Question

satellite, can you help me with this proof of Fermat's last theorem?

Answer 1

Part of Proof:
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

you are taxing my feeble brain. let me get a piece of paper

Answer 3

lol ty man

Answer 4

ok i got the first one
suppose gdc(p,q) = d > 1
i.e. d divides p and q
then d divides p + q and d divides p - q
so d divides x0 = p+ q and d divides y0 = p - q contradicting the assumption that
gcd (x0,y0)=1

Answer 5

ok looks good

Answer 6

"parity argument" is a fancy way of saying that one is even and the other is odd
that is clear yes?

Answer 7

i think so ya

Answer 8

p not≡ q (mod 2)
well there are only two possibilities mod 2, either 0 or 1. so all this is saying is that one is even and other is odd

Answer 9

if they are both even, then p + q and p - q are both even, and therefore x0 and y0 are both even, but gcd(x0,y0)= 1 so they cannot both be even

Answer 10

similarly if they are both odd, then p + q and p - q are both even, same contradiction as before
so one is even and the other is odd

Answer 11

ok, one sec brb

Answer 12

i will let you figure out why p and q are both positive. i think you have wolog x0>y0

Answer 13

and the last part is algebra. straight up compute
\[x_0^3+y_0^3=(p+q)^3+(p-q)^3=2p(p^2 + 3q^2)\]

Answer 14

ok tyvm can i ask you another question?

Answer 15

what if z0 is odd?
prove:
a) gcd(p,q) = 1
b) p not≡ q (mod 2)
c) 2p(p^2 + 3q^2) is a cube

Answer 16

last part is just algebra. same as before. you have
\[x_0^3+y_0^3=(p+q)^3+(p-q)^3=2p(p^2 + 3q^2)\] by algebra

Answer 17

and as before if z^3 is odd, and z^3 = x^3 + y^3
then one of x^3 and y^3 is even and one is odd. since the sum of an even and an odd number is odd.

Answer 18

meaning one of x,y is even and the other is odd. so as before one of p,q is even and one is odd

Answer 19

oh wait that may be wrong let me think a second

Answer 20

ok what are p and q if z^3 is odd?

Answer 21

oh i see it works the same way

Answer 22

\[x_0+y_0=2p+1\]
\[x_0-y_0=2q-1\]
\[x_0=p+q\]
\[y_0=p-q\] as before?

Answer 23

ya i think so

Answer 24

this is fermat's last theorem in the case n = 3 right? not the real one. n = 4 is easier

Answer 25

ya its case 3

Answer 26

we did case 4 in class

Answer 27

ok i think it is the same argument as before. nothing has changed.
gcd(p,q)= 1 because otherwise x and y have common factors

Answer 28

and like before if they are both even then so is x and y
and if they are both odd then x and y are both even
so it is identical to the one before if i am not mistaken. check it but i think it is right
and now i must go
good luck!

Answer 29

ty :)