Factorial Linear Forms and Denominators: Detailed Proofs and Rationality Criteria

A primitive solution on an arithmetic progression

We now separate two tasks that the basis classification alone does not combine: prescribe where support begins and make $M$ absorb every fixed denominator as the support grows. Equal spacing turns each cancellation equation into a polynomial root condition. Scaling then makes the coefficients integral and the last coefficient equal to one. Let $D,r\ge 2$ and let $\ell$ be a positive multiple of $2,\dots ,D$. The roots will be $(d!{)}^{-\ell /d}$. Put

At index $i_j$ put the coefficient

and put every other coefficient equal to zero. We verify that these coefficients are integers and that $V_2=\cdots =V_D=0$. For $2\le d\le D$, divisibility of $\ell$ by $d$ gives $\lfloor i_j/d\rfloor =\lfloor r/d\rfloor +j\ell /d$. Hence

because ${\alpha }_d^{-1}$ is a root of the displayed polynomial. Evaluating that polynomial at $1$ gives $M$

For integrality, each term of $h_j/A$ is, up to sign, the reciprocal of a product of $D-1-j$ distinct factors ${\alpha }_d$. In that denominator, each selected $d$ contributes $\ell /d$ blocks of size $d$. The total size is $(D-1-j)\ell =N-i_j$, so the multinomial coefficient shows that this denominator divides $N!/i_j!$. Thus every $c_{i_j}$ is an integer. Since $h_{D-1}=A$, the coefficient at $N$ is $1$, and the coefficients have gcd one. This does not establish minimal support. The smallest permitted step is $\ell =\mathrm{lcm}(2,\dots ,D)$; taking $D=r=\ell =2$ recovers $-6e_2+e_4$.

The stronger divisibility

follows because

is an integer: it counts partitions into one distinguished block of size $r$ and, for each $d$, an unordered collection of $\ell /d$ blocks of size $d$. Their sizes sum to $N$. Using $M=(N!/A)\prod _{d=2}^D({\alpha }_d-1)$ now gives the asserted divisibility. Put $v=max(r,\ell /2)$, an integer since $2\mid \ell$. The displayed factor divides $M$, so $v!\mid M$. Since $D\le \ell \le 2v$ and $r\le v$,

Consequently $(\lfloor \sqrt{N}/2\rfloor +1)!\mid M$. Every positive integer $q\le \lfloor \sqrt{N}/2\rfloor +1$ therefore divides $M$, as does every divisor of that factorial. In particular, $N\ge 4(q-1{)}^2$ suffices for $q\mid M$. This includes denominators growing with $N$.

The same vector also supplies the size estimate needed for a tail test. Every factor $1-1/{\alpha }_d$ lies strictly between $0$ and $1$, so $0<M<N!$ and

where $(N+1)!-1>NN!$ for $N\ge 2$. Thus the family supplies both eventual divisibility by each fixed $q$ and the size condition $M<(N+1)!-1$, without an extra hypothesis. It does not yet compare this tail with the finite part’s gap to the next integer. Both may shrink together. Before studying that arithmetic question, we show why the cofactor formula gives the same primitive vectors.