GPT 6 Astra Ultra attempt on Math-6268
mathematicsErdős problemscontributionManuscript
Contents
Erdős Problem 1214: prime supports determine the base
Statement and outcome
Suppose positive integers $x,y$ satisfy \[ p\mid x^n-1\quad\Longleftrightarrow\quad p\mid y^n-1 \qquad\text{for every prime }p\text{ and }n\ge1. \tag{1} \] Then $x=y$. We give a proof for positive rational integers, using standard Kummer theory and Chebotarev as explicit external inputs. The general number-field support theorem of Corrales-Rodrigáñez--Schoof is stronger; we do not use that target theorem as a black box.Orders and the independent case
If $x=1$, all primes divide $x^n-1=0$. If $y>1$, any prime divisor of $y$ fails to divide $y-1$, contradicting $1$. Thus this case gives $y=1$, and similarly with $x,y$ reversed. Assume henceforth $x,y>1$. For every $p\nmid xy$, the sets of positive multiples of $\operatorname{ord}_p(x)$ and $\operatorname{ord}_p(y)$ agree by $1$. Therefore \[ \operatorname{ord}_p(x)=\operatorname{ord}_p(y). \tag{2} \]Suppose the two finite prime-exponent vectors $(v_q(x))_q$ and $(v_q(y))_q$ are linearly independent over $\mathbb Q$. Some two-by-two determinant of their entries is a nonzero integer $D$. Choose an odd prime $\ell$ not dividing $xyD$, and let $K=\mathbb Q(\zeta_\ell)$. The classes of $x,y$ are independent in $K^\times/(K^\times)^\ell$. Indeed, if $x^ay^b=z^\ell$ in $K$, take valuations at primes of $K$ above the two rational primes used in $D$. They are unramified in $K$, since neither is $\ell$. Reducing the resulting two equations modulo $\ell$ and using $D\not\equiv0$ forces $a\equiv b\equiv0\pmod\ell$.
We use the following standard Kummer theorem: if a characteristic-zero field contains $\zeta_\ell$ and two classes are independent modulo $\ell$th powers, adjoining their $\ell$th roots gives Galois group $(\mathbb Z/\ell\mathbb Z)^2$, acting by independent multiplication of the roots by powers of $\zeta_\ell$. Thus in \[ L=K(x^{1/\ell},y^{1/\ell}) \] there is an automorphism $\sigma$ fixing $K$ and $x^{1/\ell}$ but sending $y^{1/\ell}$ to $\zeta_\ell y^{1/\ell}$.
The field $L$ is Galois over $\mathbb Q$: it is the splitting field of $(T^\ell-x)(T^\ell-y)$. By the Chebotarev theorem, there is an unramified rational prime $p\nmid xy\ell$ whose Frobenius conjugacy class is that of $\sigma$. Its restriction to $K$ is trivial, so $p\equiv1\pmod\ell$. Every conjugate of $\sigma$ fixes the roots of $T^\ell-x$ and acts without fixed points on the roots of $T^\ell-y$: conjugation only replaces its nonzero exponent on $y^{1/\ell}$ by another nonzero exponent. The Frobenius factorization criterion therefore says that $x$ is an $\ell$th power modulo $p$ and $y$ is not. As $\mathbb F_p^\times$ is cyclic, this means \[ x^{(p-1)/\ell}\equiv1\pmod p,\qquad y^{(p-1)/\ell}\not\equiv1\pmod p, \] contradicting $1$. This proves multiplicative dependence. The named Galois-theoretic inputs, including Chebotarev, are external; the required independence and Frobenius application have been verified here.
An elementary supply of large prime-power orders
We next prove that for every integer $t\ge2$, every prime $\ell$, and every $h\ge2$, some prime $p$ has \[ \operatorname{ord}_p(t)=\ell^h. \tag{3} \] Set $u=t^{\ell^{h-1}}$ and $C=1+u+\cdots+u^{\ell-1}$. Any prime divisor $p$ of $C$ does not divide $t$, and its order divides $\ell^h$. If that order is smaller than $\ell^h$, then $u\equiv1\pmod p$, so $C\equiv\ell\pmod p$ and $p=\ell$.Moreover the exponent of $\ell$ in $C$ is at most one. For odd $\ell$, reduction modulo $\ell$ gives $C\equiv(u-1)^{\ell-1}$, so $\ell\mid C$ forces $u=1+\ell v$. Expanding modulo $\ell^2$ gives \[ C\equiv\ell+\ell v\sum_{j=0}^{\ell-1}j\equiv\ell\pmod{\ell^2}. \] For $\ell=2$, if $t$ is even then $C$ is odd; if $t$ is odd, $h\ge2$ implies $u\equiv1\pmod8$, whence $C=u+1\equiv2\pmod8$. Since $C>\ell$, it cannot have only $\ell$ as a prime divisor. A prime $p\ne\ell$ dividing it has the exact order in $3$. This proves $3$ without importing a primitive-divisor theorem.
Dependent bases must have the same exponent
Dependence of the nonnegative prime-exponent vectors implies \[ x=t^r,\qquad y=t^s \] for an integer $t\ge2$ and positive integers $r,s$. To see the integrality, divide the vector for $x$ by the gcd of its entries, obtaining a primitive integer vector $w$. The vector for $y$ is a rational multiple of $w$; a Bezout combination of the entries of $w$ shows that this multiple is an integer. Form $t$ from the exponents $w$.If $r\ne s$, choose a prime $\ell$ with $v_\ell(r)\ne v_\ell(s)$, and take $h>\max(v_\ell(r),v_\ell(s),1)$. A prime given by $3$ satisfies \[ \operatorname{ord}_p(t^r)=\ell^{h-v_\ell(r)},\qquad \operatorname{ord}_p(t^s)=\ell^{h-v_\ell(s)}. \] These are different, contradicting $2$. Hence $r=s$ and $x=y$.