# Thread: congruence modulo twenty four

1. ## congruence modulo twenty four

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 $\displaystyle \sum_{d|n}d$

I don't need complete proofs, just some hints would be enough.

2. Hello,
(never mind what was there before)
Choose any $\displaystyle a$. We have $\displaystyle ab \equiv -1 \pmod{24}$. Therefore, $\displaystyle b \equiv -a^{-1} \pmod{24}$. Plug this result back into the second condition : $\displaystyle a + b \equiv a - a^{-1} \pmod{24}$. Now you might want to take a look at what the multiplicative inverses are in $\displaystyle \mathbb{Z}/24\mathbb{Z}$ (it is special)

3. Thanks, that proves 1, do you have a hint for 2 as well?

4. Originally Posted by EinStone
Thanks, that proves 1, do you have a hint for 2 as well?
Sorry, I haven't had time to look at the other ones and it's bed time for me now
Quickly looking, maybe proving that there are no perfect squares of the form $\displaystyle 24k - 1$ ? Maybe the prime factorization $\displaystyle 24 = 2^3 \times 3$ could help?

5. Is this correct for 2.?

Assume n ≡ −1 (mod 24) is a perfect square. Then $\displaystyle n = c^2$, thus c*c ≡ −1 (mod 24), by 1. we get c + c = 2c ≡ 0 (mod 24).

so $\displaystyle \exists M:$ 2c = M*24 => c= M*12 => $\displaystyle c^2= 144M^2$, hence $\displaystyle n = c^2 \equiv 0\mod 24$, a contradiction.

And how to show that there is an even number of divisors?

EDIT: Found 2.
Need proof for 3.