Generalization. Given two integers $ p$ and $ q$ and a natural number $ n \geq 3$ such that $ p$ is prime and $ q$ is squarefree, and such that $ p\nmid q$. Find all $ a \in \mathbb{Z}$ such that the polynomial $ f(x) = x^n + ax^{n - 1} + pq$ can be factored into 2 integral polynomials of degree at least 1. Solution. I hope the following solution is correct. It is more or less a straightforward generalization of IMO 1993 problem 1. The idea behind is an extension of Eisenstein's criterion for irreducible polynomials: Lemma 1. Let p be a prime number. If a polynomial $ A\left(x\right) = a_nx^n + a_{n - 1}x^{n - 1} + ... + a_1x + a_0$ with integer coefficients $ a_n$, $ a_{n - 1}$, ..., $ a_1$, $ a_0$ is reducible in $ \mathbb{Z}\left[x\right]$, and the prime p divides the coefficients $ a_0$, $ a_1$, ..., $ a_{n - 2}$, but does not divide $ a_n$, and $ p^2$ does not divide $ a_0$, then p does not divide $ a_{n - 1}$, and the polynomial A(x) must have a rational root. Proof of Lemma 1. Since the polynomial A(x) is reducible in $ \mathbb{Z}\left[x\right]$, we can write it in the form A(x) = B(x) C(x), where $ B\left(x\right) = b_ux^u + ... + b_1x + b_0$ and $ C\left(x\right) = c_vx^v + ... + c_1x + c_0$ are non-constant polynomials with integer coefficients $ b_u$, ..., $ b_1$, $ b_0$, $ c_v$, ..., $ c_1$, $ c_0$. Then, for any i, we have $ a_i = \sum_{k = 0}^i b_kc_{i - k}$ (this follows from multiplying out the equation A(x) = B(x) C(x)). Particularly, $ a_0 = b_0c_0$. But since the integer $ a_0$ is divisible by the prime p, but not by $ p^2$, this yields that one of the integers $ b_0$ and $ c_0$ is divisible by p, and the other one is not. WLOG assume that $ b_0$ is divisible by p, and $ c_0$ is not. Not all coefficients $ b_u$, ..., $ b_1$, $ b_0$ of the polynomial B(x) can be divisible by p (else, $ a_n = \sum_{k = 0}^n b_kc_{n - k}$ would also be divisible by p, what is excluded). Let $ \lambda$ be the least nonnegative integer such that the coefficient $ b_{\lambda}$ is not divisible by p. Then, all the integers $ b_k$ with $ k < \lambda$ are divisible by p. Hence, in the sum $ a_{\lambda} = \sum_{k = 0}^{\lambda} b_kc_{\lambda - k}$, all the summands $ b_kc_{\lambda - k}$ with $ k < \lambda$ are divisible by p, but the summand $ b_{\lambda}c_0$ (this is the summand for $ k = \lambda$) is not (since $ b_{\lambda}$ is not divisible by p, and neither is $ c_0$). Hence, the whole sum $ a_{\lambda}$ is not divisible by p. But we know that the coefficients $ a_0$, $ a_1$, ..., $ a_{n - 2}$ are all divisible by p; hence, $ a_{\lambda}$ must be one of the coefficients $ a_{n - 1}$ and $ a_n$. Thus, either $ \lambda = n - 1$ or $ \lambda = n$. If $ \lambda = n$, then it follows, since the integer $ b_{\lambda}$ is defined, that the polynomial B(x) has a coefficient $ b_n$. In other words, the polynomial B(x) has degree n. Since the polynomial A(x) has degree n, too, it follows from A(x) = B(x) C(x) that the polynomial C(x) is a constant. This is a contradiction. Thus, we must have $ \lambda = n - 1$. Hence, the integer $ a_{n - 1} = a_{\lambda}$ is not divisible by p. Also, since the integer $ b_{\lambda}$ is defined, it follows that the polynomial B(x) has a coefficient $ b_{n - 1}$. In other words, the polynomial B(x) has degree $ \geq n - 1$. Since the polynomial A(x) has degree n and A(x) = B(x) C(x), this yields that the polynomial C(x) has degree $ \leq 1$. The degree cannot be 0, since the polynomial C(x) is not constant; thus, the degree is 1. Hence, the polynomial A(x) has a linear factor, i. e. it has a rational root. Lemma 1 is proven. Now let us solve the problem: The number $ pq$ is squarefree (since $ p$ is prime and $ q$ is squarefree, and since $ p\nmid q$). Applying Lemma 1 to the polynomial $ x^n + ax^{n - 1} + pq$, using the prime p, we see that, if this polynomial can be factored into two non-constant integral polynomials, then it must have a rational root. Since it is a monic polynomial with integer coefficients, it thus must have an integer root. If we denote this root by $ r$, then $ r^n + ar^{n - 1} + pq = 0$, so that $ pq = - r^n - ar^{n - 1} = - \left(r + a\right) r^{n - 1}$ is divisible by $ r^2$ (since $ n\geq 3$ yields $ n - 1\geq 2$, so that $ r^{n - 1}$ is divisible by $ r^2$), so that $ r = 1$ or $ r = - 1$ (since $ pq$ is squarefree), so that one of the numbers $ 1$ and $ - 1$ must be a root of the polynomial $ x^n + ax^{n - 1} + pq$. Hence, we see that, if the polynomial $ x^n + ax^{n - 1} + pq$ can be factored into two non-constant integral polynomials, then one of the numbers $ 1$ and $ - 1$ must be a root of this polynomial. Conversely, if one of the numbers $ 1$ and $ - 1$ is a root of this polynomial, then it has an integer root and thus can be factored into two non-constant integral polynomials. Hence, in order to solve the problem, it remains to find all integers a such that one of the numbers $ 1$ and $ - 1$ is a root of the polynomial $ x^n + ax^{n - 1} + pq$. But in fact, $ 1$ is a root of the polynomial $ x^n + ax^{n - 1} + pq$ if and only if $ 1^n + a\cdot 1^{n - 1} + pq = 0$, what is equivalent to $ 1 + a + pq = 0$, i. e. to $ a = - 1 - pq$, and $ - 1$ is a root of the polynomial $ x^n + ax^{n - 1} + pq$ if and only if $ \left( - 1\right)^n + a\cdot\left( - 1\right)^{n - 1} + pq = 0$, what is equivalent to $ a = 1 + \left( - 1\right)^n pq$. So, the two required values of $ a$ are $ a = - 1 - pq$ and $ a = 1 + \left( - 1\right)^n pq$. The problem is thus solved.

Darij

Different way: if we reduce the polynomial in $\mathbb{Z}/p\mathbb{Z}$, it becomes $x^n+ax^{n-1}=x^{n-1}(x+a)$. Since both x and x+a are irreducible in $\mathbb{F}_p[x]$, suppose f(x)=g(x)h(x). WLOG, $g(x)=x^k+pg_1(X)$ and $h(x)=x^{n-1-k}(x+a)+ph_1(x)$, where $g_1, h_1 \in \mathbb{Z}[x]$. Multiplying these two gives $f(x)=x^{n-1}(x+a)+pg_1(x)x^{n-1-k}(x+a)+ph_1(x)x^k+p^2g_1(x)h_1(x)$. If 0<k<n-1, we get a contradiction when x=0. We know that k cannot equal zero. Thus, the only possible cases are when k is n-1. In this case, f must have an integer root. Thus, our solution is all a such that f has an integer root. EDIT: This solution is actually identical to my solution of IMO 1993 #1. In fact, I think it can be generalized to any polynomial of the form $x^n+ax^{n-1}+k$, where there exists a prime p such that $v_p(k)=1$. edit of the edit: I actually think the original problem says irreducible in Q[x]? But then it's just one extra step using gauss' lemma and the fact that it's primitive.