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'Why 24?' printed from http://nrich.maths.org/
from Collaton St Mary School, noticed something about question
$5 \times 5 = 25, 25 - 1 = 24$
$7 \times 7 = 49, 49 - 1 = 48$
$11 \times 11 = 121, 121 - 1 = 120$
I noticed that the result for the second calculation in each
example was equal to the product of the numbers on either side of
the prime number that I started with.
Masood, Akeel and Dolapo from
Wilson's School, and Rajeev from Fair Field Junior School used
algebra to explain:
$(x - 1)(x + 1) = x^2 - x +x -1 = x^2-1$
Elliott, Mark, Mark, Michael and Miriam, from Curtin Primary school
in Canberra, sent in their explanations for each part:
A) If you multiply the numbers on either side of x by each other
you get one less than the square of x. This can be represented as
:$( x + 1 ) \times ( x - 1 ) = x^2 - 1$
B) Because there is only two numbers between two consecutive
multiples of 3, every third number is a multiple of 3.
C) If you multiply coprime factors of x you get another factor of x
or x itself.
D) They are all divisible by 4 because even numbers are every
second number. Multiples of 4 are every fourth number therefore,
every second even number will be a multiple of 4. That means the
one of our numbers will be a multiple of 4.
Final Challenge They are all divisible by 24. Why? What if we put
the numbers in the formula for A? Since we are squaring the prime
$x$ and subtracting one we have $(x+1)\times(x-1)$
$x$ cannot be divisible by three so one of the numbers either side
must be - $(x+1)$ or $(x-1)$
This means that $x^2-1$ is a multiple of 3.
Every second even number is a multiple of 4 (see D) therefore every
other even number is a multiple of 2. Since $x$ cannot be divisible
by 2, then $x+1$ is divisible by 2 & $x-1$ is divisible by 4
(it could be the other way around).
We are multiplying a multiple of 2 by a multiple of 4 so $x^2-1$ is
a multiple of 8. That means that $x^2-1$ is a multiple of 8 &
3. We can't multiply normal factors but 3 & 8 are coprime and
multiplying coprime factors will result in another factor so 3 x 8
is 24. $ x^2-1$, the final number is divisible by 24.
Isabella and Ian, from
Cashmere High School in Christchurch, and Richard from Comberton
Village College, sent us similar algebraic arguments.
Hannah, from Munich
International School, Preveina, from Crest Girls' Academy tried
squaring some prime numbers and subtracting 1, and verified that
they were all divisible by 24.
Finally, well done to Patrick
from Woodbridge School, and Andrew who didn't give his school name,
who both used modular arithmetic. Here is Andrew's
The challenge asked to pick a prime number greater than 3, so let p
be such a prime.
Squaring p then subtracting one is the difference of two squares:
$p^2 - 1 =(p-1)(p+1)$
So we can look at the properties of $(p-1)$ and $(p+1)$ rather than
Since p is a prime, p is not divisible by 3. Hence it either has
the remainder 1 or the remainder -1 when divided by 3.
Case 1: $p = 1$ (mod 3) $\Rightarrow p - 1= 0$ (mod 3)
Therefore $p - 1$ is divisible by 3.
Case 2: $p = -1$ (mod 3) $\Rightarrow p + 1 = 0$ (mod 3)
Therefore $p + 1$ is divisible by 3.
We have concluded regardless of the remainder when p is divided by
3, either $p-1$ or $p+1$ is divisible by 3.
Now consider p modulo 4. p cannot equal 0 (mod 4) otherwise it
would not be prime. p also cannot equal 2 (mod 4) because it would
then be divisible by 2, and p is a prime greater than 2. Therefore
$p = 1$ or $p = 3 = -1$ (mod 4).
Case 1: $p = 1$ (mod 4) $\Rightarrow p - 1 = 1 - 1 = 0$ (mod 4)
and $p + 1 = 1 + 1 = 2$ (mod 4)
Therefore $p-1$ is divisible by 4 and $p+1$ is divisible by 2.
Case 2: $p = -1$ (mod 4) $\Rightarrow p - 1 = -1 - 1= -2 = 2$ (mod
4) and $p+1= -1 + 1 = 0$ (mod 4)
Therefore $p-1$ is divisible by 2 and $p+1$ is divisible by 4.
Therefore since either $p-1$ or $p+1$ is divisible by 4 and the
other 2, their product is divisible by 8. But it was also
established previously that $(p-1)(p+1)$ is divisible by 3.
Therefore $(p-1)(p+1)$ is divisible by 8 and 3, and therefore
This means for any prime $p > 3$, $p^2 - 1 = (p-1)(p+1)$ is
divisible by 24.