# Omni-MATH / 

task_id: 91ecd8d8-b10a-5539-927d-c38b30a16b93
task_key: test--91ecd8d8-b10a-5539-927d-c38b30a16b93
task_revision_id: 4

{"problem":"[color=blue][b]Generalization.[/b] 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$.\nFind all $ a \\in \\mathbb{Z}$ such that the polynomial $ f(x) \\equal{} x^n \\plus{} ax^{n \\minus{} 1} \\plus{} pq$ can be factored into 2 integral polynomials of degree at least 1.[/color]\r\n\r\n[i]Solution.[/i] I hope the following solution is correct. It is more or less a straightforward generalization of [url=http://www.kalva.demon.co.uk/imo/isoln/isoln931.html]IMO 1993 problem 1[/url].\r\n\r\nThe idea behind is an extension of Eisenstein's criterion for irreducible polynomials:\r\n\r\n[color=blue][b]Lemma 1.[/b] Let p be a prime number. If a polynomial $ A\\left(x\\right) \\equal{} a_nx^n \\plus{} a_{n \\minus{} 1}x^{n \\minus{} 1} \\plus{} ... \\plus{} a_1x \\plus{} a_0$ with integer coefficients $ a_n$, $ a_{n \\minus{} 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 \\minus{} 2}$, but does not divide $ a_n$, and $ p^2$ does not divide $ a_0$, then p does not divide $ a_{n \\minus{} 1}$, and the polynomial A(x) must have a rational root.[/color]\r\n\r\n[i]Proof of Lemma 1.[/i] 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) \\equal{} b_ux^u \\plus{} ... \\plus{} b_1x \\plus{} b_0$ and $ C\\left(x\\right) \\equal{} c_vx^v \\plus{} ... \\plus{} c_1x \\plus{} 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 \\equal{} \\sum_{k \\equal{} 0}^i b_kc_{i \\minus{} k}$ (this follows from multiplying out the equation A(x) = B(x) C(x)). Particularly, $ a_0 \\equal{} 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.\r\n\r\nNot all coefficients $ b_u$, ..., $ b_1$, $ b_0$ of the polynomial B(x) can be divisible by p (else, $ a_n \\equal{} \\sum_{k \\equal{} 0}^n b_kc_{n \\minus{} k}$ would also be divisible by p, what is excluded). Let $ \\lambda$ be the least nonnegative integer such that the coefficient $ b_{\\lambda}$ is [i]not[/i] divisible by p. Then, all the integers $ b_k$ with $ k < \\lambda$ are divisible by p. Hence, in the sum $ a_{\\lambda} \\equal{} \\sum_{k \\equal{} 0}^{\\lambda} b_kc_{\\lambda \\minus{} k}$, all the summands $ b_kc_{\\lambda \\minus{} k}$ with $ k < \\lambda$ are divisible by p, but the summand $ b_{\\lambda}c_0$ (this is the summand for $ k \\equal{} \\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 \\minus{} 2}$ are all divisible by p; hence, $ a_{\\lambda}$ must be one of the coefficients $ a_{n \\minus{} 1}$ and $ a_n$. Thus, either $ \\lambda \\equal{} n \\minus{} 1$ or $ \\lambda \\equal{} n$.\r\n\r\nIf $ \\lambda \\equal{} 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.\r\n\r\nThus, we must have $ \\lambda \\equal{} n \\minus{} 1$. Hence, the integer $ a_{n \\minus{} 1} \\equal{} 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 \\minus{} 1}$. In other words, the polynomial B(x) has degree $ \\geq n \\minus{} 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.\r\n\r\nNow let us solve the problem: The number $ pq$ is squarefree (since $ p$ is prime and $ q$ is squarefree, and since $ p\\nmid q$).\r\n\r\nApplying Lemma 1 to the polynomial $ x^n \\plus{} ax^{n \\minus{} 1} \\plus{} 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 \\plus{} ar^{n \\minus{} 1} \\plus{} pq \\equal{} 0$, so that $ pq \\equal{} \\minus{} r^n \\minus{} ar^{n \\minus{} 1} \\equal{} \\minus{} \\left(r \\plus{} a\\right) r^{n \\minus{} 1}$ is divisible by $ r^2$ (since $ n\\geq 3$ yields $ n \\minus{} 1\\geq 2$, so that $ r^{n \\minus{} 1}$ is divisible by $ r^2$), so that $ r \\equal{} 1$ or $ r \\equal{} \\minus{} 1$ (since $ pq$ is squarefree), so that one of the numbers $ 1$ and $ \\minus{} 1$ must be a root of the polynomial $ x^n \\plus{} ax^{n \\minus{} 1} \\plus{} pq$.\r\n\r\nHence, we see that, if the polynomial $ x^n \\plus{} ax^{n \\minus{} 1} \\plus{} pq$ can be factored into two non-constant integral polynomials, then one of the numbers $ 1$ and $ \\minus{} 1$ must be a root of this polynomial. Conversely, if one of the numbers $ 1$ and $ \\minus{} 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 $ \\minus{} 1$ is a root of the polynomial $ x^n \\plus{} ax^{n \\minus{} 1} \\plus{} pq$. But in fact, $ 1$ is a root of the polynomial $ x^n \\plus{} ax^{n \\minus{} 1} \\plus{} pq$ if and only if $ 1^n \\plus{} a\\cdot 1^{n \\minus{} 1} \\plus{} pq \\equal{} 0$, what is equivalent to $ 1 \\plus{} a \\plus{} pq \\equal{} 0$, i. e. to $ a \\equal{} \\minus{} 1 \\minus{} pq$, and $ \\minus{} 1$ is a root of the polynomial $ x^n \\plus{} ax^{n \\minus{} 1} \\plus{} pq$ if and only if $ \\left( \\minus{} 1\\right)^n \\plus{} a\\cdot\\left( \\minus{} 1\\right)^{n \\minus{} 1} \\plus{} pq \\equal{} 0$, what is equivalent to $ a \\equal{} 1 \\plus{} \\left( \\minus{} 1\\right)^n pq$. So, the two required values of $ a$ are $ a \\equal{} \\minus{} 1 \\minus{} pq$ and $ a \\equal{} 1 \\plus{} \\left( \\minus{} 1\\right)^n pq$.\r\n\r\nThe problem is thus solved.\r\n\r\n[hide=\"Old version of the solution, not generalizing the problem\"]\n\n[i]Old version of the solution (of the original problem, not of the generalization).[/i]\n\nApplying Lemma 1 to the polynomial $ x^n \\plus{} ax^{n \\minus{} 1} \\plus{} 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, and by a well-known theorem, this integer root then must be a divisor of pq. This means that the root is one of the numbers pq, p, q, 1, -pq, -p, -q, -1. Actually, none of the numbers pq, p, q, -pq, -p, -q can be a root of the polynomial $ x^n \\plus{} ax^{n \\minus{} 1} \\plus{} pq$ (in fact, every of these numbers is divisible by p or by q, and if an integer root r of the polynomial $ x^n \\plus{} ax^{n \\minus{} 1} \\plus{} pq$ would be divisible by p, then $ r^n \\plus{} ar^{n \\minus{} 1}$ would be divisible by $ p^{n \\minus{} 1}$, while $ pq$ wouldn't be because of $ n\\geq 3$, so $ r^n \\plus{} ar^{n \\minus{} 1} \\plus{} pq$ couldn't be 0, what yields a contradiction, and similarly we obtain a contradiction if an integer root would be divisible by q). Hence, only 1 and -1 remain as possible candidates for integer roots of the polynomial $ x^n \\plus{} ax^{n \\minus{} 1} \\plus{} pq$.\n\nHence, we see that, if the polynomial $ x^n \\plus{} ax^{n \\minus{} 1} \\plus{} 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 \\plus{} ax^{n \\minus{} 1} \\plus{} pq$. But in fact, 1 is a root of the polynomial $ x^n \\plus{} ax^{n \\minus{} 1} \\plus{} pq$ if and only if $ 1^n \\plus{} a\\cdot 1^{n \\minus{} 1} \\plus{} pq \\equal{} 0$, what is equivalent to $ 1 \\plus{} a \\plus{} pq \\equal{} 0$, i. e. to a = - 1 - pq, and -1 is a root of the polynomial $ x^n \\plus{} ax^{n \\minus{} 1} \\plus{} pq$ if and only if $ \\left( \\minus{} 1\\right)^n \\plus{} a\\cdot\\left( \\minus{} 1\\right)^{n \\minus{} 1} \\plus{} pq \\equal{} 0$, what is equivalent to $ a \\equal{} 1 \\plus{} \\left( \\minus{} 1\\right)^n pq$. So, the two required values of a are a = - 1 - pq and $ a \\equal{} 1 \\plus{} \\left( \\minus{} 1\\right)^n pq$.\n[/hide]\r\n\r\n  Darij"}

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