Modulo120

**dhiab** · May 24th 2010, 07:16 PM

Prove that : $\forall n\in N:9n^{5}-5n^{3}-4n\equiv 0(mod120$ )

**roninpro** · May 24th 2010, 07:26 PM

Note that $120=8\cdot 3\cdot 5$ and $9n^5-5n^3-4n=(n-1)n(n+1)(4+9n^2)$ . It suffices to show that $8, 3, \text{ and } 5$ divide $(n-1)n(n+1)(4+9n^2)$ for any $n$ .

Can you take it from here?

**simplependulum** · May 24th 2010, 07:32 PM

$9n^5 - 5n^3 - 4n = 9(n^5 - 5n^3 + 4n) - 5n^3 + 45n^3 - 4n - 36n$

$= 9n(n^2-4)(n^2-1) + 40n^3 - 40n$

$= 9(n-2)(n-1)(n)(n+1)(n+2) + 40n(n^2-1)$

We know the product of five consective integers is divisible by $5! = 120$ so we only consider $40n(n^2-1)$ .

We need to show that $n(n^2-1)$ is the multiple of $3$

$n(n^2-1) = n(n-1)(n+1) = (n-1)(n)(n+1)$ which is the multiple of $3$ .

Therefore , we prove it .

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