# Modulo120

#### dhiab

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

#### roninpro

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

Can you take it from here?

• dhiab

#### simplependulum

MHF Hall of Honor
$$\displaystyle 9n^5 - 5n^3 - 4n = 9(n^5 - 5n^3 + 4n) - 5n^3 + 45n^3 - 4n - 36n$$

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

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

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

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

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

Therefore , we prove it .

• dhiab