# Using Prime and Square Numbers - How Old Am I ?/Document

This could be a starter activity rather than a full lesson.

**Introduction**

On the board display the text:

“Last year I was square, but this year I am in my prime. How old am I?”

Ask the pupils to explain what this means. [My age last year was a square number, whereas this year, one year on, it is a prime number.]

Let them start to tackle the problem. This is a good task to begin with the pupils working individually, in silence. They can then, after a couple of minutes, share their ideas with three others and can work on it further. This allows them to have some ideas of their own but then to integrate them with other approaches and to change tack if necessary. They may discover that their initial approach was not the most efficient; explaining this is part of the learning process.

**Main part of the task**

The most efficient procedure appears to be to list the square numbers and then to check whether the number one greater than these is prime or not. They should quickly realise that, apart from 1, it is not worth considering odd square numbers because the number one greater than these is even and the only even prime is 2.

1 -> 2

4 -> 5

16 -> 17

36 -> 37

64 -> 65 (which is clearly not prime)

100 -> 101

There are, therefore, five immediate possibilities (with more above 100 too) but of these the only credible one is that I am now 37 years old.

**Next stage of the task**

Now we can turn the question around and ask “what if I make a change?”

I will now swap a couple of words so the question reads: “Last year I was in my prime, but this year I am square. How old am I?”

The pupils should be able to make a faster start this time, and again only need to consider even square numbers, subtracting one and testing their primeness.

Pupils quickly find that 4 and 3 are a pair that works, but that others are hard to come by.

Sometimes they suggest 144 (122) and 143, or maybe 324 (182) and 323.

If no-one else spots it I point out that 143 is 11x13, and 323 is 17x19.

There are no other solutions because the number we want to be prime is being generated by n2 – 1, which is the difference of two squares and which factorises to give (n+1)(n-1). This can only be prime if one of the brackets is equal to 1. This is true for n=2, which is the solution 4 and 3, but no others are possible.