Given $p(x)$ is a polynomial with integer coefficients and that $p(a)=1$ for some integer $a$ prove that $p(x)$ has no more than two integral roots.  I've attempted a proof by contradiction assuming $p(x)$ has three or more roots, but haven't gotten anywhere on this. Help would be appreciated!
8894
The question that I am trying to answer is :  Suppose is $p(x)$ is a polynomial with integer coefficients. Show that if $p(a) = 1$ for some integer a then $p(x)$ has at most two integer roots.  I have no idea how to get started. Any help would be awesome!
44175
Let $p(x)$ be a non-zero polynomial in $F[x]$, $F$ a field, of degree $d$. Then $p(x)$ has at most $d$ distinct roots in $F$.   Is it possible to prove this without using induction on degree? If so, how can I do this?
73433
Suppose $z_1, .....z_{m+1}$ are distinct elements of $F$ and $w_1,....,w_{m+1} \in F$. Prove that there exists a unique polynomial $p \in P_m(F)$ such that $p(z_j) = w_j$ for j=1,...m+1.  Any ideas on where to start/go with this one. I'm having quite some trouble with it.
70665
This is my hypothesis and I am not sure if this is a valid one. I need a proof for this. My point is that infinite number of minimas are needed if this is valid, but I am lack of knowledge to prove it.   Any one has a proof ?   Hypotehis:  I want to show that a non-constant polynomial $P(x)$ where $x \in \mathbb{N}$ can not have infinite number of the same constant values in its range.   i.e. for some infinite set of distinct $x_i$  values, where $i \in \mathbb{N}$ and $c$ is a constant, $P(x_i) \ne c $
73938
I've got following problem. Let $z$ be a complex number. Let $P(z)$ be a polynomial of degree $n$. Suppose that $|P(z)| \le M$ for $|z| \le 1$. Prove that for $|z| \ge 1$ we have. $|P(z)| \le M \cdot |z|^n$. Can somebody help?
105975
Let: $P(x) = a_nx^n + a_{n-1}x^{n-1} + ..... + a_1x + a_0$ where $a_0a_n &lt; 0$ I have to prove that the Polynomial $P(x)$ has at least one positive root how can I prove it? Any ideas?
55241
Doing a discrete math review and am trying to solve problem 1.6 in the text found here:  - I believe I've gotten parts (a) and (b) correctly, but (c) is a bit tricky for me. Would appreciate a review of a/b and a hint for c if possible. Restatement here:  For n = 40, the value of polynomial $p(n)::=n^2 + n + 41$ is not prime, as noted in Section 1.1. But we could have predicted based on general principles that no nonconstant polynomial can generate only prime numbers.  In particular, let $q(n)$ be a polynomial with integer coefﬁcients, and let $c::=q(0)$ be the constant term of q.  (a) Verify that $q(cm)$ is a multiple of c for all $m \in Z$  Proof: Since q is a polynomial, it will be of the form: $q(x) = a_nx^n+a_{n-1}x^{n-1} + ... + a_1x + c$  If $x = cm$, then notice that cm will be inside every term of the polynomial, and since the final term is c, c can be factored out of every term. Hence, q(cm) is a multiple of c for all $m \in Z$.  b) Show that if q is nonconstant and c > 1, then as n ranges over the nonnegative integers, $N$, there are infinitely many $q(n) \in Z$ that are not primes.  Proof: Continuing from the result found in (a), it is easy to see that for all $m \in Z$, there will be an infinite number of multiples of c that will be generated by q. Since c > 1, these multiples are guaranteed not to be prime.  c) Conclude that for every nonconstant polynomial, q, there must be an $n \in N$ such that $q(n)$ is not prime. Hint: Only one easy case remains.  c = 0 is trivially easy to prove using the (b) above.  I assume the "easy" case is referring to c = 1, but in this case I'm not sure how to continue, since the result of (a) and (b) don't apply: I can't use them since adding 1 to any even number may make it prime. If the result of the terms not including c is odd, then adding 1 to that result makes it even and it is not prime. However, if the terms add up to an even number, I don't see a way of using the knowledge I have so far to prove conclusively that that number + 1 will NOT in fact be prime.
6319
Can a non-zero polynomial in one variable have infinitely many roots ? Can a non-zero polynomial in one variable have uncountably many roots ?  Motivation: over $\mathbb Z/12\mathbb Z$, $X^2-4$ has 4 roots. When it comes to polynomials with coefficients over an integral domain, the answer is clearly negative (in that case, a polynomial can't have more roots than its degree). What happens with a ring that has zero divisors ?
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Does there exist a polynomial $f$ with integer coefficients such that for every natural number $n$, $f(0) , f(1) ... f(n) $ are all distinct powers of 2 ?   I have no clue about how should I start thinking about this problem but I felt that it is very interesting.
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Given a polynomial $f(x)$ with integral coefficients and $f(x) = 37$ has 5 distinct integral roots, find the number of integral roots of $f(x) = 41$?  My Approach: Say $f(x) = (x-r_1)(x-r_2)(x-r_3)(x-r_4)(x-r_5)g(x) + 37$, where $r_i$ are the distinct integers.   Now for $f(x) = 41$ we have $(x-r_1)(x-r_2)(x-r_3)(x-r_4)(x-r_5)g(x) = 4$, so the factors can be $\pm 1, \pm2$ or $\pm 4$. Given $r_i$ are distinct at most two of them will give $\pm1$, then there can be both of $\pm 2$ or one of $\pm 4$. This is where I get lost, since even if I use all of $\pm1, \pm2$, I will be still be left with one $x-r_i$ factor. What about that? Does it matter that 37 and 41 are primes, or is it just a coincidence?  Thanks in advance.
19912
This is a question from Putnam and Beyond, and I have a question about the proof. The question is:  Show $x^{p-1} + 2x^{p-2} + 3x^{p-3} + \dots + (p-1)x + p$ is irreducible over $\mathbb{Z}[X]$.  The proof first shows that the polynomial $P(x) = x^{p-1} + 2x^{p-2} + 3x^{p-3} + \dots + (p-1)x + p$ only has zeros of absolute value $&gt; 1$. Then, it assumes it can be factored into two polynomials, namely, $P(X) = Q(X)R(X)$, and since $p = P(0) = Q(0)R(0)$, one of $Q(0), R(0) = \pm 1$, so WLOG $Q(0)$ has a zero of absolute value $&gt; 1$ which contradicts the fact that $P(X)$ only has zeros of absolute value $&gt; 1$. I don't see why $Q(0) = \pm 1$ means that it has a zero with absolute value greater than one. (The following is the proof in question.) &quot;We will show that all the zeros of $P (x)$ have absolute value greater than $1$. Let $y$ be a complex zero of $P(x)$. Then $0 = (y − 1)P(y) = y^p + y^{p−1} + y^{p−2} +···+ y − p.$ Assuming $|y| ≤ 1$, we obtain $p = |y^p + y^{p−1}+ y^{p−2}+···+ y| ≤ \sum_{i=1}^p |y|^ i ≤  \sum_{i=1}^p 1 = p.$ This can happen only if the two inequalities are, in fact, equalities, in which case $y = 1$. But $P (1) &gt; 0$, a contradiction that proves our claim. Next, let us assume that $P (x) = Q(x)R(x)$ with $Q(x)$ and $R(x)$ polynomials with integer coefficients of degree at least $1$. Then $p = P (0) = Q(0)R(0)$. Since both $Q(0)$ and $R(0)$ are integers, either $Q(0) = ±1$ or $R(0) = ±1$. Without loss of generality, we may assume $Q(0) = ±1$. This, however, is impossible, since all zeros of $Q(x)$, which are also zeros of $P (x)$, have absolute value greater than $1$. We conclude that $P(x)$ is irreducible.&quot;
33610
Let $P(x)$ be a nonconstant polynomial whose coefficients are positive integers. If $P(n)$ divides $P(P(n) - 2015 )$ for every natural number n, then prove that $$P (-2015) = 0$$  I've observed that the proof depends on  $ P(n)$ divides $ P(P(n)-2015)-P(-2015)\,$ but I cannot see how to prove that. Could someone please elaborate on how to proceed.
64532
I have $P(x)$ a polynomial with degree $n$ ,$P(x) \ge 0$ for all $x \in$ real.  I have to prove that:     $f(x)=P(x)+P'(x)+P"(x)+......+P^{n}(x) \ge 0$ for all $x$.   I tried different methods to solve it but I got stuck.Any suggestion or advice is welcomed.
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Consider the root of the polynomial $p(x) = x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots+a_1x -1$.    Suppose that $p(x)$ has no roots in the open unit disc in a complex plane and $p(-1)=0$. Show that $p(1)= 0$.
105120
$\displaystyle ax^3+bx^2+cx+d$ is a polynomial with integer coefficients. It is given that $ad,\,bc$ are odd and even respectively. Then prove that not all roots of the polynomial are rational.  It is easy to see that none of the roots are integer. But how to tackle the rational case? Any help is appreciated.
76413
A polynomial with integer coefficients is an expression of the form: $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$  where $a_n$, $a_{n-1}, \ldots, a_1, a_0$ are integers and $a_n$ is not equal to $0$.  a zero of the polynomial is a $c \in \mathbb{R}$ such that $f(c)=0$  A real number is said to be algebraic if it is a zero polynomial with integer coefficients  1) Show that every rational number is algebraic  2) Show that if $a$, $b$ and $k$ are positive integers, then the $k$-th root of $a/b$ is algebraic  I don't even know where to start on this.  What is a zero of a polynomial with integer coefficients?
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Let $A$ be a field and let $p(X)$ be a polynomial over $A$. Let $a\in A$.   Want to show:     $p(X)$ is irreducible if and only if $p(X+a)$ is irreducible.    I suspect that I should use the substitution principle somehow, but that's as far as I've come. Completely stumped.