Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

On the right track....

Answer 5

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?

Answer 6

a prime..

Answer 7

1 prime then ..

Answer 8

i havent come across an algorithm for primes to math against

Answer 9

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.

Answer 10

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

Answer 11

Getting there...

Answer 12

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

Answer 13

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

Answer 14

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.

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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.

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

alright, i can see that.

Answer 25

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

Answer 26

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

Answer 27

As an aside, this question is somewhat easier

http://openstudy.com/groups/mathematics/updates/4e578a740b8b9ebaa896cfbd

Answer 28

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

Answer 29

right right.

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

Actually, makes more sense to put it here

Show that any prime greater than 3 must be of form 6k+1 or 6k+5

Answer 35

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.

Answer 36

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.

Answer 37

Yes, using the same principle as here

6q,6q+1...etc

3 are even and 1 is divisible by 3...

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

Sorry...:-)

Answer 42

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

Answer 43

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

Answer 44

heheh

Answer 45

class starts at 9:30 :)