Euler's proof of Fermat's Last Theorem for cubes I'm working through Richard Freidburg's An Adventurer's Guide to Number Theory, and on p. 149, chapter 7, he presents Euler's proof of Fermat's Last Theorem for cubes.
He starts with $z^3 = x^3 + y^3$, where $x, y, z$ have no factors in common. This implies that $z$ is even and that $x$ and $y$ are odd.
He doesn't state explicitly, but I think his proof assumes $x > y$.
Because $x$ and $y$ are both odd, $x + y$ is even and $x - y$ is also even.
He defines $p = (x + y)/2$, $q = (x - y)/2$, therefore $x = p + q$ and $y = p - q$.
By substitution, binomial expansion and factoring, he converts $z^3 = x^3 + y^3$ to $z^3 = 2p(p^2+3q^2)$
He then presents 4 'remarks', that the reader should prove. I think I've got 1, 2, and 4 figured out & need help with 3:
Remark 1: $p$ and $q$ have no common factor
Proof: Suppose $p$ and $q$ have a common factor $a$, then $x = a(p\prime + q\prime)$ and $y = a(p\prime - q\prime)$, therefore $a$ is a common factor of both $x$ and $y$, but $x$ and $y$ have no common factors
Remark 2: $p^2 + 3q^2$ is odd
Proof: $x = p + q$, $x$ is odd, therefore either $p$ is odd or $q$ is odd. Both cannot be odd, nor can both be even. If $p$ is odd then $q$ is even and $p^2$ is odd, $q^2$ is even and thus $3q^2$ is also even, and $p^2 + 3q^2$ is an odd number plus an even number, hence an odd number. Conversely, if $p$ is even, then $p^2$ is also even, and $q$ is odd, as is $q^2$, thus $3q^2$ is also odd. In this case $p^2 + 3q^2$ is an even number plus an odd number, hence an odd number.
Remark 3: $q$ is not divisible by 3
(Hints provided by the author: For this you must use the fact proven earlier in this chapter, that every cube is congruent to 0, 1 or -1 modulo 9. If $q$ is divisible by 3, then $x \equiv y$ modulo 3, and hence $x^3 \equiv y^3$ modulo 9.)
Remark 4: $p$ and $p^2 + 3q^2$ have no prime factor in common, except possibly 3.
Proof: Follows from remark 1. $p$ and $q$ have no common factors, therefore $p^2$ and $q^2$ have no common factors. However if 3 is a factor of $p$, then $p$ and $3q^2$ share 3 as a common factor.

Well, if $x^3$ and $y^3$ are congruent to $0$ mod $9$, then so is $z^3$, which contradicts "no common factors". If they are both $1$ (resp $-1$) then $z^3$ is $2$ (resp $-2$) which contradicts that "cube should be congruent to $0,1,-1$".
