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  "body": "\n## Proof\n\n1. Let $f = a_n x^n + \\dots + a_0 \\in \\Z[x]$ and the [[prime|0SN0MAEKYX]] $p$ satisfy the three hypotheses. Since $p \\nmid a_n$, we have $a_n \\ne 0$, so $\\deg f = n \\ge 1$.\n2. The ring $\\Z$ is a [[Euclidean domain|000JFQGDXG]], hence a [[principal ideal domain|0QKVE1GD9A]], hence a [[unique factorisation domain|055HT85T3S]]. Its [[field of fractions|05K3J8J8T2]] is $\\Q$.\n3. Let $c \u003e 0$ be the greatest common divisor of $a_0, \\dots, a_n$, and let $f_0 = f / c \\in \\Z[x]$. Write $f_0 = \\sum_i a'_i x^i$ with $a'_i = a_i / c$.\n4. The polynomial $f_0$ is [[primitive|0SP14DZN0G]]. Indeed, if an integer $r$ with $\\abs{r} \\ge 2$ divides every $a'_i$, then $c \\abs{r}$ divides every $a_i$, which contradicts the choice of $c$.\n5. Since $c \\mid a_n$ and $p \\nmid a_n$, we have $p \\nmid c$. For $i \u003c n$, $p \\mid a_i = c \\, a'_i$, so $p \\mid a'_i$ by [[Euclid's lemma|051VYKA1EX]].\n6. Also $p \\nmid a'_n$, since $a'_n$ divides $a_n$. And $p^2 \\nmid a'_0$, since otherwise $p^2$ would divide $c \\, a'_0 = a_0$. So $f_0$ satisfies the three hypotheses.\n7. In $\\Q[x]$, $f = c f_0$ and $c$ is a [[unit|019NFGFPR5]]. So $f$ is [[irreducible|00X2A9G9W3]] in $\\Q[x]$ if and only if $f_0$ is.\n8. By [[Gauss's lemma|0QFV5Y1Z5X]] and Step 4, it is enough to show that $f_0$ is irreducible in $\\Z[x]$. We argue by contradiction.\n9. The units of $\\Z[x]$ are $\\pm 1$. Indeed, if $u v = 1$, then $\\deg u = \\deg v = 0$ by the [[degree of a product|0T233AZ1GE]], and $u, v \\in \\Z$ with $u v = 1$.\n10. Since $\\deg f_0 = n \\ge 1$, the polynomial $f_0$ is nonzero and not a unit. Suppose that $f_0 = g h$ with $g, h \\in \\Z[x]$ not units.\n11. If $g$ were a constant $r$, then $r$ would divide every coefficient of $f_0$. By Step 4, $r = \\pm 1$, a unit. So $\\deg g \\ge 1$, and in the same way $\\deg h \\ge 1$.\n12. Write $g = b_s x^s + \\dots + b_0$ and $h = c_t x^t + \\dots + c_0$ with $b_s c_t \\ne 0$. Then $s, t \\ge 1$ and $s + t = n$, so $s \u003c n$. Set $b_j = 0$ for $j \u003e s$ and $c_j = 0$ for $j \u003e t$.\n13. We have $a'_0 = b_0 c_0$. Since $p \\mid a'_0$, Euclid's lemma gives $p \\mid b_0$ or $p \\mid c_0$. Both cannot hold, because $p^2 \\nmid a'_0$. Say $p \\mid b_0$ and $p \\nmid c_0$.\n14. We show by induction on $i$ that $p \\mid b_i$ for $0 \\le i \\le s$. The case $i = 0$ is Step 13.\n15. Let $1 \\le i \\le s$ and assume $p \\mid b_j$ for all $j \u003c i$. Comparing coefficients of $x^i$ gives $a'_i = b_i c_0 + \\sum_{j=0}^{i-1} b_j c_{i-j}$.\n16. Since $i \\le s \u003c n$, we have $p \\mid a'_i$ by Step 5. Each term $b_j c_{i-j}$ with $j \u003c i$ is divisible by $p$. So $p \\mid b_i c_0$.\n17. Since $p \\nmid c_0$, Euclid's lemma gives $p \\mid b_i$. This completes the induction.\n18. In particular $p \\mid b_s$. The leading coefficient of $f_0$ is $a'_n = b_s c_t$, so $p \\mid a'_n$. This contradicts Step 6.\n19. So $f_0$ is irreducible in $\\Z[x]$. By Steps 7 and 8, $f$ is irreducible in $\\Q[x]$.\n\n## Idea\n\nAfter Gauss's lemma, a factorisation over $\\Q$ becomes one over $\\Z$. The prime $p$ divides exactly one of the two constant terms. Comparing coefficients one degree at a time pushes $p$ through all coefficients of that factor, up to its leading coefficient. Then $p$ would divide the leading coefficient of $f$, which is excluded.\n",
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