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Source. Aron Bhalla, A conditional note on an Erdős problem on large prime factors of polynomial products, Lemma 2.1, physical pp. 2--3, in the five-page PDF held by its library source card, Bhalla (2026).

Standing. This is an author-recorded reconstruction of the displayed conditional route. It is not an independent review and does not change Problem 976's status or assign a verification tier. The standard Gauss lemma cited by the source is used as an external algebraic input for the final $\mathbb Z[x]$ irreducibility transfer; its cited book was not reread here.

Statement

Let $f\in\mathbb Z[x]$ be irreducible of degree $d\ge2$. Replace $f$ by $-f$ if necessary so that its leading coefficient is positive. Define

$$D=\gcd\{f(m):m\in\mathbb Z\}.$$

Then there are integers $a,M$ and a polynomial $h\in\mathbb Z[x]$ such that

$$M\ge1,\qquad 0\le a<M,\qquad h(x)=\frac{f(a+Mx)}{D},$$

and $h$ is irreducible in $\mathbb Z[x]$, has degree $d$, has positive leading coefficient, and has no prime divisor common to all of its values.

Reconstruction

Since $f$ is irreducible and has degree at least two, it has no integer root. Thus every $f(m)$ is nonzero and $D$ is a positive integer. If $D=1$, take $a=0$, $M=1$, and $h=f$; all assertions are immediate. Assume from now on that $D>1$ and write

$$D=\prod_{p\mid D}p^{e_p}.$$

For each $p\mid D$, the integer $e_p=v_p(D)$ is the minimum of the nonnegative integers $v_p(f(m))$. Choose an integer $b_p$ with

$$v_p(f(b_p))=e_p.$$

The moduli $p^{e_p+1}$ are pairwise coprime. The Chinese remainder theorem therefore gives an integer $a_0$ satisfying

$$a_0\equiv b_p\pmod {p^{e_p+1}}\qquad(p\mid D).$$

Set

$$M=\prod_{p\mid D}p^{e_p+1}$$

and replace $a_0$ by its representative $a$ with $0\le a<M$. Then $D\mid M$ and $D\mid f(a)$, the latter because $D$ divides every value of $f$.

Define the polynomial identity

$$h(x)=\frac{f(a+Mx)}{D}.$$

It remains to establish that this quotient has the claimed integral and irreducible structure.

Integrality, degree, and sign

For an integer polynomial $f$, every nonconstant coefficient of $f(a+Mx)-f(a)$ is divisible by $M$. Hence

$$f(a+Mx)-f(a)\in M\mathbb Z[x].$$

Because $D\mid M$ and $D\mid f(a)$, every coefficient of $f(a+Mx)$ is divisible by $D$. Thus $h\in\mathbb Z[x]$. If $L_f>0$ is the leading coefficient of $f$, then $h$ has degree $d$ and leading coefficient

$$L_h=\frac{L_fM^d}{D}>0.$$

No fixed prime divisor

First let $p\mid D$. Polynomial evaluation preserves congruences modulo $p^{e_p+1}$, so

$$f(a)\equiv f(b_p)\pmod {p^{e_p+1}}.$$

The right side has $p$-adic valuation exactly $e_p$. The congruence therefore gives $v_p(f(a))=e_p$, and hence

$$v_p(h(0))=v_p\left(\frac{f(a)}D\right)=0.$$

So no prime dividing $D$ divides every value of $h$.

Now let $q$ be a prime with $q\nmid D$. If $q$ divided $h(t)$ for every $t\in\mathbb Z$, then the identity $f(a+Mt)=Dh(t)$ would imply

$$q\mid f(a+Mt)\qquad(t\in\mathbb Z).$$

Every prime divisor of $M$ divides $D$, so $q\nmid M$. The progression $a+Mt$ therefore runs through every residue class modulo $q$. It follows that $q\mid f(u)$ for every integer $u$, which would imply $q\mid D$, a contradiction. Thus no prime divides all values of $h$.

Irreducibility

The substitution $x\mapsto a+Mx$ is an automorphism of $\mathbb Q[x]$, with inverse $x\mapsto(x-a)/M$. It preserves irreducibility, so $f(a+Mx)$ is irreducible over $\mathbb Q$. Dividing by the nonzero constant $D$ does not change irreducibility over $\mathbb Q$, and therefore $h$ is irreducible over $\mathbb Q$.

If a prime divided every coefficient of $h$, it would divide every value $h(t)$, contrary to the preceding paragraph. Thus $h$ is primitive. Gauss's lemma, the standard result cited in the source, transfers its $\mathbb Q[x]$ irreducibility to irreducibility in $\mathbb Z[x]$.

This proves all six parts of Lemma 2.1. The construction depends only on the fixed polynomial $f$ and is independent of the later endpoint $n$.

Boundary. No prime-value assertion is used in this lemma. The retained source's citation to S. Lang, Algebra, revised third edition, Chapter IV, is the external source boundary for Gauss's lemma. The next page supplies the separate prime-values hypothesis and its conditional application.