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Finding an expression for the sum of n tems of the series $1^2+2^2+3^2+...+n^2$
I know that if you have a non-arithmetic or geometric progression, you can find a sum S of a series with the formula $S=f(n+1)-<!--\; -\; -->f(1)$ where the term $u_n$ is $u_n=f(n+1)-<!--\; -\; -->f(n)$. Then you can prove that with induction.
What I don't understand is how I should go about finding the function f(n). For example if I want to calculate the sum to n terms of the series $1^2+2^2+3^2+...+n^2$ then, according to my textbook, my f(n) function should be a polynomial with degree one more than the degree of a term in my sequence - so because the nth term in the sequence is $P(n)=n^2$ then the function f(n) should be $f(n)=a{n}^3+b{n}^2+cn+d$. But how did they know that it should look like that and how do I gain some intuition into finding that function to help me solve similar problems in the future?

Galois theory in a different setting
Suppose that instead of wanting to express the roots of a polynomial equation with arithmetic operations and radicals we instead wanted expressed it with arithmetic operations and sin(x) ?
What (if anything) does Galois theory say about this situation?

Do all monic polynomials factor into monic irreducibles?
Do all monic polynomials in $\mathbb{Z}/n\mathbb{Z}[x]$ where n is prime factor into monic irreducibles?
Just by thinking about these monic polynomials and how they only act on a finite number of elements makes me think that the assertion is true, because if d is a root of a monic polynomial f(x) then $f(x)=(x-<!--\; -\; -->d)g(x)$ where g(x) is monic, and you could apply the same rule iteratively until you exhaust all roots. But I am not sure how to really prove this, it sounds really similar to one part of the fundamental theorem of arithmetic, but I can't find how to quantify a polynomial the same way you would an integer for an inductive proof. How would I go about starting a proof of this?

Why is the triangle inequality equivalent to $a^4+b^4+c^4\le <!--\; \le \; -->2(a^2{b}^2+b^2{c}^2+c^2{a}^2)$
Consider the existential problem of a triangle with side lengths $a,b,c\ge <!--\; \ge \; -->0$. Such a triangle exists if and only if the three triangle inequalities
$\begin{array}{}\text{(0)}& a+b\ge <!--\; \ge \; -->c,b+c\ge <!--\; \ge \; -->a\text{and}c+a\ge <!--\; \ge \; -->b\end{array}$
are all satisfied.
Alternatively, if ${\mathrm{\ell <!--\; \ell \; -->}}_1\le <!--\; \le \; -->{\mathrm{\ell <!--\; \ell \; -->}}_2\le <!--\; \le \; -->{\mathrm{\ell <!--\; \ell \; -->}}_3$ are the values of a,b and c ordered in ascending order, then the triangle exists iff ${\mathrm{\ell <!--\; \ell \; -->}}_1+{\mathrm{\ell <!--\; \ell \; -->}}_2\ge <!--\; \ge \; -->{\mathrm{\ell <!--\; \ell \; -->}}_3$.
Interestingly, the three triangle inequalities can be recast into a single quartic polynomial inequality. Let $0,x,y\in <!--\; \in \; -->{\mathbb{R}}^2$ be the three vertices of the triangle, with $\Vert <!--\; \Vert \; -->x\Vert <!--\; \Vert \; -->=a,\Vert <!--\; \Vert \; -->y\Vert <!--\; \Vert \; -->=b$ and $\Vert <!--\; \Vert \; -->x-<!--\; -\; -->y\Vert <!--\; \Vert \; -->=c$. Then $c^2=\Vert <!--\; \Vert \; -->x-<!--\; -\; -->y{\Vert <!--\; \Vert \; -->}^2=\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}^2-<!--\; -\; -->2⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->+\Vert <!--\; \Vert \; -->y{\Vert <!--\; \Vert \; -->}^2=a^2+b^2-<!--\; -\; -->2⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->$
Therefore $x^{T}y=⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->=\frac{1}{2}(a^2+b^2-<!--\; -\; -->c^2)$ and
$\begin{array}{}\text{(1)}& \left(\begin{array}{c}{x}^T\\ y^T\end{array}\right)\left(\begin{array}{cc}x& y\end{array}\right)=\frac{1}{2}\left(\begin{array}{cc}2a^2& a^2+b^2-<!--\; -\; -->c^2\\ a^2+b^2-<!--\; -\; -->c^2& 2b^2\end{array}\right).\end{array}$
The RHS of (1) must be positive semidefinite because the LHS is a Gram matrix. Conversely, if the RHS is indeed PSD, it can be expressed as a Gram matrix. Hence we obtain x and y and the triangle exists.
As $2a^2$ and $2b^2$ are already nonnegative, the RHS of (1) is positive semidefinite if and only if $(2a^2)(2b^2)-<!--\; -\; -->(a^2+b^2-<!--\; -\; -->c^2{)}^2\ge <!--\; \ge \; -->0$, by Sylvester's criterion. That is, the triangle exists if and only if
$\begin{array}{}\text{(2)}& -<!--\; -\; -->(a^4+b^4+c^4)+2(a^2{b}^2+b^2{c}^2+c^2{a}^2)\ge <!--\; \ge \; -->0.\end{array}$
This polynomial inequality can be derived by more elementary means. See circle-circle intersection on Wolfram MathWorld. The geometric explanation for the necessity of (2) is given by Heron's formula, which states that the square root of the LHS is four times the area of the triangle.
Since both (0) and (2) are necessary and sufficient conditions for the existence of the required triangle, the two sets of conditions must be equivalent to each other. Here are my questions. Is there any simple way to see why (0) and (2) are equivalent? Can we derive one from the other by some basic algebraic/arithmetic manipulations?

Modular arithmetic with polynomial
Given $n=pq$ where $p\mathrm{\&<!--\; \&\; -->}q,p\ne <!--\; \ne \; -->q$ are primes, P(x) is polynomial and $z\in <!--\; \in \; -->{\mathbb{Z}}_n$
I need to prove that:
$p(z)\equiv <!--\; \equiv \; -->0(\mathrm{mod}n)⟺<!--\; ⟺\; -->p(z(\mathrm{mod}p))\equiv <!--\; \equiv \; -->0(\mathrm{mod}p)\wedge <!--\; \wedge \; -->p(z(\mathrm{mod}q))\equiv <!--\; \equiv \; -->0(\mathrm{mod}q)$
If i could prove the more general case: $P(z)\mathrm{mod}n\equiv <!--\; \equiv \; -->P(z\mathrm{mod}p)\mathrm{mod}p$ (q is the same) then i could also prove what i need of course. from my understanding so far, the above is true, but i can't figure out a way to prove this.
Back to the original question, i tried to see why the fact that p divides P(z mod p) and q divides P(z mod q) implies that pq divides P(z) and vice versa, but i didn't have much success with this.

Prove that given projective curve has genus 1.
Show that
$X=\{((x_0:x_1),(y_0:y_1)):(x_0^2+x_1^2)(y_0^2+y_1^2)=x_0{x}_1{y}_0{y}_1)\}\subseteq <!--\; \subseteq \; -->{\mathbb{P}}^1\times <!--\; \times \; -->{\mathbb{P}}^{\mathbb{1}}$
is a smooth curve of genus 1.
I can prove it with the following reasoning.
Using the Segre Embedding, the curve consists on all the elements $(x:y:z:w)\in <!--\; \in \; -->{\mathbb{P}}^3$ satisfying the equations
$x^2+y^2+z^2+w^2-<!--\; -\; -->xw=0,xw-<!--\; -\; -->yz=0,$
i.e. the vanishing set of I=$I=(x^2+y^2+z^2+w^2-<!--\; -\; -->xw,xw-<!--\; -\; -->yz)$. I can use the Jacobian criterion to prove it's smooth; no issue.
To find the genus, since I'm better at it, I decided to compute the arithmetic genus. To do that, I proved that
$\{x^2+y^2+z^2+w^2-<!--\; -\; -->xw,xw-<!--\; -\; -->yz\}$
is in fact a Gröbner basis (Using GRevLex ordering), so
$LT(I)=(x^2,yz).$
Then every minimal free resolution of the quotient over a monomial ideal (I think I can remove the hypothesis of it being monomial; I'm not completely sure, but in such case I don't need to compute LT(I) or even prove that I have a Gröbner basis) generated by two elements has the form $0\to <!--\; \to \; -->S\to <!--\; \to \; -->S^2\to <!--\; \to \; -->S\to <!--\; \to \; -->0$, in particular, in this case it has the form
$0\to <!--\; \to \; -->S(-<!--\; -\; -->4)\to <!--\; \to \; -->S(-<!--\; -\; -->2{)}^2\to <!--\; \to \; -->S\to <!--\; \to \; -->0$
which allows me to compute the Hilbert polynomial of X, by the method in the first section of The Geometry of Syzygies from Eisenbud (I remember it's also used in Cox's), and with the Hilbert polynomial I also have that the arithmetic genus is 1.
But can I prove it without computing the Hilbert polynomial?

"Let $\mathrm{\alpha <!--\; \alpha \; -->}$ be an algebraic integer and p(x) be its minimal (monic) polynomial
$p(x)=\underset{i=0}{\overset{n}{\sum <!--\; \sum \; -->}}{a}_i{x}^i$
Such that $p(\mathrm{\alpha <!--\; \alpha \; -->})=0$ and $a_i\in <!--\; \in \; -->\mathbf{\text{Z}}$ (Z is the set of integers) and $a_n=1$.
"The extension of a ring A by the element a is the set $A[\mathrm{\alpha <!--\; \alpha \; -->}]$ of all complex numbers of the form
$\underset{j=0}{\overset{n-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}{c}_j{\mathrm{\alpha <!--\; \alpha \; -->}}^j$
such that $c_j\in <!--\; \in \; -->A$, with all the operations inherited from A.
"The degree of the extension is the degree of the polynomial."
I completely understand the 'algebraic integer' and the 'minimal polynomial' and the concept of set/ring extension, at least I think. My issue is mostly with the middle sentence; when it says 'all the complex numbers of form', but isn't there only one minimal polynomial so there's only one element in the set $A[\mathrm{\alpha <!--\; \alpha \; -->}]$? Is that single element basically just $p(\mathrm{\alpha <!--\; \alpha \; -->})-<!--\; -\; -->x^n$? Or are the $c_j$ related to the $a_i$ at all? The same n is mentioned twice. If not then why introduce p(x) at all? And what is ring A? Is that an initially empty set? And what does 'all the operations inherited from A' mean? Honestly, I can't find an explanation online for the $A[\mathrm{\alpha <!--\; \alpha \; -->}]$ sort of notation or any of my other questions online?

How to find explicit formula for a sequence seems to be geometric but isn't?
I recently came across this interesting problem, where the progression was neither arithmetic nor geometric. The problem asks for an explicit formula for the following recursive formula:
$a_1=2$
$a_n=5(a_{n-<!--\; -\; -->1}+2)$
So the first five terms are 2,20,110,560,2810.
I tried to distribute the 5 in the second equation to get $5a_{n-<!--\; -\; -->1}+10$ and then tried to apply the formula for a geometric sequence, which is $a_1(r{)}^{n-<!--\; -\; -->1}$, but got only $2(5{)}^{n-<!--\; -\; -->1}$, which clearly doesn't work for the sequence.
I also tried using finite differences on the first five terms to see if it was just a polynomial rule, but that didn't work.
How would you go about doing a problem like this?

Need help understanding finite fields / modulo for polynomials
I'm taking a class in finite fields and have not been able to conceptualize how modulo + finite fields works in polynomial space. I understand the basic premises of modular arithmetic, but can't work out how to actually generate a finite field of polynomials.
For example:
Find all f(x) and g(x) in ${\mathbb{Z}}_3[x]$:
$(x^3+x+1)f(x)+(x^2+x+1)g(x)=1$
I know conceptually how to solve this sort of equation when the coefficients are integers and f(x),g(x) are simple variables, but I don't know how to generate fields in ${\mathbb{Z}}_3[x]$ and then how exactly to use them to solve this sort of equation for polynomials once I have their gcd in ${\mathbb{Z}}_3[x]$.

Prove that f(x) is differentiable $U(x)=x^2+f(x)$ is differentiable for all x
Here's my proof please let me know if it's OK:
$x^2$ is differentiable for all x because it's a polynomial. Thus if f(x) is differentiable for all x then U(x) is differentiable for all x. Let U(x) be differentiable for all x. Thus according to addition arithmetic $U(x)-<!--\; -\; -->x^2$ is differentiable for all x.
But $U(x)-<!--\; -\; -->x^2=x^2+f(x)-<!--\; -\; -->x^2=f(x)$. Thus the addition that differentiable itself equals f(x) therefore f(x) is differentiable as well.
I have to admit that there was a very similar exercise in my class and from which I used the "addition arithmetic" however I'm not sure what it would refer to. Does this refer to the fact that according to derivatives arithmetic the result of addition of 2 function which are differentiable themselves is differentiable? But we didn't actually derive these functions.

Potential such that the time-independent Schroedinger equation has an explicit solution
Consider the time-independent Schroedinger equation
${\mathrm{\varphi <!--\; \varphi \; -->}}^{″}(x)+V(x)\mathrm{\varphi <!--\; \varphi \; -->}(x)=\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{\varphi <!--\; \varphi \; -->}(x)$
For testing my numerics, I would like to now:
For which choices of V are there explicit expressions for the solutions $(\mathrm{\varphi <!--\; \varphi \; -->},\mathrm{\lambda <!--\; \lambda \; -->})$?
For example, if $V(x)=x^2$, then the solutions are products of Hermite polynomials and $exp(-<!--\; -\; -->x^2)$ (up to correct scales) and arithmetic progressions (the same works in higher dimensions with the analogue potential).

Prove that there exists an integer greater than x such that any polynomial f(x) will be strictly non-negative and get large?
I am taking a number theory class and so far I have been proving modular congruences, modular arithmetic, and prime properties. There is this theorem that came up in the textbook and apparently it does not involve any modular arithmetic. The theorem is as follows
Suppose $f(x)=a_n{x}^n+a_{n-<!--\; -\; -->1}{x}^{n-<!--\; -\; -->1}+\cdots <!--\; \cdots \; -->+a_0$ is a polynomial of degree $n>0$. Then there exists an integer k such that if $x>k$ then $f(x)>0$.
I feel like this would come up in real analysis, but I have not come that far in my studies. I have an idea of applying induction and using the ceiling function somehow, but I have no idea how to start off this proof. Any help will do and thank you

A general theory of convolution product
in my childhood, I learned about convolution products for function over R (1). For quite a while now, I have played with polynomial rings, where also, the product is sometime called a convolution product (2). I am now discovering convolution products for arithmetic function and multiplicative functions [Apostol, Introduction to Analytic Number Theory] (3).
In case (1), the domain is R, an ordered abelian group. In case (2), seeing a ring of polynomial as a group ring Z⟨G⟩, the domain is an abelain group, without ordering. In case (3), the domain seems to be the monoid $(\mathbb{Z},\times <!--\; \times \; -->)$, with an ordered by the divisibilty relation.
I don't have anything very formal, but it seems to me that there should be a general theory of convolution I don't know yet about ? Is the monoid structure the most general domain, or maybe something less structured as an acyclic graph ? Would you have lectures notes on such a theory ?

If a polynomial is evaluated over an interval (by the use of interval arithmetic and Horner's method), what does the result say about the polynomial's roots?
I mean, if the resulting interval contains zero, does it mean there must be one more more roots inside the original interval? Could it happen that there might be no zero?
If the resulting interval contains no zero (i.e. both of its end-points have the same sign), does it mean there is no polynomial root inside the original interval? What if one of the end-points of the result is zero?

A polynomial whose roots form an arithmetic progression
Let f be a fourth degree polynomial whose roots form an arithmetic progression. Prove that f′'s roots also form an arithmetic progression.
I didn' t make much progress, I just wrote $f(x)=a(x-<!--\; -\; -->b-<!--\; -\; -->r)(x-<!--\; -\; -->b-<!--\; -\; -->2r)(x-<!--\; -\; -->b-<!--\; -\; -->3r)(x-<!--\; -\; -->b-<!--\; -\; -->4r)$ and I tried to differentiate, which obviously doesn't help too much.

What is the remainder on division of $z^{400}+z^{303}+1$ by $z^4-<!--\; -\; -->1$?
I am asked to determine the remainder when the polynomial $f(z)=z^{400}+z^{303}+1$ is divided by the polynomial $g(z)=z^4-<!--\; -\; -->1$.
I expressed f as $f(z)=h(z)(z^4-<!--\; -\; -->1)+r(z)$ where r(z) is a polynomial
Realising that $z^4-<!--\; -\; -->1=(z^2-<!--\; -\; -->1)(z^2+1)=(z-<!--\; -\; -->1)(z+1)(z^2+1)$, I expressed f as
$f(z)=h(z)(z-<!--\; -\; -->1)(z+1)(z^2+1)+r(z)$
I also wrote f as $f(z)=q(z)(z-<!--\; -\; -->1)+3$, since $f(1)=3$, and q is some polynomial.
So I had $f(z)=q(z)(z-<!--\; -\; -->1)+3=h(z)(z-<!--\; -\; -->1)(z+1)(z^2+1)+r(z)$
Hence, for some constant k, $q(z)=h(z)(z+1)(z^2+1)+k$
We know $f(-<!--\; -\; -->1)=1$
From $f(z)=q(z)(z-<!--\; -\; -->1)+3$, we have $f(-<!--\; -\; -->1)=q(-<!--\; -\; -->1)(-<!--\; -\; -->2)+3=1$, which implies $q(-<!--\; -\; -->1)=1$.
Then $q(-<!--\; -\; -->1)=h(-<!--\; -\; -->1)(0)+k=1$. Thus, $k=1$
We then have $q(z)=h(z)(z+1)(z^2+1)+1$, and from this, we have $f(z)=q(z)(z-<!--\; -\; -->1)+3=(h(z)(z+1)(z^2+1)+1)(z-<!--\; -\; -->1)+3$. Which simplifies to $f(z)=h(z)(z^4-<!--\; -\; -->1)+z+2$
And gives $r(z)=z+2$.
But I am given the answer is $r(z)=z^3+2$. What is going wrong here? Can I use modular arithmetic instead? How?

Can all programs be modeled as operations of elementary arithmetic operations on inputs?
In mathematics and computabiltiy theory, we treat all inputs and intermediate results and final outputs as natural number. While algorithms/programs themselves are considered natural numbers, here we treat these programs/functions/algorithms as just computable functions.
The question is, when the function operates on an input to produce an output, can we consider the operation of function as using only a number of arithmetic operations (addition, subtraction, multiplication and division) on an input? Or does the use of if/else make the aforementioned not true?
If this is true, is the number of arithmetic operations polynomially proportional to the lowest time complexity bound possible for solving a problem? (That is, if the lowest time complexity is O(whatever), then the number of arithmetic operations is ${\text{O(whatever}}^k)$ where k is some rational number.)

Evaluation of polynomial modulo in GF(2)
I'm new to this forum and I have a question about about modular arithmetics. My background is in computational chemistry and I work now as a software developer. For my work I need to learn about cryptography and I'm reading the book "Cryptography Made Simple" by Nigel P. Smart.
In paragraph 1.2 on Finite Fields, the author gives an example of polynomial modulo in ${\mathbb{F}}_2(x)/f(x){\mathbb{F}}_2(x)$ as such:
$(1+x+x^2)(x+x^3)(\mathrm{mod}{x}^4+1)=x^2+1$
However, when I do the evaluation myself I get a different result, which is confirmed by using the PolynomialMod function in Mathematica. I will show the steps I took below. First of all I expand the multiplication and evaluate modulo 2:
$x^5+x^4+2x^3+x^2+x(\mathrm{mod}2)=x^5+x^4+x^2+x$
Then:
$x^5+x^4+x^2+x(\mathrm{mod}{x}^4+1)$
$x(x^4+1)+x^4+x^2(\mathrm{mod}{x}^4+1)$
$x(x^4+1)+(x^4+1)+x^2-<!--\; -\; -->1(\mathrm{mod}{x}^4+1)$
Which finally results in:
$(x+1)(x^4+1)+x^2-<!--\; -\; -->1(\mathrm{mod}{x}^4+1)=x^2-<!--\; -\; -->1$
I put in attachment a screenshot 1 from Mathematica to confirm it matches my result. Initially I thought it was an error in the book, but in a following section the author gives another example:
$(x^3+1)(x^4+1)(\mathrm{mod}{x}^7+x+1)=x^4+x^3+x$
Again, evaluating the expression manually I obtain:
$(x^3+1)(x^4+1)(\mathrm{mod}{x}^7+x+1)=x^4+x^3-<!--\; -\; -->x$
Which is confirmed once more by Mathematica, see screenshot 2.
Does anyone know if there is a reason behind the difference in the results? Any help would be much appreciated.

Assume we have a computable, nonstandard model of Robinson arithmetic consisting of the set of integer-coefficient polynomials with positive leading coefficient, plus the zero polynomial, with their usual arithmetic.
Consider the formula
$\underset{i=0}{\overset{n}{\sum <!--\; \sum \; -->}}i=\frac{n^2+n}{2}$
Is this formula valid in this model? Let |2x| be the nonstandard natural number represented by the polynomial 2x. In this model, is the sum of all natural numbers from |0| through |2x| equal to $|2x^2+x|$? If so, does this mean some hyper-finite sums are undefined? For example, the sum of |0| through |x| would be $\frac{x^2+x}{2}$, which doesn't exist in this model.

Nonstandard models of Presburger Arithmetic
I have a question about nonstandard models of Presburger Arithmetic. I read that an example of a nonstandard model is the set of polynomials with rational coefficients with positive leading coefficient and a (positive) integer constant coefficient. It is obvious that all axioms other than the induction axiom schema are satified; but how can one prove that the induction axiom holds?