2105/049) For what positive integral values of x is $3^x-x^2$ is divisible by 5

As 3 and 5 are coprimes

So $3^4 = 1$ mod 5
$3^{4k+1} = 3$ mod 5
$3^{4k+2}= 4$ mod 5
$3^{4k+3} = 2$ mod 5

Now $(5k+m)^2$ mod 5 = 0 if m= 0, 1 if m= 1 or 4, 4 if m = 2 or 3 mod 5

now $3^x= x^2$ mod 5 if

x = 4k that is 0 mod 4 and 1 mod 5 ( in both cases remainder 1)
or x = 4k and 4 mod 5( in both cases remainder 1)
or x = 4k+2 or 2 mod 4 and 2or 3 mod 5
x = 0 mod 4 and 1 mod 5 => x= 16 mod 20
or x = 4k and 4 mod 5 => x = 4 mod 20

or x =2 mod 5 and 2 mod 4 => x = 2 mod 20

or x = 2 mod 4 and 3 mod 5 => x= 18 mod 20

x = a mod 4 ,b mod 5 can be solved by Chinese remainder theorem
I have not detailed steps