Easy number theory.

If the gcd of the two numbers is $24$ , then you know both numbers have to be divisible by $24$ . And if the lcm of the two numbers is $144$ , and not $24$ , then at least one of the numbers is greater than $24$ , and therefore a multiple of $24$ .

$\frac{144}{24}=6$ . This means that the two numbers must each have a $24$ in their factorization, and a unique part of $6$ . This ensures that the lcm will be $144$ and the gcd will be $24$ .

$a=24\times 6=144 \text{ }\text{ }\text{ }\text{ }b=24 => a+b=168\neq 120$

$a=24\times 3=72 \text{ }\text{ }\text{ }\text{ } b=24\times 2=48 => a+b=120$

So $a=72$ and $b=48$ . This means that $a^2-b^2=72^2-48^2=5184-2304=2880$

Let, $a=pg$ and $b=qg$

So, their lcm, gcd and sum can be written as ,

$pqg=144$ ;

$g=24;$

$g(p+q)=120;$

Now, $\frac{g(p+q)}{g}$ = $p+q=5.$

And,

$\frac{pqg}{g}$ = $pq=6.$

$(p-q)²=(p+q)²-4pq=5²-4×6=1$

$\Rightarrow$ $p-q=1.$

From this, we get $p=3$ $and$ $q=2.$

So, $a=pg=3×24=72$ $and$ $b=qg=2×24=48.$

So, $a²-b²=72²-48²=$ $\boxed {2880}$