Finding Largest N for n^5-5n^3+4n Divisibility

[ehrenfest]

[SOLVED] larson 3.3.19b

Homework Statement

What is the largest number N for which you can say that n^5-5n^3+4n is divisible by N for every positive integer N.

EDIT: change the last N to n

Homework Equations

The Attempt at a Solution

I have just been plugging in things for n and seeing what happens. If n=N-2,N-1,N,N
+1,N+2, then n^5-5n^3+4n is divisible by N because -2,-1,0,1,2 are the roots of that equation. If n=N+3, we get that 120 = -120 must equal 0 mod N. So, N=3 is a lower bound. So N must be a factor of 120. Should I just keep keep plugging in numbers for n and setting them equal to 0 mod N? It seems like that will give me a solution but that won't prove that this particular N works for all values of n.

 

[morphism]

I suppose you meant to say "What is the largest number N for which you can say that n^5-5n^3+4n is divisible by N for every positive integer n."

Now, n^5-5n^3+4n=(n-2)(n-1)n(n+1)(n+2) is the product of 5 consecutive integers hence divisible by 5!=120. I guess this is what you said, but I didn't really follow your reasoning. Next note that for n=3, the expression is 120.

 

[ehrenfest]

Very nice! I was just beating around that solution...I just could quite hit it on the head.