1.

In the first case,

In the second case, because is a factor of a.

2. You have (2,3) = 1 but 2+3=1

Also (17,19) = 1

The statement I not true as I understand it.

Results 1 to 6 of 6

- Jan 14th 2009, 03:43 PM #1
## GCD Problem

I've been working on these two problems for the last few days, can't seem to get it. Any help would be very much appreciated.

1. Prove that a|bc iff a/(a,b)|c

So far I know that the denominator has to be a common multiple of both a & b, but I cannot prove that it has to be the gcd of a & b.

2. Show that if (a,b)=1, then (a+b,a2-ab+b2)=1 or 3

So far this is what I've come up with, using the theorems:

a) (a,m)=1, (b,m)=1, then (ab,m)=1

b) (a,b)=(a,b+na) that:

(a(a+b),b) = (a2+ab,b) = 1

(a,b(b-2a)) = (a,b2-2ab)=1

but I can't seem to relate the two.

Again, thanks for any help in advance.

- Jan 14th 2009, 05:44 PM #2

- Jan 14th 2009, 06:02 PM #3
(1) Since it's an if and only if statement, we have to show the statement forward and its converse is true.

(a)

Let . For some , we have:

Since divides both terms on the left hand side (justify it), we can conclude .

(b)

So we have:

Since by definition, then .

Put those 2 lines together and we have our conclusion.

_______________________________

You have (2,3) = 1 but 2+3=1

Also (17,19) = 1 (19-17)^2 = 4

The statement I not true as I understand it.

For example, and . Then:

And another: and . Then:

- Jan 14th 2009, 07:08 PM #4

- Jan 14th 2009, 08:11 PM #5

- Jan 15th 2009, 09:17 AM #6