Show that $\gcd\left(n^2+20,(n+1)^2+20\right)$ divides $81$ for all natural number $n.$

Question

Question:

$$a_n=n^2+20$$ where $n$ is a natural number. $d_n$ denotes the greatest common divisor of $a_n$ and $a_{n+1}$. Show that $d_n$ divides $81$ for all $n$.

I tried the question and got some results but could not prove what is asked to. I got $d_n$ divides $2n+1$ and $n(n-40)$ also I tried by checking congruences modulo $3$ and other numbers but ended up nowhere. I also came across a solution using Chinese remainder theorem while searching but did not understand it.

Answer 1

Hint : $d | n(n-40)$ $\Rightarrow$ $d | 2n^2 - 80n$

Answer 2

As you noted, $d_n\mid 2n+1$. So $d_n\mid (2n+1)^2=4n^2+4n+1$.

Also $d_n\mid n^2+20$. So $d_n\mid 4(n^2+20)=4n^2+80$

Then $d_n\mid (4n^2+4n+1)-(4n^2+80)=4n-79$.

From this and $d_n\mid 2n+1$, you should be able to get your result.

Answer 3

We have $$d_n=\gcd(a_n,a_{n+1})=\gcd(n^2+20,(n+1)^2+20)=\gcd(n^2+20,2n+1),$$ so $d_n\mid2n+1$ and $$d_n\mid4(n^2+20)=(2n)^2+80=(2n+1-1)^2+80=(2n+1)^2-2(2n+1)+81,$$ which leads to $d_n\mid81$.

Answer 4

If $d$ divides both $a_n$ and $a_{n+1}$, it also divides $(2\,n+3)\,a_n-(2\,n-1)\,a_{n+1}=81$.