ And then I substituted them for the $p+n$:

$$1(p-n) = 12\\ 12(p-n) = 12\\ 2(p-n)=12\\ 6(p-n)=12\\ \cdots$$

Then I got the solutions that matched the options: $1$, $2$, and $4$ and so I choose (E), but I found out that it wasn't the answer. What did I do wrong?

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Multiples of $12$ are $12, 24, 36, \ldots$ – Ross Millikan Oct 11 '12 at 16:29
Just out of curiosity... did you just post a GRE question? And is that a good thing? – kcrisman Oct 31 '12 at 1:29
What is GRE? This is a question from my SAT practice exam. – David Oct 31 '12 at 11:42

Well, you have the three pairs of factors of 12. The larger of the two numbers is $p+n$ while the smaller one is $p-n$

(You can be certain that that's true as $p>n>0$.)

Now, if $p+n = 12$ and $p-n = 1$, you get: $p=\frac{13}{2}$ and $n=\frac{11}{2}$

If you use 6 and 2 you get: $p = 4$ and $n = 2$

And for 4 and 3 you get: $p = \frac{7}{2}$ and $n = \frac{1}{2}$

Obviously only $p = 4$ and $n = 2$ are integers. so $p - n = 2$

Hint: $p+n$ also depends on $p,n$... For a proper solution, notice that if $(p-n)(p+n)$ is even, then either both $p+n$ and $p-n$ are even, and that $p+n-(p-n)$ is at least two if $n>0$.
Your set is factorizations of $12$. It is exactly what you want to do. Then you have three possibilities: $(p+n=12, p-n=1), (p+n=6,p-n=2), (p+n=4,p-n=3)$. Each one is a pair of simultaneous equations you can solve for $p$ and $n$ and see what the possibilities for $p-n$ are.