ziyuen wrote:
If m and n are non-zero integers, is \(m^6-m^5-m^4\) divisible by n?
1) \(m^2-1=m+n^2\)
2) \(m^4-2m^3+m^2=n^2+11\)
OFFICIAL SOLUTION
If you modify the original condition and the question,\(m^6-m^5-m^4 = nt\)? (t = any non-zero integer), so you get \(m^4(m^2-m-1)=nt\)?. If you look at con 1), from \(m^2-m-1=n^2\), if you substitute it to \(m^4(m^2-m-1)\), you get \(m^4(m^2-m-1)=m^4*n^2=n(m^4n)\), hence yes, it is sufficient. For con 2), it is difficult to approach the question, but if you apply CMT 4 (B: if you get A or B too easily, consider D), it is also sufficient. Therefore, the answer is D. If you actually solve it, for con 2), you get \((m^2)^2-2m(m^2)+m^2=(m^2-m)^2=n^2+11\), \((m^2-m)^2-n^2=(m^2-m-n)(m^2-m+n)=11\),
① When it becomes m^2-m-n=1 and m^2-m+n=11, from m^2-m-n=1, you get m^2-m-1=n, so m^4(m^2-m-1)=m^4*n, hence yes.
② When it becomes m2-m-n=11 and m^2-m+n=1, if you add the two equations, you get 2(m^2-m)=12, m^2-m=6 and n=-5, then m^2-m-1=6-1=5=(-1)(-5)=-n, so from m^4(m^2-m-1)=m^4(-n), it is also yes.
③ When it becomes m^2-m-n=-1 and m^2-m+n=-11, if you add the two equations, from 2(m^2-m)=-12, m^2-m=-6 and n=-5, integer m that satisfies this does not exist, so it is out of scope.
④ When it becomes m^2-m-n=-11 and m^2-m+n=-1, if you add the two equations, from 2(m^2-m)=-12, m^2-m=-6 and n=5, integer m that satisfies this does not exist, so it is out of scope.
Therefore, as ①, ② shown above, it always becomes yes, hence it is sufficient. Therefore, the answer is D.
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