Need of Proof

[googol]

After messing around with equations a found 2 things. I'm not sure if they've already been discovered by I would like to see if it is possible to prove them:

If A is a whole number greater that 2 and B is the sum of every positive integer reletively prime and less than a, then B (mod a) = 0

and

If A is a whole number greater that 2 and B is the product of every positive integer reletively prime and less than a, then B is congruent to -1 modulo a.

I'm not sure if these are always true but oh well...

[HallsofIvy]

The first, "If A is a whole number greater that 2 and B is the sum of every positive integer reletively prime and less than A, then B (mod A) = 0", is true and, I THINK, is a consequence of Euler's theorem: "If a is relatively prime to n and phi(n) is the number of integers less than n and relatively prime to it, then a^(phi(n)) is congruent to 1 mod n".

The second, "If A is a whole number greater that 2 and B is the product of every positive integer reletively prime and less than A, then B is congruent to -1 modulo A", is not true. In particular, if A= 8, then the numbers relatively prime to 8 are 1, 3, 5, 7. Their product is 105 which is congruent to 1 mod 8, not -1. Similarly, if A= 12, then the numbers relatively prime to A are 1, 5, 7, 11. Their product is 385 which is congruent to 1 mod 12, not -1.

[Metal70]

The first one is certainly true for a when a is prime. Assume a is prime. Then where N={n|n is less than, and relatively prime to, a}={1,2,3, ..., a-1}. Then B=sum(n), can be ordered as [(1+[a-1])+(2+[a-2])+ . . .] = a + a + a . . . =na. If B=na, then B(mod a)=0.

I'll think of the other two proofs tomorrow . . . I'm tired, lol.

[Metal70]

When a is a nonprime odd integer or even, we must show that (n,a)=1 => (a-n,a)=1. Then, again, we can order the elements of the set of numbers less than and relatively prime to a so that we have n+(a-n)=a for all n in the set. Then, again, B=a+a+a+ ... +a=0(mod a).

Proof by contradiction: Assume (n,a)=1 and (a-n,a)=k for some inteteger k. Then k|(a-n) and k|a. if k|a, then a=kl for some integer l. If k|(a-n), a-n=rk for some integer r and a=rk+n. Thus, setting a's equal, kr+n=kl => n=kl-kr => n=k(l-r), which implies that k|n. This contradicts our assumption that (a,n)=1.