Let $d=GCD(n^2+5,n^3-5n^2+6n)$ show that $d|630$ 
Let $d=GCD(n^2+5,n^3-5n^2+6n)$ show that $d|630$

My work: $d|n^3+5n,n^3-5n^2+6n\implies\ d|5n^2-n,5n^2+25\implies d|n+25 $
Now I can't go further!!

Let $\,f(n) = n^2+5.\,$ Then $\,d\mid \color{#c00}{f(n),n\!+\!25}\,\Rightarrow\, d\mid  \overbrace{{f(n)\ \rm mod}\ (n\!+\!25) = f(-25)}^{\textstyle\!\! \color{#c00}{f(n)} - (\color{#c00}{n\!+\!25})q(n) = r(n)} = 630\, $ where we used the Remainder theorem $\ f(n)\equiv f(a)\pmod{n-a}$
Remark $\ $ It is simpler to work modulo the common divisor $\,d,\,$ namely
$ n^2\!+5\equiv 0\,\Rightarrow\, \color{#c00}{n^2\equiv -5}\,\Rightarrow\, 0\equiv n(\color{#c00}{n^2})\!-5\color{#c00}{n^2}\!+6n\equiv (\color{#c00}{-5})n-5(\color{#c00}{-5})+6n\equiv \color{#0a0}{n+25} $ 
thus $\ \color{#0a0}{n\equiv -25}\,\Rightarrow\, 0\equiv \color{#0a0}n^2+5\equiv (\color{#0a0}{-25})^2+5\equiv 630,\ $ therefore $\ d\mid 630$.

Alternatively $\, 25\!+\!n\equiv 0\,\Rightarrow\ 0 \equiv (25 + n)\,(25 - n)\equiv 25^2\!-n^2\equiv 625-(-5)\equiv 630$
essentially $\,\ 25\!+\!\sqrt5\equiv 0\,\Rightarrow\, 0\equiv (25\!+\sqrt5)(25\!-\!\sqrt5)\equiv 625,\, $ like taking a norm in $\,\Bbb Z[\sqrt 5]$
