# Question about difference of powers related to FLT

I am wondering about differences of powers and some idea about it. It is related to FLT in that if the idea is correct something good could be done with it. I will give an example for prime $3$ but it could easily work for any odd prime. My question is: is the idea correct?

Let the following equation hold:

$$x^3 - y^3 = 3y^2(x - y) + 3y(x - y)^2 + (x - y)^3 = z^3$$

for some integers $x,y,z$.

We can obtain this by means of the binomial expansion of $((x - y) + y)^3 = x^3$ and subtracting $y^3$ to get $z^3$.

Rewrite the equation as follows(by recursively factoring out $x - y$):

$$(x - y)(3y^2 + (x - y)(3y + (x - y))) = z^3$$

We will prove: $p \mid x - y \implies p^3 \mid x - y$ for any prime other than $3$:

Assume $\gcd(x,y,z) = 1$,

Let $p \ne 3$ be some prime for which $p \mid x - y$,

$\quad\quad p \mid x - y \implies p \mid z \implies p^3 \mid z^3$

So we get:

$\quad\quad (x - y)(3y^2 + (x - y)(3y + (x - y))) \equiv z^3 \equiv 0 \pmod{p^3}$

Suppose now $p^3 \nmid x - y$, then:

$\quad\quad 3y^2 + (x - y)(3y + (x - y)) \equiv 0 \pmod{p}$

And so:

$\quad\quad (x - y)(3y + (x - y)) \equiv -3y^2 \pmod{p}$

$\quad\quad \implies p \mid 3y^2$ which is impossible because of $p \ne 3$ and $\gcd(x,y,z) = 1$.

We may conclude that for every prime $p \ne 3$, $p \mid x - y$ we have $p^3 \mid x - y$

The same idea should work for prime-divisors of $x - z$.

• I neither know what FLT nor I completely understand what your question, so sorry in advance. If $x - y = p$ then $p | x- y$ and $p^2 \nmid x - y$. Surely there are infinitely many integers with $x - y = p$ Dec 6, 2017 at 14:17
• Yes but there should be none for which the equation holds
– user507181
Dec 6, 2017 at 14:22
• So you assume from the start that $(x - y),z \equiv 0 \pmod{5}$ Then make the conjecture/assumption that $(x^5 - y^5)=z^5$ which implies $(x^5 - y^5)\equiv 0 \pmod{5^5}$. It seems both assumptions cannot be true. However you do not show this is the only possible pairing of assumptions in regard to FLT? What about other assumptions like $(x - y)^5,z \equiv 0 \pmod{5}$? Since $(x^5-y^5)-(x-y)^5=5xy(x-y)(x^2+xy+y^2-2xy(x+y))$, $(x-y)^5$ need only be divisible by single prime factor of $5$. Dec 6, 2017 at 14:22
• I don't really follow this but if $x - y \equiv z \equiv 0 \pmod{5}$ then $(x - y)^5 \equiv 0 \pmod{5^5}$
– user507181
Dec 6, 2017 at 14:35
• Sorry got mixed up. $(x^5-y^5)-(x-y)^5=5xy(x-y)(x^2+xy+y^2-2xy(x+y))$ proves that $(x - y) \equiv 0 \pmod{5}$ if we assume $(x^5-y^5) \equiv 0 \pmod{5}$ Dec 6, 2017 at 14:50

Assuming we have $z^3=x^3-y^3=(x-y)(x^2+xy+y^2)$ and then \begin{align} \gcd(x-y,x^2+xy+y^2) &= \gcd(x-y,x^2+xy+y^2)\\ &= \gcd(x-y,2xy+y^2)\\ &= \gcd(x-y,3xy). \end{align} But because you assume also $\gcd(x,y,z)=1$, it follows $\gcd(x-y,x^2+xy+y^2)=\gcd(x-y,3)$.

In other words, $x-y$ and $x^2+xy+y^2$ have no common factor apart from $3$, and so of course, if you find prime $p \mid x-y$, $p\neq 3$ then indeed $p^3 | x-y$.

But that is not new, same applies for generic $n$. If you have $$z^n=x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dots+xy^{n-2}+y^{n-1})$$ then you can show similarly as before (assuming $\gcd(x,y,z)=1$ again) that $$\gcd(x-y,x^{n-1}+x^{n-2}y+\dots+xy^{n-2}+y^{n-1})=\gcd(x-y,n)$$ and so if there is a prime $p$ such that $p\nmid n$ and $p|x-y$, then it follows $p^n\mid x-y$.

So the idea is correct, question is whether it is of any use, which I do not know.

• Thanks, I suspect(for case $n = 3$) one of $x - y$,$x - z$ should be a $3$-th power but am not sure yet.
– user507181
Dec 10, 2017 at 18:01
• Yes, since at least one of the $x-y$ or $z-x$ cannot be divisible by $3$ (because otherwise both $z^3$ and $y^3$ would be divisible by $3$, which is not possible). But if for example $x-y$ is not divisible by $3$, then $(x-y,x^2+xy+y^2)=1$ and $(x-y)(x^2+xy+y^2)=z^3$ implies that $x-y=a^3$ and $x^2+xy+y^2=b^3$.
– Sil
Dec 10, 2017 at 18:06