Rooted Inequalities

${(\sqrt{n} - \sqrt{23 \times 24})}^{2} < 1$

${(\sqrt{n} - \sqrt{552})}^{2} <1$

${(\sqrt{n} - 23.49...}^{2} < 1$

Now for ${a}^{2} <1$ , a < 1 or -1

Hence, $\sqrt{n} \approx 22.5 or 24.48$ ( $\approx$ should mean "around")

First take the case of 22.5,

hence, $n \approx 506.25$ , therefore, least value of n has to be 507

Now, taking the case of 24.48,

hence, $n \approx 599.27$ , therefore, maximum value of n has to be 599(because $\sqrt{600}$ will become more than 24.49

As a result,

all numbers from 507 to 599 can all be the values of n. So, there are in total $\boxed{93}$ numbers

(n^(1/2) - (23*24)^(1/2))^2 <1

sq.rt on both sides we get

(n^(1/2) - (23 24)^(1/2)) <1 and (n^(1/2) - (23 24)^(1/2)) >-1

n^(1/2)<1 + (23 24)^(1/2) and n^(1/2) >-1+ (23 24)^(1/2)

square on both sides

n<(1 + (23 24)^(1/2) )^2 and n >(-1+ (23 24)^(1/2) )^2

now range of n is [(1 + (23 24)^(1/2) )^2 ,(-1+ (23 24)^(1/2) )^2]

n (max) -n (min) = 4 ((23 24)^(1/2) ) =93.979 i.e no.of integers that satisfy inequalities are 93

It's trivial to see that $(m\pm1)^2=m^2\pm2m+1$ . Note since $\sqrt{(m+1)^2}-\sqrt{m^2}=1$ it's obvious that $\left|\sqrt{n+k}-\sqrt{n}\right|<1$ if and only if $-2\sqrt{n}+1<k<2\sqrt{n}+1$ .

In our case, $n=23\cdot24$ so clearly $23<\sqrt{n}<24$ and it follows that for integer $k$ it must be that $-2\cdot23+1\le k\le2\cdot23+1$ giving a total of $4\cdot23+1=93$ solutions.