# primes and unique factorization

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• Jan 20th 2010, 12:37 PM
empressA
primes and unique factorization
Prove or disprove each of the following statements:

a. if p is prime and p divides (a^2 + b^2) and p divides (c^2 + d^2), then p divides (a^2 - c^2).

b. if p is prime and p divides (a^2 + b^2) and p divides (c^2 + d^2), then p divides (a^2 + c^2).

c. If p is prime and p divides a and p divides (a^2 + b^2), then p divides b.

help plz..dont know where to start..
• Jan 20th 2010, 12:40 PM
Drexel28
Quote:

Originally Posted by empressA
Prove or disprove each of the following statements:

a. if p is prime and p divides (a^2 + b^2) and p divides (c^2 + d^2), then p divides (a^2 - c^2).

b. if p is prime and p divides (a^2 + b^2) and p divides (c^2 + d^2), then p divides (a^2 + c^2).

c. If p is prime and p divides a and p divides (a^2 + b^2), then p divides b.

help plz..dont know where to start..

What do you know about primes? Let's start with that.
• Jan 20th 2010, 12:46 PM
empressA
I know that p is a prime only if its divisors are positive neg. 1 and pos.neg. p..

i also know that if p is a prime and q is a prime then p divides q

and if p is a prime number that divides bc then it p divides b and p divides c..
• Jan 20th 2010, 12:49 PM
Drexel28
Quote:

Originally Posted by empressA
I know that p is a prime only if its divisors are positive neg. 1 and pos.neg. p..

i also know that if p is a prime and q is a prime then p divides q

and if p is a prime number that divides bc then it p divides b and p divides c..

That second statement is scarily wrong. $\displaystyle 2,3$.
• Jan 20th 2010, 12:50 PM
Dinkydoe
This question should rather be posted in the number-theory section.

(a) is false: let p= 5, a= 1, b= 2, c= 3, d = 1

Have you tried a few examples?

(b) Find a counter-example for (b).

(c) Is correct. If $\displaystyle a\equiv 0$ mod p then $\displaystyle a^2+b^2\equiv b^2$ mod p. Thus $\displaystyle (a^2+b^2)\equiv 0$ mod p $\displaystyle \Rightarrow b \equiv 0$ mod p.
• Jan 20th 2010, 12:52 PM
empressA
thanks dinky doe..

drexel 128..this is the notes my lecturer gave..lol..at least that second and third line
• Jan 20th 2010, 01:01 PM
empressA
Quote:

Originally Posted by Dinkydoe

(c) Is correct. If $\displaystyle a\equiv 0$ mod p then $\displaystyle a^2+b^2\equiv b^2$ mod p. Thus $\displaystyle (a^2+b^2)\equiv 0$ mod p $\displaystyle \Rightarrow b \equiv 0$ mod p.

we haven't reached congruence modular arithmetic
• Jan 20th 2010, 01:01 PM
Dinkydoe
Empress. I can hardly believe that's what your lecturer gave.

The very definition of prime-numbers is that they can't be divided by numbers other then themself and 1.

Thus if p,q are primenumbers and $\displaystyle p|q \Rightarrow p=q$. I think that's what your lecturer meant.
• Jan 20th 2010, 01:04 PM
empressA
yea..thats what i have..but i took it as p / q..because he didn't differentiate between the sign between the p and the q..but i c now
• Jan 20th 2010, 01:08 PM
Dinkydoe
Quote:

we haven't reached congruence modular arithmetic
Then try to prove the following:

If $\displaystyle p|a$, this means $\displaystyle a = np$ for some $\displaystyle n\geq 0$. So $\displaystyle a^2 = n^2p^2$.
And if $\displaystyle p|(a^2+b^2)\Rightarrow a^2+b^2 = pm$

Thus if $\displaystyle a^2+b^2 = n^2p^2+ b^2 = pm$. What does this say about $\displaystyle b^2$?
• Jan 20th 2010, 01:18 PM
empressA
Thus if $\displaystyle a^2+b^2 = n^2p^2+ b^2 = pm$. What does this say about $\displaystyle b^2$?[/quote]

that b is zero
• Jan 20th 2010, 01:22 PM
Dinkydoe
No:

Observe that from the above equation follows: $\displaystyle b^2 = p(m-n^2p)\Rightarrow p|b^2$

Now use the definition of prime-numbers.
• Jan 20th 2010, 01:24 PM
empressA
that would imply that p divides a^2 + b^2
• Jan 20th 2010, 01:26 PM
Dinkydoe
Quote:

that would imply that p divides a^2 + b^2
No, that's what we assumed in the first place. We also assumed that p divides a.

In the last post I showed you that from these assumptions follows that p divides $\displaystyle b^2$.

The definition of prime-numbers gives you that p divides b.
• Jan 20th 2010, 01:33 PM
empressA
ok..i understand where you're coming from now. so since p / b^2 it also divides b..
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