Hello, can anyone help me with these two problems?? Thank you so much in advance.
1) Prove: If x ≡ y (mod m), then (x, m) = (y, m)
2) Show that if n > 4 is not prime, then (n-1)! ≡ 0 (mod n).
Let .
I presume it means .
We know that,
thus, for some .
Let .
Let .
We will prove it by trichtonomy.
Assume .
But that cannot be because,
And and .
Thus, thus . And we also know that .
Thus, because . And hence it is not "greatest".
By similar reasoning. We can show leads to contradiction.
Thus, by trichtonomy,
There are two possibilities.
1)n is not a square.
2)n is a square.
If n is not a square then it has a non-trivial proper factorization where .
Where are distinct.
Thus among the factors of :
We can find its factors .
When is a square there are 2 possibilities.
1)n is not a square of a prime.
2)n is a square of a prime.
If n is not a square of a prime we know that where because it is not a prime. Thus, it has a factorization in the form where are distinct and the same argument applies.
If n is a square of a prime then we have a minor problem. For example if n=4=2^2 it does not work. Which is why the initial conditions of the problem says n>4. Thus we know that and in no other way. We know that contains one factor. But what about other another factor of ? It turns out that when the factor appears among . Thus, there is another number that has a factor of . The reason why is because for thus, and is hence among one of the factors.
Hmm, you have:
2 <= a,b <= n - 1, shouldn't it be: 1 < a < b < n - 1? Maybe you got it mixed up for when it is a square. The proof I was given was:
Either n is a perfect square, n = a^2 in which case 2 < a < 2a <= n−1 and hence a and 2a are among the numbers {3,4, . . . ,n−1} or n is not a perfect square, but still composite, with n = ab, 1 < a < b < n−1.
Does this work? It contradicts a few of your results.
Sorry to be a pain, but I clearly understand your proof now TPF. What I don't understand is how the proof I was given (below) proves the claim; how does that tie it all together showing that n is composite if n > 4 and that n will then divide (n - 1)!
Proof:
Either n is a perfect square, n = a^2 in which case 2 < a < 2a <= n−1 and hence a and 2a are among the numbers {3,4, . . . ,n−1} or n is not a perfect square, but still composite, with n = ab, 1 < a < b < n−1.