Let x = 2m and y = 2n
Then
4(m^2 - n^2) = 7389746232
m^2 - n^2 = 1847436558 = 2(923718279)
(m-n)(m+n) = 2(923718279)

The above expression means that 1847436558 can be expressed as a product of 2 factors, (m-n) and (m+n)
Sum of factors = 2m -> even.
This is however not possible since 1847436558 = 2(923718279) has only 1 even factor, which is 2. And even + odd = odd.

Hence no solutions

All right, I'll give various ways to solve this problem, take your pick.

Method 1.
All primes (greater than 3) when squared are divisible by 24.
All primes can be expressed as 6k+/-1 -> k can vary from 1 to 16. As also the integer 1 satsifies.
So total 33 values.

Method 2.
We need x^2Mod24 = 1
Let x = (12k + n) -> n can take values from 0 to 11
x^2 = 144k^2 + 24k + n^2
All terms on RHS are divisible by 24 except n^2.
So x^2Mod24 = n^2Mod24
Values of n that satsify our condition are - 1,3,5,7
Total
12k+1 - (k varies from 0 - 8) - 9 values
12k+3 - (k varies from 0 - 8) - 9 values
12k+5 - (k varies from 0 - 7) - 8 values
12k+7 - (k varies from 0 - 7) - 8 values

Total = 34 (but remove the number 3 of the form 12k + 3 for k = 0)
So total = 33

Method 3
24 = 8 * 3
Charmichael number for both 3 and 8 is 2.
What that means is that if k is coprime to 24, then k^2Mod24 = 1
So all no.s less than 100 coprime to 24 satisfy.
Uptill 96, we hav 96(1-1/2)(1-1/3) = 32 such numbers.
From 97 - 100, 97 also satisfies.
So total 33 numbers.

E[23] = 22
13 repeated 22 times is divisible by 23
13 repeated 22(18) = 396 times is divisible by 23.

@giyan Say a number is abcd
This can be represented as 1000a + 100b + 10c + d
Now,
(1000a + 100b + 10c + d)Mod9 = (a + b + c + d)Mod9 => A number that comes down to a single digit between 1-9

So finding remainder when dividing a number by 9 gives you the sum of its digits, down to a single digit.