a) José learned a method for calculating the product of two numbers more quickly based on factorization:

$$
(n-k)(n+k)=n^{2}-k^{2}
$$

To calculate 23 $\cdot 17$, he chooses $n=20, k=3$ and calculates:

$$
23 \cdot 17=20^{2}-3^{2}=400-9=391
$$

Determine, without using a calculator, the value of $\sqrt{1001 \cdot 1003+1}$.

b) Verify that $(n(n+3)+1)^{2}=n(n+1)(n+2)(n+3)+1$.

c) Determine, without using a calculator, the value of:

$$
\sqrt{(2014)(2015)(2016)(2017)+1}
$$
Solution

a) It suffices to choose $n=1002$ and $k=1$, since

$$
\begin{aligned}
\sqrt{1001 \cdot 1003+1} & =\sqrt{1002^{2}-1^{2}+1} \\
& =\sqrt{1002^{2}} \\
& =1002
\end{aligned}
$$

b)

$$
\begin{aligned}
(n(n+3)+1)^{2} & =n^{2}(n+3)^{2}+2 n(n+3)+1 \\
& =n(n+3)[n(n+3)+2]+1 \\
& =n(n+3)\left[n^{2}+3 n+2\right]+1 \\
& =n(n+3)[(n+1)(n+2)]+1 \\
& =n(n+1)(n+2)(n+3)+1
\end{aligned}
$$

c) Using the previous item and choosing $n=2014$, we have

$$
\begin{aligned}
\sqrt{(2014)(2015)(2016)(2017)+1} & =\sqrt{(2014 \cdot 2017+1)^{2}} \\
& =2014 \cdot 2017+1 \\
& =4062239
\end{aligned}
$$