The sequence $$\langle a_{n}\rangle_{n\ge 0}$$ is defined by $$a_{0}=1$$, $$a_{1}=-4$$ and $$a_{n+2}=-4a_{n+1}-7a_{n}$$, for $$n\ge 0$$. Find the number of positive integer divisors of $$a_{50}^{2}-a_{49}a_{51}$$.

Define the auxiliary quantity
$$\Delta_n \;=\;a_{n}^{\,2}\;-\;a_{n-1}\,a_{n+1},\qquad n\ge 1.$$

Step 1 : Establish a recurrence for $$\Delta_n$$
Using the given relation $$a_{n+2}=-4a_{n+1}-7a_{n}\,,$$ put $$n\to n-1$$ to get
$$a_{n+1} \;=\;-4a_{n}\;-\;7a_{\,n-1}$$ $$-(1)$$

Compute $$\Delta_{n+1}$$:
$$\Delta_{n+1}=a_{n+1}^{\,2}-a_{n}\,a_{n+2}.$$
Substitute $$a_{n+2}=-4a_{n+1}-7a_{n}$$:
$$\Delta_{n+1}=a_{n+1}^{\,2}-a_{n}\bigl(-4a_{n+1}-7a_{n}\bigr)$$ $$\qquad =a_{n+1}^{\,2}+4a_{n}a_{n+1}+7a_{n}^{\,2}$$ $$\qquad =a_{n+1}\bigl(a_{n+1}+4a_{n}+7a_{\,n-1}\bigr)\,+\,7\bigl(a_{n}^{\,2}-a_{\,n-1}a_{n+1}\bigr)$$ (using $$-(1)$$ for the term in parentheses). But $$a_{n+1}+4a_{n}+7a_{\,n-1}=0$$ by $$-(1)$$, so

$$\boxed{\;\Delta_{n+1}=7\,\Delta_n\;}\qquad\text{for all }n\ge 1.$$

Step 2 : Initial value of $$\Delta_n$$
First find $$a_2$$:
$$a_2=-4a_1-7a_0=-4(-4)-7(1)=16-7=9.$$
Hence
$$\Delta_1=a_{1}^{\,2}-a_{0}a_{2}=(-4)^{2}-1\cdot9=16-9=7.$$

Step 3 : Closed form for $$\Delta_n$$
From the recurrence $$\Delta_{n+1}=7\Delta_n$$ and the initial value $$\Delta_1=7$$, we get by induction
$$\Delta_n=7^{\,n}\qquad(n\ge 1).$$

Step 4 : Required expression
For $$n=50$$,
$$a_{50}^{\,2}-a_{49}a_{51}=\Delta_{50}=7^{50}.$$

Step 5 : Number of positive divisors
If $$N=p^{k}$$, the number of positive divisors of $$N$$ is $$k+1$$.
Here $$7^{50}$$ has prime factorisation with exponent $$50$$, so the divisor count is
$$50+1=51.$$

Answer : 51