The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field.

# Practice Paper 1 Question 15

Does $$30$$ divide $$n^5-n$$ for all positive integers $$n$$?

The above links are provided as is. They are not affiliated with the Climb Foundation unless otherwise specified.

## Warm-up Questions

1. Factorise $$n^2-2n-3$$.
2. Is $$n^2+n$$ always even when $$n$$ is an integer?
3. Does $$6$$ divide $$100002$$? Try to reason about the divisors of both numbers.

## Hints

• Hint 1
How do you split $$30$$ into a product of prime factors?
• Hint 2
Can you factorise $$n^5 - n$$?
• Hint 3
You should obtain $$n(n-1)(n+1)(n^2+1)$$. What can you say about the product of 3 consecutive numbers in terms of divisibility by $$2$$ and $$3$$?
• Hint 4
If you assume that $$5$$ divides $$n^5-n$$, how can you prove that 5 divides $$(n+1)^5-(n+1)$$?

## Solution

First notice that $$30=2\cdot3\cdot5$$. Since $$2$$, $$3$$ and $$5$$ are all coprime we will prove that each of them divides $$n^5-n$$ and hence conclude that their product does too. We now factorise $$n^5-n$$: \begin{align} n^5-n &= n(n^4-1) \\ &= n(n^2-1)(n^2+1)\\ &= n(n-1)(n+1)(n^2+1) \end{align} The product $$(n-1)n(n+1)$$ is of 3 consecutive numbers, hence necessarily both 2 and 3 must divide it. We must now prove divisibility by $$5$$. There are several approaches to do this, but here we shall use induction. It's easy to verify the base case for $$n=0$$ or $$n=1$$. Then do the inductive step: \begin{align} (n+1)^5 - (n+1) &= n^5 + 5n^4 + 10n^3 + 10n^2 + 5n - n \\ &= n^5-n + 5(n^4+2n^3+2n^2+n) \end{align} First part is divisible by $$5$$ owing to the induction hypothesis, while the rest is obviously a multiple of 5.

If you have queries or suggestions about the content on this page or the CSAT Practice Platform then you can write to us at . Please do not write to this address regarding general admissions or course queries.