Number theory test pending

Q1. Does there exist a natural number “(n)” such that (n^2 + n + 1) is divisible by (1955)? If “Yes”, provide an example and if “No”, provide a valid proof.
S1.
n^2 + n + 1 = 0 mod 5 (if it's div by 1955, it's also div by 5)
n^2 + n = 4 mod 5
Let's check all possible remainders of n with 5.
n = 0 mod 5
No.
n = 1 no.
n = 2
4 + 2 = 1 mod 5
n = 3
9 + 3 = 2 mod 5
n = 4
16 + 4 = 0 mod 5

So nothing works.
"No".

Q3.
Using the properties of congruences, prove that
89 | 2^44 - 1
and
97 | 2^48 - 1

S3.
2^44 - 1= 0 mod 89
(2^11 - 1)(2^11 + 1)(2^22 + 1) = 0 mod 89
Since 89 is a prime, one of these factors has to be a multiple of 89.
Let's test one by one.
2^11 = 1 mod 89.
2^5.2^6.3 = 3 mod 89
96.64 = 3 mod 89 = 7.64 = 448 = 3 mod 89 H.P.

Part 2:
2^48 - 1 = 0 mod 97 = (2^6 - 1)(2^6 + 1)(2^12 + 1)(2^24 + 1)
First 2 factors are small.
Next:
2^12 + 1 = 0 => 2^12 = -1 mod 97 = 96 mod 97 = 32.128 => 128 = 3 mod 97 which is wrong

Next:
2^24 = -1 = 96 mod 97
2^19 = 3 mod 97
2^19.3 = 9 mod 97 = 2^14.96 = -2^14
=> 2^14 = -9  mod 97
2^14*3 = -27 = 2^9*-1 => 2^9 = 27 mod 97
=> 512 = 27 mod 97
which is correct
H.P.

Q4.

Prove that the number (6n^3 + 3) cannot be a perfect sixth power of an integer for any natural number (n).
S4.

Proof by contradiction.
6n^3 + 3 = a^6 = (3k)^6
since a has to be multiple of 3
=>
2n^3 + 1 = 3^5.k^6 = 0 mod 9
Any integer cube mod 9 is -1,0,1
And one of them works here.
H.P.

Q10.

Find the pairs of integers ((x, y)) for which 4^y - 615 = x^2.

S10.
2^(2y) - x^2 = 615
(2^y - x)(2^y + x) = 615
Prime factors of 615 are 3.41.5

Case 1:
2^y + x = 615
2^y - x = 1
=> 2^(y+1) = 616, no go

Case 2:
2^y + x = 205
2^y - x = 3
=> 2^(y+1) = 208, no go

Case 3:
2^y + x = 123
2^y - x = 5
=> 2^(y+1) = 128, => y = 6, x = -+59

Case 4:
2^y + x = 41
2^y - x = 15
=> 2^(y+1) = 56, no go

Final answer:
y = 6, x = -+59