The set $\{1,4, n\}$ has the property that when any two distinct elements are chosen and 2112 is added to their product, the result is a perfect square. If $n$ is a positive integer, the number of possible values for $n$ is
(A) 8
(B) 7
(C) 6
(D) 5
(E) 4
According to the requirements, the three quantities $4+2112=2116, n+2112$, and $4 n+2112$ must be perfect squares.

Now $2116=46^{2}$ (a perfect square), so the conditions say that we must have $n+2112=x^{2}$ and $4 n+2112=y^{2}$ for some positive integers $x$ and $y$. Since $n$ is a positive integer, then each of $x$ and $y$ must be greater than 46 .

We now have 3 variables, but can easily eliminate the $n$ by manipulating the two equations to obtain $4 n+4(2112)=4 x^{2}$ and $4 n+2112=y^{2}$, and thus $4 x^{2}-y^{2}=3(2112)$.

We can factor both sides of this equation to get $(2 x+y)(2 x-y)=3(11)(192)=2^{6} 3^{2} 11$.

So we must now determine the possibilities for $x$ and $y$, and hence the possibilities for $n$. We could do this by direct trial and error, but we will try to reduce the work we have to do somewhat by analyzing the situation.

Suppose $2 x+y=A$ and $2 x-y=B$, with $A B=2^{6} 3^{2} 11$. Then solving for $x$ and $y$, we obtain $x=\frac{A+B}{4}$ and $y=\frac{A-B}{2}$.

Since $A B$ is even, then at least one of $A$ and $B$ is even.

Since $y=\frac{A-B}{2}$ is an integer and one of $A$ and $B$ is even, then both of $A$ and $B$ are even.

Since $A$ and $B$ contain in total 6 powers of 2 , then at least one of them contains 3 powers of 2 , ie. is divisible by 8 (and thus 4 ).

Since $x=\frac{A+B}{4}$ is an integer and one of $A$ and $B$ is divisible by 4 , then both $A$ and $B$ are divisible by 4 .

So let $A=4 a$ and $B=4 b$, and so $x=a+b$ and $y=2 a-2 b$ and $a b=2^{2} 3^{2} 11$.

[We may assume that $y$ is positive, because if $y$ is negative this reverses the role of $a$ and $b$.] Since $y$ is positive, then $a>b$. How many possibilities for $a$ and $b$ are there given that $a b=2^{2} 3^{2} 11$ ? The integer $2^{2} 3^{2} 11$ has $(2+1)(2+1)(1+1)=18$ factors, so there are 9 possibilities for the pair $(a, b)$, namely
$(396,1),(198,2),(132,3),(99,4),(66,6),(44,9),(36,11),(33,12),(22,18)$

Of these, only the first 7 give values for $x$ and $y$ bigger than 46 (ie. positive values for $n$ ). Therefore, there are 7 possible values for $n$.

[For completeness, it is worth determining these values:

$$
n=x^{2}-2112=(a+b)^{2}-2112
$$

so the corresponding values of $n$ are $155497,37888,16113,8497,3072,697$, and 97.]