You may also like

Building Tetrahedra

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

Ladder and Cube

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?

Bendy Quad

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.

Square Product

Age 14 to 16 Short
Challenge Level

Answer: 8

We make use of two key facts:
First, $n^2 -1 = (n-1)(n+1)$.
Second, when a square number is factorised, each prime factor appears an even number of times.
 
Now $2^2 -1 = 1\times 3$.
We next get a prime factor $3$ with $4^2-1= 3\times 5$.
We next get a prime factor $5$ with $6^2-1= 5\times 7$.
We next get a prime factor $7$ with $8^2-1= 7\times 9$.
 
As $9=3^2$, it does not require any further factors. Hence we need $n\geq 8$. Checking the product $n=8$, we get
$(2^2 -1)(3^2 -1)(4^2 -1)(5^2 -1)(6^2 -1)(7^2 -1)(8^2 -1)$
$$\eqalign{
&=1\times 3\times2 \times4 \times3 \times5 \times4\times6 \times5\times7 \times6\times8 \times7\times9 \cr
&=2\times8 \times 3 \times 3\times4\times4\times5\times5\times6\times6\times7\times7\times3\times3 \cr
&=4\times4\times3\times3\times4\times4\times5\times5\times6\times6\times7\times7\times3\times3 \cr
&= (4\times3\times4\times5\times6\times7\times3)^2}$$
So in fact $n=8$ is sufficient, and is thus the minimum.
 
 
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.