### Root to Poly

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

### Common Divisor

Find the largest integer which divides every member of the following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.

### 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?

# Two Cubes

##### Stage: 4 Challenge Level:

We had a good crop of solutions to this one, using different methods.

Two cubes have integral side lengths, and the sum of their volumes is exactly the sum of their edge lengths. What are their edge lengths? Let the lengths of the edges of the two cubes be $a$ and $b$, with $a \geq b$. Thus $a^3 + b^3 = 12a + 12b.$

Soh Yong Sheng, age 12, Tao Nan School, Singapore noted that $a$ and $b$ have to be quite small or else $a^3 + b^3$ is too big, and as it increases much faster than $12a + 12b$, the gap gets wider, also that $a$ and $b$ cannot be an odd number and an even number because then the total of the two volumes would be odd whereas it has to be a multiple of 12. Testing the smallest few pairs (3,1), (5,1), (4,2), (5,3) ... the only one of these satisfying the conditions is $a = 4$ and $b=2$. For larger values of $a$ and $b$ the value of $a^3 + b^3$ is already too big so there cannot be any other possibilities.

Alex Fletcher, age 17, King Edward and Queen Mary School, Lytham solved a quadratic equation to find $a$ in terms of $b$: $$a^3 + b^3 = (a + b)(a^2 - ab + b^2) = 12 (a + b).$$ As $a$ and $b$ are positive integers, $a + b \neq 0$. Dividing by $(a + b)$ gives $$a^2 - ab + b^2 = 12.$$ Hence $$a = {b\pm \sqrt{b^2 - 4(b^2 - 12)}\over 2}= {b\pm \sqrt {48 -3b^2}\over 2}.$$ Since $a$ and $b$ are both positive integers, $b\leq 4$ and the only values satisfying the equation are $a = 4$ and $b = 2$. Clearly the original equation is symmetric but we took $a> b$.

Kerwin Hui reasoned as follows. We have
\begin{eqnarray} \\ a^2 - ab + b^2 &= 12 \\ (a - b)^2 + ab &= 12. \end{eqnarray}

Checking the parity of $a$ and $b$ we have $a$, $b$, $(a - b)^2 + ab$, (Odd Odd Odd), (Odd Even Odd), (Even Odd Odd), (Even Even Even). So $a$ and $b$ must both be even. Let $a=2c$ and $b=2d$, we have \eqalign{ (2c - 2d)^2 + (2c)(2d) &= 12 \cr (c - d)^2 +cd = 3.} As $c$ and $d$ are positive, $cd> 0$ so $(c - d)^2 \leq 3$ so $(c - d)= 0,1$ [it can't be negative because $a> b$]. The case $c=d$ gives $c^2=3$ which has no integer solutions so $(c - d)=1.$ This gives $cd = 2$ hence $c - 2/c = 1$ which gives the equation $$c^2 - c - 2 = 0.$$ The solutions are $c=2$ and $d=1$, or $c=-1$ which is rejected as $c$ is positive. Therefore the only non-trivial solution for the lengths of the edges is 4 and 2.

Sergio Moya, Ling Xiang Ning and Andaleeb Ahmed also sent in good solutions.