For how many odd integers $k$ between 0 and 100 does the equation

$$
2^{4 m^{2}}+2^{m^{2}-n^{2}+4}=2^{k+4}+2^{3 m^{2}+n^{2}+k}
$$

have exactly two pairs of positive integers $(m, n)$ that are solutions?
(A) 17
(B) 20
(C) 19
(D) 18
(E) 21
Step 1: Using parity and properties of powers of 2 to simplify the equation

We note that if $2^{x}=2^{y}$ for some real numbers $x$ and $y$, then $x=y$.

This is because $2^{x}=2^{y}$ implies $\frac{2^{x}}{2^{y}}=1$ or $2^{x-y}=1$, and so $x-y=0$ or $x=y$.

We examine equations of the form $2^{a}+2^{b}=2^{c}+2^{d}$ where $a, b, c$, and $d$ are integers.

(This is more general than the given equation, but allows us to determine what is possible.)

We may assume without loss of generality that $a \leq b$ and $c \leq d$ and $a \leq c$. (We can always switch the variable names to make these true.)

We factor the equation as $2^{a}\left(1+2^{b-a}\right)=2^{c}\left(1+2^{d-c}\right)$, and then divide both sides by $2^{a}$ to obtain $1+2^{b-a}=2^{c-a}\left(1+2^{d-c}\right)$.

We show that $c=a$ by contradiction:

If $c \neq a$, then $c \geq a$ gives $c>a$.

If $c>a$, then $c-a>0$, so $c-a \geq 1$, since $c-a$ is an integer.

Therefore, the right side has a factor of $2^{c-a}$, so the right side is even.

Thus, the left side is even too, which means that $2^{b-a}$ must be an odd integer.

For $2^{b-a}$ to be an odd integer, we must have $2^{b-a}=1$ and so $b-a=0$ or $b=a$.

In this case, the left side equals 2 and the right side is greater than 2 , since $2^{c-a} \geq 2$ and $1+2^{d-c}>1$. This is a contradiction.

Therefore, $c=a$.

Since $a=c$, then $2^{a}+2^{b}=2^{c}+2^{d}$ becomes $2^{b}=2^{d}$ and so $b=d$.

Therefore, if $2^{a}+2^{b}=2^{c}+2^{d}$ with $a, b, c, d$ integers, then either $a=b=c=d$ or $a=c$ and $b=d$ (with $a \neq b$ ) or $a=d$ and $b=c$ (with $a \neq b$ ).

We examine these three possibilities in the given equation, noting that $m, n$ and $k$ are all positive integers:

- Case 1: $4 m^{2}=m^{2}-n^{2}+4=k+4=3 m^{2}+n^{2}+k$

From the last equality, we obtain $3 m^{2}+n^{2}=4$.

Since $m, n$ are positive integers, then $m^{2} \geq 1$ and $n^{2} \geq 1$.

Since $3 m^{2}+n^{2}=4$, then it must be that $m=n=1$.

Thus, $4 m^{2}=k+4$ implies $4=k+4$ or $k=0$.

But $k>0$, so this case is not possible.

- Case 2: $4 m^{2}=k+4$ and $m^{2}-n^{2}+4=3 m^{2}+n^{2}+k$ and $4 m^{2} \neq m^{2}-n^{2}+4$

From the second equality, we obtain $2 m^{2}+2 n^{2}+k=4$, which is not possible since $m, n, k>0$, and so $2 m^{2}+2 n^{2}+k \geq 5$.

Therefore, this case is not possible.

- Case 3: $4 m^{2}=3 m^{2}+n^{2}+k$ and $m^{2}-n^{2}+4=k+4$ and $4 m^{2} \neq m^{2}-n^{2}+4$

The first equality rearranges to $m^{2}-n^{2}=k$.

The second equality also rearranges to $m^{2}-n^{2}=k$.

The last statement is equivalent to $3 m^{2}+n^{2} \neq 4$. As we saw in Case 1 , this means that $(m, n)$ cannot be the pair $(1,1)$, which is consistent with $m^{2}-n^{2}=k$ and $k>0$.

Therefore, having examined all of the cases, we have reduced the original problem to finding the number of odd integers $k$ between 0 and 100 for which the equation $m^{2}-n^{2}=k$ has exactly two pairs of positive integers $(m, n)$ that are solutions.

Step 2: Connecting solutions to $m^{2}-n^{2}=k$ with factorizations of $k$

We can factor the left side of this equation to give $(m+n)(m-n)=k$.

Since $m, n$ and $k$ are positive integers, then $m+n>0$ and $k>0$ so $m-n>0$, or $m>n$.

Since $k$ is odd and each of $m+n$ and $m-n$ is an integer, then each of $m+n$ and $m-n$ is odd (since if either was even, then their product would be even).

Also, we note that $m+n>m-n$ since $n>0$.

Suppose that $(m, n)$ is a solution of the equation $m^{2}-n^{2}=k$ with $m+n=a$ and $m-n=b$ for some odd positive integers $a$ and $b$ with $a>b$.

Then $a b=k$, so $a b$ is a factorization of $k$.

Therefore, the solution $(m, n)$ corresponds to a specific factorization of $k$.

Now suppose that we start with a factorization $k=A B$ where $A$ and $B$ are odd positive integers with $A \geq B$.

If we try setting $m+n=A$ and $m-n=B$, then we can add these equations to give $2 m=A+B$ (or $m=\frac{1}{2}(A+B)$ ) and subtract them to give $2 n=A-B$ (or $n=\frac{1}{2}(A-B)$ ). Note that since $n>0$, then $A>B$.

Therefore, every factorization of $k$ as the product of two odd positive integers $A$ and $B$ with $A>B$ gives a solution to the equation $m^{2}-n^{2}=k$.

Since each solution gives a factorization and each factorization gives a solution, then the number of solutions equals the number of factorizations.

Therefore, we have reduced the original problem to finding the number of odd integers $k$ between 0 and 100 which have exactly two factorizations as the product of distinct odd integers $a$ and $b$ with $a>b$.

Step 3: Counting the values of $k$

Since $k$ is odd, then all of its prime factors are odd.

Since $k11^{2}=121$, which is larger than 100 .

Therefore, $p$ can only be 3,5 or 7 .

If $p=3$, there are 9 possible values for $q$ (primes from 5 to 31 ).

If $p=5$, there are 5 possible values for $q$ (primes from 7 to 19).

If $p=7$, there are 2 possible values for $q$ (11 and 13).

Thus, there are $9+5+2=16$ values of $k$ of this form that work.

If $k=p^{r} q^{s}$ with $r$ and $s$ positive integers and at least one of $r$ or $s$ is larger than 1 , then $k$ will have at least three factorizations. (For example, if $r>1$, then $k=1 \cdot p^{r} q^{s}=p \cdot p^{r-1} q^{s}=p^{r} \cdot q^{s}$ and all of these are distinct.)

If $k=p$ or $k=p^{2}$ with $p$ an odd prime, then $k$ has only one factorization as the product of distinct factors $\left(1 \cdot p\right.$ and $1 \cdot p^{2}$, respectively). Thus, $k$ cannot be of this form.

If $k=p^{3}$ with $p$ an odd prime, then the divisors of $k$ are $1, p, p^{2}, p^{3}$, so it has exactly two factorizations of the desired type (namely $1 \cdot p^{3}$ and $p \cdot p^{2}$ ).

Since $k100$ ).

Thus, there is 1 value of $k$ of this form that works.

If $k=p^{4}$ with $p$ an odd prime, then the divisors of $k$ are $1, p, p^{2}, p^{3}, p^{4}$, so it has exactly two factorizations of the desired type (namely $1 \cdot p^{4}$ and $p \cdot p^{3}$ ). In this case, $k$ has a third factorization, but it is of the wrong type since the two factors will be equal.

Since $k100$ ).

Thus, there is 1 value of $k$ of this form that works.

If $k$ has more than 4 factors of $p$, then $k$ will have at least three factorizations of the desired type, so $k$ cannot be of this form. (In particular, if $k=p^{n}$ and $n>4$, then $k=1 \cdot p^{n}=$ $p \cdot p^{n-1}=p^{2} \cdot p^{n-2}$ and these are all distinct since $n-2>2$.)

Having examined all of the possible forms, we see that there are $16+1+1=18$ values of $k$ that work, and so there are 18 positive integer solutions to the original equation.