Hello,
(never mind what was there before)
Choose any . We have . Therefore, . Plug this result back into the second condition : . Now you might want to take a look at what the multiplicative inverses are in (it is special)
1. if ab ≡ −1 (mod 24), then a + b ≡ 0 (mod 24).
2. If n ≡ −1 (mod 24), then it is not a perfect square, and the number of divisors is even.
3. If n ≡ −1 (mod 24), then 24 divides
I don't need complete proofs, just some hints would be enough.
Hello,
(never mind what was there before)
Choose any . We have . Therefore, . Plug this result back into the second condition : . Now you might want to take a look at what the multiplicative inverses are in (it is special)
Is this correct for 2.?
Assume n ≡ −1 (mod 24) is a perfect square. Then , thus c*c ≡ −1 (mod 24), by 1. we get c + c = 2c ≡ 0 (mod 24).
so 2c = M*24 => c= M*12 => , hence , a contradiction.
And how to show that there is an even number of divisors?
EDIT: Found 2.
Need proof for 3.