# Tag Info

### Interesting third degree polynomial

The problem is the same as showing that there exist integers $x$ such that $$P(x)^2 \not \equiv P(x) \pmod{10}.$$ For the time being, I propose a bad solution. I believe that there exists a much ...
• 2,030
1 vote

• 15.1k

### $f$ is irreducible if the polynomial reduced $p$ is irreducible and the degrees are the same

This is a standard argument. I assume you mean $f\in\Bbb{Z}[x]$, otherwise it makes no sense to talk about reducing modulo $p$. Let us proceed by contraposition: If $f$ is reducible in $\Bbb{Q}[x]$, ...
• 2,955
Accepted

### Probability of choosing a constant polynomial

Let us use only whole numbers between $0$ & $(m-1)$ : Each Co-efficient has $m$ choices. Total number of Polynomials : $m^n$ Total number of Constant Polynomials : $m$ Probability : $P=1/m^{n-1}$ ...
• 9,960
1 vote

### How to factor $(x-y)^5 + (y-z)^5 + (z-x)^5$

Alternate solution/idea that should be helpful in the future. Let $a,b,c$ such that $a+b+c = 0.$ Then, they are the roots of the cubic: $$P(x) = (x-a)(x-b)(x-c) = x^3+x(ab+bc+ca)-abc = x^3+qx-r.$$ ...
• 13.9k
Accepted

### How to factor $(x-y)^5 + (y-z)^5 + (z-x)^5$

OP arguments (from the question and the comments) are valid and lead to a quick result. First of all, considering the given polynomial $$P(x,y,z)=(x-y)^5 + (y-z)^5 + (z-x)^5$$ as a polynomial in $x$...
• 33.5k

• 175
1 vote

### Upper bound of depressed quartic

The following result has not been officially recognized by the mathematical community, and all I can do is guarantee that it is absolutely correct. If the depressed quartic $x^4+ax^2+bx+c=0$ has four ...
1 vote

Use SymPy: ...
• 104k

### Find all integers such that $2n^2+1$ divides $n^3+9n-17$

If $2n^2 +1$ divides $n^3 + 9n -17$, then it also divides $2\cdot \left(n^3 + 9n -17\right)$, and thus, it also divides $2\cdot \left(n^3 + 9n -17\right) - n \cdot \left(2n^2 + 1\right)$ which is ...
• 324

• 1,282
1 vote
Accepted

• 27.6k
1 vote
Accepted

### If $f\in \mathbb{Q}[x,y]$ factors in some field extension of $\mathbb{Q}$, then it also factors in some finite extension of $\mathbb{Q}$?

I think it is true. For every field extension $E$ of $\mathbb{Q}$, we still have that $E[x,y]$ is an UFD. Thus there are only finitely many monic factors of $f\in\mathbb{Q}[x,y]$ in $E[x,y]$. Thus by ...
1 vote
Accepted

• 529

### Galois Group of $\sqrt{2+\sqrt{2}}$ over $\mathbb{Q}$

This is not a new answer, but an addition to Tim Ratigan's answer. If you solve the system of equations  $ax_{1}^3$  + $bx_{1}^2$ + $cx_{1}$  +  $d$ = $x_{2}$  $ax_{2}^3$  + $bx_{2}^2$ + $cx_{2}$  +  ...
• 313
1 vote
Accepted

### The projection of the polynomial onto U parallel to V

Since $V$ is one dimensional, you can write $f(x)=g(x)+a(x^3+x^2+x+1)$ where $g(x)\in U$. Evaluating at $1$ and using the fact that $g(1)=0$ by definition, we have $f(1)=4a$, so $a=f(1)/4$. So to ...
• 24.2k
1 vote
Accepted

### Definitions of polynomials $F(x):=f(x+c)$ and $G(x):=f(cx)$ if $c\in Z(R)$

Essentially, to have a nice polynomial ring, you want your indeterminates to commute with the coefficients. If this is not true, you get a dichotomy between polynomials as elements of a ring and their ...
• 2,955

### Why is the Discriminant always an integer?

As the earlier answers and the counter example by @jyrki-lahtonen tells that discriminant according to your definition need not be an integer always, if your polynomial is not monic. But one can ...
• 443

### When are two polynomials with similar roots identical?

The statement is false, unless it takes into account the multiplicities of the roots as well. A simple counterexample is given by @ThomasAndrews in the comments (consider $x^2(x-1)$ and $x(x-1)^2$). ...
• 4,695
Whenever I see a progression where the change itself changes (think two degrees of change), I think quadratic. Quadratics only need 3 points to be unique. Let $f(n):=a_n$ for notation purposes. We ...