$k$
are there between $1$
and $1000$
inclusive that can be expressed as a difference of two squared integers?

 Factorise
$a^3b^3.$
 If
$a$
is divisible by$2$
and$b$
is divisible by$4,$
would$ab$
be divisible by$8?$
 If
$m$
is divisible by$2$
and$4,$
would$m$
be divisible by$8?$
 Factorise

Have you expressed the question statement in mathematical terms and simplified it?

Pay close attention to the two divisors of
$k.$

… in particular, their oddeven parity.

Could they be odd or even independently? What are the possible combinations?

You should find that they have to be both odd or both even, as a necessary condition. Is it a sufficient condition?

Try giving an expression for
$a$
and$b$
in terms of$k$
for each scenario.

Let’s assume that
$a^2b^2=k,$
for some$k \in \{1,2,3,\ldots,1000\}.$
Now, we can write$(a+b)(ab) = k.$
Notice that$(a+b)$
and$(ab)$
are both odd or both even. We can look at the two cases separately.When
$(a+b)$
and$(ab)$
are both odd,$k$
must be an odd integer with two odd divisors. Every odd integer between$1$
and$1000$
has at least two odd divisors that are$1$
and itself. Hence, when$k$
is odd, we can take$a=\frac{(k+1)}{2}$
and$b=\frac{(k1)}{2}.$
Indeed, we can show that any odd number can be written as the difference between two perfect squares:$2n+1=(n+1)^2n^2$
for$n \in \mathbb{N}.$
When
$(a+b)$
and$(ab)$
are both even,$k$
must be divisible by$4.$
We can take$a=\frac{k}{4}+1$
and$b= \frac{k}{4}1.$
Since there are
$250$
integers divisible by$4$
and$500$
odd integers between$1$
and$1000,$
the answer is$750.$