Question

Proof help

Answer 1

Consider two cases, z0 even and z0 odd. If z0 is even, then x0 and y0 are odd. But then x0 + y0 and x0 − y0 are both even. Let 2p = x0 + y0 and 2q = x0 − y0. Then x0 =p+q and y0 = p-q. It follows from gcd(x0, y0) = 1 that
gcd(p, q) = 1. Parity arguments show that p not≡ q (mod 2).
Lastly, z0^3 = 2p(p^2 + 3q^2). Summarizing, there exist p and q so that:
(a) p and q are positive
(b) gcd(p, q) = 1
(c) p not≡ q (mod 2)
(d) 2p(p^2 + 3q^2) is a cube

Answer the following:
1. Use gcd(x0, y0) = 1 to prove that gcd(p, q) = 1.
2. Prove that p not≡ q (mod 2).
3. Prove that z0^3 = 2p(p^2 + 3q^2).
Note x0 = x not, y0 = y not, z0 = z not

Answer 2

where is the problem

Answer 3

at the bottom 1, 2 and 3

Answer 4

which course is it?

Answer 5

first year algebra, university

Answer 6

q (mod 2) what deos it mean?

Answer 7

if you dont know what it means then you probably cant help, not too be rude :)

Answer 8

okay lets get started from the beginning it say when your z0 is even you x0,y0 are odd
and vice versa right

Answer 9

ok

Answer 10

you have 2q=x0-y0 and 2p=x0+y0
and you are already given that (x0,y0)= 1

Answer 11

you do not have a restriction on that you know what I mean
on other words it doesn't matter whether your (x0,Y0) are multiplied, added , or substracted

Answer 12

so setting (x0+y0)=1 and (x0-Y0)=1 and adding them together

Answer 13

r u there

Answer 14

ya how does that help with the question?

Answer 15

since you 2q= (x0-Y0) and 2p=(X0+Y0) substituting 1 in (X0,Y0) and adding the equations together which you will end up with q+p=1

Answer 16

hmm ok

Answer 17

you see what i did and this should be right cuz (p,q) are positive

Answer 18

ya

Answer 19

so do understand what i did , didn't you ?

Answer 20

ya

Answer 21

ty

Answer 22

which part do u want other than this

Answer 23

?

Answer 24

any of the other two

Answer 25

i am glad i can help but you were rushed in the beginning

Answer 26

Prove that p not≡ q (mod 2). Is this an equal or denoted ?

Answer 27

hello are you there?

Answer 28

hello

Answer 29

hi its kinda hard to explain that part

Answer 30

i got it i think it is modular of congruence isn't it ?

Answer 31

ya

Answer 32

so this mean if their a difference P0-q should be a multiple of your modular

Answer 33

its actually p is not congruent to q (mod 2)

Answer 34

and your modular in this case is 2

Answer 35

is what i meant the the "not"

Answer 36

ya

Answer 37

so having said that do u know how to use induction

Answer 38

yes

Answer 39

could please reminds of the condition of induction

Answer 40

im not sure if you understand exactly whats happening here

Answer 41

let assume we have c=2 and the we will have P0^n=q^n(modC) by using induction we prove this function is true for n=1.
Plugging n=1 we have the condition P0^1=q^1(modC) which gives P0=q^(modC) which is true

Answer 42

are you still here