I'm trying to show that if $n+2\mid(n^2+n+1)(n^2+n+2)$, then $n+2 \mid 12$ for all $n\in \mathbb{N}$.
Asked
Active
Viewed 86 times
2
-
1Please show us what you've tried so far :) – Shaun Jul 06 '14 at 10:22
-
Q: What do you get when you substitute $x=-2$ into $$f(x)=(x^2+x+1)(x^2+x+2)?$$ A: Same answer as timbuc, but with less calculations. – Jyrki Lahtonen Jul 06 '14 at 13:41
3 Answers
6
Hint
$$(n^2+n+1)(n^2+n+2)=n^4+2n^3+4n^2+3n+2$$
and
$$ n^4+2n^3+4n^2+3n+2=(n+2)(n^3+4n-5)+12$$
Timbuc
- 34,191
3
Hint $\ {\rm mod}\,\ n\!+\!2\!:\,\ n\equiv -2\,\Rightarrow\, f(n)\equiv f(-2)\,$ for all polynomials $\,f\in\Bbb Z[x],\,$ a special case of the Polynomial Congruence Rule
Bill Dubuque
- 272,048
1
Observe that $\displaystyle n^2+n=n(n+2)-(n+2)+2=(n+2)(n-1)+2$
$\displaystyle\implies (n^2+n+1)(n^2+n+2)=\{\underbrace{(n+2)(n-1)+2}+1\}\{\underbrace{(n+2)(n-1)+2}+2\}\equiv(2+1)(2+2)\pmod{(n+2)(n-1)}\equiv12\pmod{(n+2)} $
lab bhattacharjee
- 274,582
-
-
Remark that this method is implicitly using congruence arithmetic,, i.e. $,f(n) = g(n)h(n),\Rightarrow, {\rm mod}\ n!+!2!:\ f(n)\equiv f(-2)\equiv g(-2)h(-2)\equiv 3\cdot 4\ \ $ – Bill Dubuque Jul 06 '14 at 14:03