Wieferich Criterion proof verification

Question

Possible proof of Wieferich criterion:

Theorem: Let $p$ be an odd prime and $$ \gcd(x,y,z) = 1$$ $$xyz \not \equiv 0 \pmod p$$ $$x^p = y^p + z^p$$ Then $p$ is Wieferich prime.

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Proof:

We use the congruence:

$$\frac{x^n-y^n}{x-y} \equiv nx^{n-1} \pmod{x-y}$$

which we can prove by induction on $n$ and in which we divide first.

Let $n = p$.

Since:

$$ \gcd(x - y,\frac{x^n-y^n}{x-y})$$

$$ = \gcd(x - y,px^{p - 1})$$

$$ = \gcd(x - y,p)$$

$$ = \gcd(x - z,p)$$

$$ = 1$$

it follows that:

$$ x - y = r^p$$

$$ (x^p - y^p)/(x - y) = s^p$$

$$ x - z = t^p$$

$$ (x^p - z^p)/(x - z) = u^p$$

$$ rs = z$$

$$ tu = y$$

for some $r,s,t,u$ with $\gcd(r,s) = \gcd(t,u) = 1$.

The following also holds for $x - z,t,u$:

$$ s \equiv 1 \pmod p \implies s^p \equiv 1 \pmod {p^2}$$

Now let $n=p+1$:

$$\frac{x^{p+1}-y^{p+1}}{x-y} = (p+1)y^p+a(x-y)$$ $$\frac{x^{p+1}-z^{p+1}}{x-z} = (p+1)z^p+b(x-z)$$

$$\implies (p+1)(y^p-z^p) = b(x-z)-a(x-y) = bt^p -ar^p$$

$$\implies p(y^p-z^p) = t^p(b-u^p)-r^p(a-s^p) \equiv t(b-1)-r(a-1) \equiv 0 \pmod{p}$$

Now let $q_p(x)$ denote the Fermat-quotient of $x$ modulo $p$, then

$$q_p(x) \equiv q_p(y) \equiv q_p(z) \equiv 0 \pmod{p}$$

which is provable by other means and so we have:

$$\frac{x^{p+1}-y^{p+1}}{x-y} = (p+1)y^p+a(x-y) \equiv x+y \pmod{p^2}$$ $$\frac{x^{p+1}-z^{p+1}}{x-z} = (p+1)z^p+b(x-z) \equiv x+z \pmod{p^2}$$

$$\implies by \equiv bt \equiv az \equiv ar \equiv x \pmod{p}$$

$$\implies t(b-1)-r(a-1) \equiv r-t \equiv 0 \pmod{p}$$

$$\implies r \equiv t \equiv y \equiv z \pmod{p}$$

$$\implies x \equiv y+z \equiv 2y \pmod{p}$$

$$\implies x^p \equiv y^p+z^p \equiv 2y^p \equiv (2y)^p \pmod{p^2}$$

$$\implies 2^p \equiv 2 \pmod{p^2}$$

Answer 1

Shorter definite proof of the Wieferich-criterion.

Theorem

Let:

$$x^p = y^p+z^p$$ $$p \not | xyz$$ $$gcd(x,y,z)=1$$

Then:

$$2^{p-1} \equiv 1 \pmod{p^2}$$

Proof

Let:

$$az \equiv x-py \pmod{p^3}$$ $$by \equiv x-pz \pmod{p^3}$$ $$a \equiv \frac{x}{z}+rp+sp^2 \pmod{p^3}$$ $$b \equiv \frac{x}{y}+tp+up^2 \pmod{p^3}$$

for some $r,s,t,u$.

Note:

$$az \equiv x+prz+p^2sz \equiv x-py \pmod{p^3}$$ $$by \equiv x+pty+p^2uy \equiv x-pz \pmod{p^3}$$

$\implies$

$$y+rz+psz \equiv 0 \pmod{p^2}$$

$\implies$

$$\frac{y}{z} \equiv -(r+ps) \equiv (a-1)^p \pmod{p^2}$$

$\implies$

$$(a-1)^p \equiv a^{p}-1 \equiv -(r+ps) \equiv -\left(\frac{a-\frac{x}{z}}{p}\right) \pmod{p^2}$$

$\implies$

$$pz(a^p-1) \equiv x-az \pmod{p^2}$$

$\implies$

$$az(pa^{p-1}+1) \equiv az(1+p) \equiv x+pz \pmod{p^2}$$

By symmetry:

$$az(1+p) \equiv x+pz \pmod{p^2}$$ $$by(1+p) \equiv x+py \pmod{p^2}$$

$\implies$

$$(1+p)(by-az) \equiv p(y-z) \equiv by-az \pmod{p^2}$$

$\implies$

$$by-az \equiv p(y-z) \equiv 0 \pmod{p^2}$$

$\implies$

$$y \equiv z \pmod{p}$$

$\implies$

$$y^p \equiv z^p \pmod{p^2}$$

$\implies$

$$x^p \equiv y^p+z^p \equiv 2y^p \pmod{p^2}$$

$\implies$

$$2^{p-1} \equiv 1 \pmod{p^2}$$

Answer 2

Wieferich criterion:

Theorem

Let:

$$x^p = y^p+z^p$$ $$p \not | xyz$$ $$gcd(x,y,z)=1$$

Then:

$$2^{p-1} \equiv 1 \pmod{p^2}$$

Proof

Assume:

$$(yz)^{p-1}-1 \equiv y^{p-1}+z^{p-1}-2 \equiv 0 \pmod{p^3}$$

Let:

$$az \equiv x-py \pmod{p^3}$$ $$by \equiv x-pz \pmod{p^3}$$

Note:

$$a^{p^2} \equiv \left(\frac{x}{z}\right)^p \equiv (a-1)^{p^2}+1 \pmod{p^3}$$ $$b^{p^2} \equiv \left(\frac{x}{y}\right)^p \equiv (b-1)^{p^2}+1 \pmod{p^3}$$

Note:

$$\left(\frac{y}{z}\right)^{p-1}+1-\frac{2}{z^{p-1}} \equiv \left(\frac{y}{z}\right)^{p-1}+1-2y^{p-1} \equiv 0 \pmod{p^3}$$

$\implies$

$$\left(\frac{2y}{z}\right)(y^{p-1}-1) \equiv \left(\frac{y}{z}\right)^p -\frac{y}{z} \equiv \left(\frac{y}{z}\right)^p+1-\left(\frac{y}{z}-1\right) \equiv\left(\frac{x}{z}\right)^p-\frac{y+z}{z} \pmod{p^3}$$

By symmetry:

$$2y(y^{p-1}-1) \equiv za^{p^2}-(y+z) \pmod{p^3}$$ $$2z(z^{p-1}-1) \equiv yb^{p^2}-(y+z) \pmod{p^3}$$

$\implies$

$$za^{p^2}-yb^{p^2} \equiv 2(y^p-z^p)-2(y-z) \pmod{p^3}$$

$\implies$

$$x^p\left(\frac{1}{z^{p-1}}-\frac{1}{y^{p-1}}\right) \equiv x^p(y^{p-1}-z^{p-1}) \equiv 2y(y^{p-1}-1)-2z(z^{p-1}-1) \equiv (y+z)(y^{p-1}-z^{p-1}) \pmod{p^3}$$

$\implies$

$$y(y^{p-1}-1)-z(z^{p-1}-1) \equiv zy^{p-1}-yz^{p-1} \pmod{p^3}$$

$\implies$

$$y^{p-1}(y-z)-z^{p-1}(z-y) \equiv y^{p-1}(y-z)+z^{p-1}(y-z) \equiv y-z \pmod{p^3}$$

Suppose $p \not | y-z$:

$$y^{p-1}+z^{p-1} \equiv 1 \pmod{p}$$

which is impossible.

Conclusion:

$$y \equiv z \pmod{p}$$

$\implies$

$$y^p \equiv z^p \pmod{p^2}$$

$\implies$

$$x^p \equiv (y+z)^p \equiv y^p+z^p \equiv (2y)^p \equiv 2y^p \pmod{p^2}$$

$\implies$

$$2^{p-1} \equiv 1 \pmod{p^2}$$