### Building Tetrahedra

Can you make a tetrahedron whose faces all have the same perimeter?

A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?

Four rods are hinged at their ends to form a convex quadrilateral. Investigate the different shapes that the quadrilateral can take. Be patient this problem may be slow to load.

# Primes and Six

##### Age 14 to 16 ShortChallenge Level

Using just $p$ and $q$
$$\begin{split}pq+1&=p(p+2)+1\\ &=p^2+2p+1\\ &=(p+1)^2\end{split}$$ $p$ is odd so $p+1$ is even
$p,q\gt 3$ so
\text{not multiples of 3}\\ \begin{align} &\downarrow & &\downarrow\\ &p &p+1\hspace{5mm} &q\\ & &\uparrow\hspace{10mm}& \end{align}\\ \text{a multiple of 3}\\ \text{(since 1 in every 3 numbers is a multiple of 3)} So $p+1$ is a multiple of $2$ and $3$
$\therefore p+1$ is a multiple of $6$
$\therefore (p+1)^2$ is a multiple of $36$, and $(p+1)^2=pq+1$

Using another letter $k$
Let $k=p+1$. We note that $k$ is even because any prime $p>3$ is odd.

Moreover, neither $p$ nor $q$ can be divisble by three because the only prime number divisible by three is $3$ itself. On the other hand, if you have three consecutive integers exactly one of them must be divisible by three. Thus, $k$ is also a multiple of $3$.

So, we can write $k$ in the form $k=6n$ for some integer $n$. We deduce further that
$$pq = (6n-1)(6n+1) = 36n^2-1$$
which implies that $pq+1 = 36n^2$ is indeed divisible by $36$.
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.