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!!
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Hamid Reza Ebrahimi
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2If $\ f(n) = n^2+5\ $ then $,d\mid f(n),n!+!25,\Rightarrow, d\mid {f(n)\ \rm mod}\ (n!+!25) = f(-25) = 630 \ \ $ – Bill Dubuque Aug 09 '16 at 14:16
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@BillDubuque Please explain more about this: $f(n) \mod (n+25)=f(-25)$ – Hamid Reza Ebrahimi Aug 09 '16 at 14:24
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1By Division $, f(n) = q(n) (n - a) + r,,$ so $,f(a) = r = f(n)\ {\rm mod}\ n!-!a\ $ (Remainder Theorem) $\ \ $ – Bill Dubuque Aug 09 '16 at 14:38
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@BillDubuque Thank you so much:) – Hamid Reza Ebrahimi Aug 09 '16 at 14:41
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I added a link to Wikipedia. You could of course do the division if you were not aware of the remainder theorem, i.e. do the final step of the Euclidean algorithm. – Bill Dubuque Aug 09 '16 at 14:43
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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 $\,\Bbb Z[\sqrt 5]$
Bill Dubuque
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