jmc

Notice that $(a+1)(a-5)=a^2-4a-5$, so $a^2-4a+1=(a+1)(a-5)+6$.

Also, we know that by the Euclidean algorithm, the greatest common divisor of $a+1$ and $a-5$ divides $6$: $$\begin{array}{rl}\text{gcd}(a+1,a-5)& =\text{gcd}(a+1-(a-5),a-5)\\ & =\text{gcd}(6,a-5).\end{array}$$As $10508$ is even but not divisible by $3$, for the sum of the digits of $10508$ is $1+5+8=14$, it follows that the greatest common divisor of $a+1$ and $a-5$ must be $2$.

From the identity $xy=\text{lcm}(x,y)\cdot \text{gcd}(x,y)$ (consider the exponents of the prime numbers in the prime factorization of $x$ and $y$), it follows that $$\begin{array}{rl}(a+1)(a-5)& =\text{lcm}(a+1,a-5)\cdot \text{gcd}(a+1,a-5)\\ & =2\cdot 10508.\end{array}$$Thus, the desired answer is $2\cdot 10508+6=\boxed{21022}.$

With a bit more work, we can find that $a=147$.