1. ## Number theory(Primes)

*Determine whether the following assertions are true or false. if true, prove the result, and if false, give a counter example

2. If p is a prime and p/(a^2+b^2) and p/(b^2+c^2) then p/(a^2-c^2).

2. Hint : if $p|a,\: p|b$ then $p|(a\pm b)$

3. ## help me pls

oh bruno, i don't really get it..can u help me

4. If $p|a,\: p|b$ then $a=pm,\: b=pn$. Then $a \pm b = pm \pm pn = p(m \pm n)$, so $p|(a \pm b)$.

Now how can you get $a^2-c^2$ from $a^2+b^2$ and $b^2+c^2$?

5. ## primes

i'm still blurred...sorry i still don't get it. oh..is it true? is it the final answer or do i need to elaborate it further?

6. Exactly one minute has gone by between my post and your call for more help. Perhaps taking five minutes to try to understand wouldn't do any harm.

If you can't even judge by yourself whether this is the answer or not, you might want to go back to your textbook. Do you even understand the statement of the problem?

7. Maybe this will help:

We will use some substitution, let $a^2=x, \ b^2=y$ and $c^2=z$

So we have $p|x+y$ and $p|y+z$

Note that $a^2 - c^2= x-z$, so we are trying to see if $p|x-z$

In case Bruno's proof confused you because he used $a$ and $b$ also, I'll put it here again with different letters, and with abit more detail.

If $p|g$ and $p|h$ then there exists $m$ and $n$ such that $g=pm$ and $h=pn$. So $g \pm h = pm \pm pn = p(m \pm n)$.

Note that the $\pm$ just means that it works for both addition and subtraction.

Now, $m \pm n$ is just an integer, which we can designate as $l$. So, $g \pm h = p(m \pm n) = pl$. Since $l$ exists, by definition $p|g \pm h$

What Bruno is asking is that if we let $x+y=g$ and $y+z=h$, is there a way we can add or subtract $g$ or $h$ so that we get $x-z$