Factorial Linear Forms and Denominators: Detailed Proofs and Rationality Criteria

Exact identities for shared prime powers

Let $I\subseteq \{2,3,\dots \}$ be finite. We first compute the prime-power levels common to two summand denominators. We then compare those exponents with the ones already supplied by a factorial factor.

Write $D(I)={\mathrm{lcm}}_{i<j,i,j\in I}gcd(d_i,d_j)$, with $D(I)=1$ for an empty or singleton family. For each prime $r$, its valuation is the second-largest valuation among the $d_i$, counting missing values as zero; the denominator lcm has the largest valuation. This proves $D(I){\mathrm{lcm}}_{i\in I}{d}_i\mid \prod _{i\in I}{d}_i$ prime by prime.

The integer $D(I)$ can be updated exactly when one index is added. For the positive factorial-gap denominators, adjoining a new index $a\ge 2$, $a\notin I$, gives

since finite-family gcd and lcm distributivity collapses the lcm of all pairwise gcds against $d_a$ to a single gcd. For a fixed positive integer $F$, take the lcm with $F$ on both sides:

Thus one need only retain the denominator lcm and $\mathrm{lcm}(F,D(I))$ at each step. The prescribed factor $F$ is not an extra summand denominator.

There is also an exact bound by the product of the denominators divided by their least common multiple. For a positive integer $F$, define $\tilde{C}(I)=\mathrm{lcm}(F,D(I))/F=D(I)/gcd(F,D(I))$. Since $\tilde{C}(I)$ divides $D(I)$, the preceding divisibility gives

For the factorial block this specialises to

an upper bound for the shared factor in terms of the denominator product and lcm. The bound alone does not establish (29).

Dividing by $F_p$ subtracts its prime exponents; it does not remove every prime that occurs in $F_p$. With $F_p=(p-1)!$, $D_p=D(I_p)$ and $C_p=\mathrm{lcm}(F_p,D_p)$, as in the main record,

For $e>0$, $r^e\mid {\tilde{C}}_p$ exactly when $D_p$ is divisible by $r^{e+v_r(F_p)}$, which in the block forces two distinct denominators to be divisible by that higher power, by the second-largest-valuation formula. If $i<j$ are two such indices and $f=v_r(F_p)$, then $r^{e+f}\mid j!/i!-1$. Since $r\mid j!-1$ forces $j<r$, $0<j!/i!-1<r^{j-i}$, hence $e+f<j-i$. For a prime $r\mid {\tilde{C}}_p$, this bounds the surviving valuation by $v_r({\tilde{C}}_p)+v_r(F_p)<r$. Since $v_r({\tilde{C}}_p)\ge 1$ and $v_r(F_p)\ge \lfloor (p-1)/r\rfloor$, we obtain $\lfloor (p-1)/r\rfloor \le r-2$, hence $p-1<r(r-1)<r^2$. Consequently ${\tilde{C}}_p$ is coprime to $k!$ whenever $k(k-1)\le p-1$: a prime $r\le k$ dividing both would give $p-1<r(r-1)\le k(k-1)\le p-1$. This excludes the small primes but does not bound the product of the remaining prime powers.

If a prime $r$ divides a denominator $n!-1$ with $n\ge p$, then $r\nmid F_p$ and $r^e\mid {\tilde{C}}_p⟺r^e\mid C_p$ for every $e>0$. This has an exact incidence-count form,

so a bound of at most one such index implies $v_r({\tilde{C}}_p)<e$. Counting the exponents gives

and for a prime $r>2p-1$ the range can be restricted to $e\in [1,2p-4]$. The same equivalence holds under the exact condition $r\nmid F_p$ in place of $r>2p-1$, hence in particular for every prime $r\ge p$. At $e=2$, an upper bound of one on the number of indices gives $v_r({\tilde{C}}_p)\le 1$.

A surviving prime power also forces two indices to be far apart. If $r$ is prime, $e>0$ and $r^e\mid {\tilde{C}}_p$, there are $i<j$ in $I_p$ with $r^{e+v_r(F_p)}\mid i!-1$, $r^{e+v_r(F_p)}\mid j!-1$ and $r^{e+v_r(F_p)}\le j^{j-i}$; consequently $(2p-1{)}^d<r^{e+v_r(F_p)}$ forces $d<j-i$. The required spacing follows directly: any two indices $i<j$ with $r^e\mid i!-1$ and $r^e\mid j!-1$ satisfy $e<j-i$, because $r\mid j!-1$ already forces $j<r$ and the inequality for $j!/i!-1$ then forces the strict separation. More generally, for any two indices $2\le i<j$ and any prime power $r^e$ with $e>0$,

so successive solutions of $i!\equiv 1(\mathrm{mod}{r}^e)$ are at least $e+1$ apart, and

The resulting bound on the exponent is

for every prime $r$ and $p\ge 2$. Thus the exponent already present in $F_p$ uses part of the same bound $2p-3$. To apply (25), one still needs sufficiently strong bounds for the product of the shared prime powers and for the gap ${\rho }_p/R_p$.

The available squarefreeness evidence is finite. An exhaustive modular scan through $r\le \mathrm{2,000,000}$ and $n\le 240$ found four pairs $(r,n)$ with $r^2\mid n!-1$ and no prime occurring in two such pairs. All $\mathrm{498,501}$ pairs $2\le a<b\le 1000$ have squarefree $gcd(a!-1,b!-1)$. An aggregate scan through $p=499$ also reports a ratio below $0.374$, but does not specify its logarithmic normalisation; that ratio is not used as a quantitative premise here. These are reported finite computations, not an asymptotic bound or a proof of squarefreeness at all indices.

A finite version of the argument for first prime occurrences compares the product of a chosen set of primes, each at least $5$, with $\prod _{2\le k\le B}(k!-1)$; if the prime product is larger, at least one chosen prime $q$ divides no $k!-1$ with $2\le k\le B$, while Wilson’s theorem bounds its first occurrence by $q-2$. Wilson reflection limits what a linear-size divisor can be assumed to be: if $n$ is odd, $n<q$, and $q\mid n!-1$, then $q\mid (q-n-1)!-1$. Suppose also that $p\le n$, both indices lie in $I_p$, and the reflected index is earlier, equivalently $q<2n+1$. The prime $q$ then divides two denominators in the block. Since $q>n\ge p$, it does not divide $F_p=(p-1)!$, so $q\mid {\tilde{C}}_p$: removing the factorial factor does not remove this shared prime.

Stewart states that for every $\epsilon >0$ there are infinitely many odd $n$ whose least prime factor $q$ of $n!-1$ satisfies

estimate (9) of Theorem 1 being stated for $n!+1$ [8] and transferred to $n!-1$ in the text [8]. This controls $q$ relative to the specified index $n$, not to its first occurrence. It does not by itself force a repeated divisor. For the minus sign, the bound is already met at all sufficiently large Wilson indices $n=q-2$. For every prime $q\ge 5$, Wilson gives $q\mid (q-2)!-1$, and $q$ is its least prime factor. Indeed, every prime divisor exceeds $q-2$, and $q-1$ is even; moreover $q/(q-2)\to 1$. The reflected index is then $1$, outside the denominators of $S$.

For example, modulo $11$ the only solution of $k!\equiv 1$ with $2\le k\le 9$ is $k=9$, although $11/9<(\sqrt{145}-1)/8$. A small ratio $q/n$ therefore supplies neither a second admissible index nor membership of a reflected index in the chosen block. The earlier reflection argument needs both block-membership hypotheses and distinct indices. Neither the required estimate at first occurrences nor a bound for the total shared part follows from the transferred least-prime-factor estimate alone.

For a selected prime $q\mid R_p$, the factors $1$ and $q$ are coprime, and the associated moduli $R_p$ and $R_p/q$ have lcm $R_p$. Thus this specialisation requires no second selected prime.

The elementary fact behind the projection argument is as follows. Let $Z,T,B$ be nonnegative integers, let $R>0$, and suppose $Z\equiv T(\mathrm{mod}R)$. Every positive divisor $Q$ of $R$ with $Z\le B<Q$ satisfies $TmodQ=Z$, since $Z$ is already the least nonnegative representative. Consequently, unequal residues modulo two divisors exceeding $B$ rule out such a $Z$. More generally, if $Tmod{Q}_1\ne Tmod{Q}_2$, then $min(Q_1,Q_2)\le T$: otherwise both residues would equal $T$.