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Mathematics
OpenStudy (anonymous):

Show that any integer of form 4k+3 is divisible by a prime of same form.

OpenStudy (amistre64):

4k+3 is odd; it has the same structure as 2k+1; or rather 2(2k) is even, and any even + an odd turns odd... but thats the answer to a different question aint it

OpenStudy (amistre64):

3,6,9,... has a characteristic about it that i can vaguely recall

OpenStudy (amistre64):

4(0) + 3 = 3, is prime; 4(1) + 3 = 7, is prime; 4(2)+3 = 11 is prime; ..15 is not prime;

OpenStudy (anonymous):

On the right track....

OpenStudy (amistre64):

15/3 is good; but are we saying it divisible by 1 prime of the same form? or by all primes of the same form?

OpenStudy (anonymous):

a prime..

OpenStudy (amistre64):

1 prime then ..

OpenStudy (amistre64):

i havent come across an algorithm for primes to math against

OpenStudy (anonymous):

im thinking i can rephrase the question like this: If\[m \equiv 3\]mod 4, then at least one prime that divides m must also be equivalent to 3 mod 4.

OpenStudy (amistre64):

\[\frac{4k+3(factored?)}{4k+3}\]

OpenStudy (anonymous):

Getting there...

OpenStudy (amistre64):

if you use "\ ( \ )" remove spaces, you form inline latex \(\cfrac{top}{not.the.top}\) like this

OpenStudy (anonymous):

Any integer is one of 4q,4q+1,4q+2 or 4q+3 (what Joe said).

OpenStudy (anonymous):

Which has to be true, because using the Fundamental Theorem of Arithmetic to prime factor m gives: \[m \equiv 3 \iff p_1p_2p_3\ldots p_n \equiv 3\] Each p is going to be equivalent to either 0, 1, 2, or 3. None of them can be 0, otherwise m would be divisible by 4. If each p was only 1 and 2, then the product of primes would be divisible by some power of 2, which has no way of being congruent to 3 mod 4. If each p was just 1, when the product would be equivalent to 1. So there must be at least one prime equivalent to 3 mod 4.

OpenStudy (anonymous):

4q and 4q+2 are even....

OpenStudy (anonymous):

i thought we were only concerned with integers in the form 4q+3?

OpenStudy (anonymous):

So any number of either of those forms is composite (excluding 2)

OpenStudy (anonymous):

The way im doing the problem is a proof by contradiction. Im showing if m was in the form 4k+3, and none of its primes were in that form, then we get a contradiction. If m is in the form 4k+3, then m is equivalent to 3 mod 4. If none of its primes are equivalent to 3 mod 4, then there is no way the product of the primes can be equivalent to 3 mod 4.

OpenStudy (anonymous):

So at least one prime has to be equivalent to 3 mod 4.

OpenStudy (anonymous):

That's fine, what about the divisibility...

OpenStudy (amistre64):

reading "the things they carried"-tom obrien; gotta talk about our feelings in phi101

OpenStudy (anonymous):

i used the Fundamental Theorem of Arithmetic. Those primes have to divide m.

OpenStudy (anonymous):

4k+3 is either prime or a a product of 2+ primes of form 4q+1 or 4q+3

OpenStudy (anonymous):

alright, i can see that.

OpenStudy (anonymous):

If u (4x+1)(4y+1) the answer is of form 4(z+1)

OpenStudy (anonymous):

it will actually be in the form 4z+1 because: \[(4x+1)(4y+1) = 16xy+4x+4y+1 = 4(4xy+x+y)+1 = 4z+1\]

OpenStudy (anonymous):

As an aside, this question is somewhat easier http://openstudy.com/groups/mathematics/updates/4e578a740b8b9ebaa896cfbd

OpenStudy (anonymous):

So if, 4k+3 was divisible only by primes of type 4q+1 it would itself be form 4q+1

OpenStudy (anonymous):

right right.

OpenStudy (anonymous):

thats what i said above! in a roundabout way probably, my bad >.<

OpenStudy (anonymous):

Just a question of dotting i's and crossing t's...

OpenStudy (anonymous):

"If each p was just 1, when the product would be equivalent to 1" it was that statement.

OpenStudy (anonymous):

I think u can easily manage the other one....

OpenStudy (anonymous):

Actually, makes more sense to put it here Show that any prime greater than 3 must be of form 6k+1 or 6k+5

OpenStudy (anonymous):

im a huge fan of "Proof by contradiction" For the other, i would show that if we had a prime in the form: \[6k+r, 0\leq r\leq 5\] that if r wasnt a 1 or a 5, you could factor our an integer, which shows it wasnt prime.

OpenStudy (anonymous):

Like if r was 3, then the "prime" would be: \[6k+3 = 3(2k+1)\]this shows that the number 6k+3 is divisible by 3, and therefore isnt prime.

OpenStudy (anonymous):

Yes, using the same principle as here 6q,6q+1...etc 3 are even and 1 is divisible by 3...

OpenStudy (anonymous):

i cant wait for stuff like this in my class. its very interesting lol

OpenStudy (anonymous):

thank you for stopping me from doing my homework <.< lololol. its not due till monday anywhos.

OpenStudy (anonymous):

It's hard to get ur head around it being all in integers when u are used to dealing with real numbers,lol

OpenStudy (anonymous):

Sorry...:-)

OpenStudy (anonymous):

I will put up 1 or 2 more , maybe....

OpenStudy (anonymous):

i should leave! otherwise i'll be late for class because im on here lol

OpenStudy (anonymous):

heheh

OpenStudy (amistre64):

class starts at 9:30 :)

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