# The equation $x^4 − y^4 = z^p$

In Darmon's paper The equation $x^4 − y^4 = z^p$, he says:

"Factorizing the left hand side of $a^4 − b^4 = c^p$, the assumption $\gcd(a, b) = 1$ forces the three factors $a + b, a − b, a^2 + b^2$ to be $p$th powers up to powers of $2$."

How does $\gcd(a,b)=1$ force this? What does he mean by "$p$th powers up to powers of $2$"?

Also, He states that $a^2-b^2$ is a $p$th power up to powers of $2$. This occurs in the discriminant and is also the product of the first two factors of the factorization of the left hand side of his equation.

Whatever "up to the powers of $2$" means, if two of the factor $a+b$ and $a-b$ have it then their product has it. The thing is, Darmon cites this as a reason that he is able to do Ribet's level lowering on part of his solution, so it thereby becomes critical to know what he means by "up to the powers of $2$".

Darmon states, "In any case, ord$_l(\Delta) \equiv 0 \pmod p$, since $a^2 − b^2$ is a $p$th power up to powers of $2$."

But why should the "since" and the appeal to "up to the powers of $2$" mean ord$_l(\Delta) \equiv 0 \pmod p$?

• $a+b\ \$ and $\ \ a-b\ \$ can both have prime factor $2$ , explaining "up to" Aug 19, 2018 at 17:42

A number $n$ is a $p$th power up to powers of 2 if and only if it can be expressed in the form $2^km^p$ for some integers $k,m$. Note that this is equivalent to saying that, in the prime factorization of $n$, all primes appear a multiple of $p$ times, except possibly for the prime 2.
For the first half of the question, one can use Euclid's algorithm to check that if $a$ and $b$ are relatively prime, then the three factors $a-b,a+b,a^2+b^2$ of $a^4-b^4$ are pairwise relatively prime up to a factor of 2. Thus, any prime factor $l$ of one of those three factors is not a factor of any of the others unless $l=2$. So, since $l$ must appear a multiple of $p$ times in the prime factorization of $a^4-b^4$ (as $a^4-b^4$ is a $p$th power), it must appear a multiple of $p$ times in the factorization of that factor alone.
For the second half, ord$_l(\Delta)$ is used to mean the number of times $l$ appears as a factor of $\Delta$. So, this is really saying that each prime other than 2 (because we are in the case $l\ne2$) appears a multiple of $p$ times in the factorization of $\Delta$. So, if we write $a^2-b^2$ as $2^k*m^p$, then $\Delta=2^6*c^{2p}*(a^2-b^2)^{2}=2^6*c^{2p}*(2^k*m^p)^{2}=2^{2k+6}*(c^2*m^2)^{p}‌​$, from which the claim follows.
• So, because they are factors of $a^4-b^4$ then obviously they are powers of $p$. No problem there. Since $gcd(a,b)=1$ means for example $a+b=2+3=5$, it is confusing to see exactly how that fact forces anything regarding a power of $2$. But, maybe I'm just having a senior moment. Aug 19, 2018 at 18:54
• Not powers of $p$ ($p^0,p^1,\dots$) but $p$th powers ($1^p,2^p,\dots$). Aug 19, 2018 at 20:09
• Actually, a better example would be a $p$th power such as in $49+76=125$ which is $7^2+4*19=5^3$ where $gcd(49,76)=1$ and $p=3$. So now the question can be precise. How does the fact that $49$ and $76$ are relatively prime affect the value of $p$ being "up to powers of $2$"? I guess it still is the "up to" that is bothering me. (Sorry I said "powers of $p$" in my 1st comment when I meant $p$th powers) Aug 19, 2018 at 20:22
• Consider the following simpler claim: if $xy$ is a square and gcd$(x,y)=1$, then $x$ and $y$ are squares. Can you see why this is true? Aug 19, 2018 at 20:31
• Example of the simpler claim: $4^2*9^2=36^2$. In this, $xy$ is a square and $gcd(x,y)=1$ and $x$ and $y$ are squares, but $6^2*6^2=36^2$ and in this expression $xy$ is a square and $gcd(x,y)\ne1$ and $x$ and $y$ are squares Aug 19, 2018 at 20:53