### Two Cubes

Two cubes, each with integral side lengths, have a combined volume equal to the total of the lengths of their edges. How big are the cubes? [If you find a result by 'trial and error' you'll need to prove you have found all possible solutions.]

### Root to Poly

Find the polynomial p(x) with integer coefficients such that one solution of the equation p(x)=0 is $1+\sqrt 2+\sqrt 3$.

### Janine's Conjecture

Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. Does this always work? Can you prove or disprove this conjecture?

# Common Divisor

##### Stage: 4 Challenge Level:

The problem was to find the largest integer which divides every member of the following sequence:

$1^5-1,\ 2^5-2,\ 3^5-3,\cdots\ n^5-n.$

The solution depends only on a little algebra and some clear mathematical thinking.

Pierre, Tarbert Comprehensive, Ireland, Prateek, Riccarton High School, Christchurch, New Zealand and Vassil from Lawnswood Sixth Form, Leeds started by taking small values of $n$, usually a good way to begin. This solution comes from Arun Iyer, S.I.A High School and Junior College, India. They all found the answer which is $30$.

Given the sequence $1^5-1,\ 2^5-2,\ 3^5-3,\cdots \ n^5-n$ we see that

$n^5 - n = n(n^4 - 1) = n(n - 1)(n + 1)(n^2 + 1)$

and it is quite easy to see that $n(n-1)(n+1)(n^2+1)$ is divisible by $2$, $3$ and $5$ for all values of $n$. As $n$, $(n-1)$ and $(n+1)$ are three consecutive integers their product must be divisible by $2$ and by $3$. If none of these numbers is divisible by $5$ then $n$ is either of the form $5k+2$ or $5k+3$ for some integer $k$ and in both of these cases we can check that $n^2 + 1$ is divisible by $5$. Since $2$, $3$ and $5$ are coprime therefore $n^5 - n$ is divisible by $2 \times 3 \times 5$ i.e by $30$.

Since the second term of the sequence is $2^5-2 = 30$ therefore the divisor cannot be greater than $30$. Therefore $30$ is the largest number that d ivides each member of the sequence.