We have to find all possible integer values of a and b where the difference of the squares of $a$ and $b$ is $924$ and then we have to discover which numbers CANNOT be written as the difference of two perfect squares?

The use of a spreadsheet lends itself to systematic exploration of number, in this case squares and differences of squares. Students from Bourne Grammar School who worked hard on this question would have gone further had they used spreadsheets.

We know that

$$(a^2 - b^2) = (a - b)(a + b)$$

It is always true that $(a - b)$ and $(a + b)$ must both be even or both odd. If they are both even then the product is a multiple of $4$. If they are both odd then the product is odd. So even numbers which are not multiples of $4$ cannot be written as the difference of two squares.

As $924 = 2 \times2 \times3 \times7 \times11$ we know that $(a - b)$ and $(a + b)$ must both be even. From this we can write down and solve pairs of simple equations to find $a$ and $b$ using all the combinations of prime factors of $924$ which give pairs of even factors.

The solution below from Soh Yong Sheng, Raffles Institution, Singapore uses a different method based on the fact that the square of any number is the sum of consecutive odd numbers

$$n^2 = 1 + 3 + 5 + ... (2n - 1)$$

and so the difference of two squares is the sum of consecutive odd numbers.

Many numbers can be expressed as differences of 2 squares, but exactly which? It is known that

$1=1 (1^2)$
$1+3=4 (2^2)$
$1+3+5=9 (3^2)$
$1+3+5+7=16 (4^2)$
$\ldots$
$1+3+\ldots+n= ((n+1)/2)^2 $

Using this, we try to find the sums of consecutive odd numbers that add up to $924$ which will tell us $a$ and $b$.

Case of $2$ odds
$924 = 461 + 463$
$a= 464/2 = 232$, $a^2 = 53824$, $b= 230$, $b^2 = 52900$. One possible solution.

Case of $4$ odds
$924 = 229 + 231 + 231 + 233$ so this is impossible.

Case of $6$ odds
$924=149+151+153+155+157+159$
$a = (159 + 1)/2 = 80$, $a^2 = 6400$, $b = (149-1)/2 = 74$, $b^2 = 5476$. A possible solution.

Cases of $8$, $10$ and $12$ odds impossible

Case of $14$ odds
$924=53+55+57+59+61+63+65+67+69+71+73+75+77+79$
$a= 80/2 = 40$, $a^2 = 1600$, $b = (53-1)/2= 26$, $262 = 676$. Another solution.

Cases of $16$, $18$ and $20$ odds are impossible

Case of $22$ odds

$924=21+23+25+27+29+31+33+35+37+39+41+43+45+47+49+51+53+55+57+59+61+63$
$a = 64/2 =32$, $a^2 = 1024$, $b = (21-1)/2$, $b^2 = 100$ Another solution.

This gives the pairs $(232, 230)$, $(80, 74)$, $(26, 40)$ and $(64,10)$

The solutions can also be negative integers $a = \pm232$ and $b = \pm 230$ etc. giving sixteen solutions in all.