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10

Your logic argument for $-1$ being the only negative prime: $-3$, for example, could not be prime because it is equal to $3\cdot(-1)$ $-1$, however, is prime because it is divisible only by itself and by $1$ According to this logic, $+1$ is the only positive prime, since: $3$, for example, could not be prime because it is equal to $(-3)\cdot(-1)$


7

You are confusing computer arithmetic with mathematical arithmetic. In a computer, integer arithmetic is exact as long as you don't overflow and $5(n^2-3)$ will always yield an exact integer if $n$ is an integer and the computation is done using integer representations. When you go out of the integers, you use floats, which have a limited number of bits of ...


5

If your computer system does not give an integer when computing $(n-\sqrt3)(n+\sqrt 3)$, then it is because of the floating point arithmetic involved in computing with the irrational number $\sqrt3$. See the section in https://en.m.wikipedia.org/wiki/Floating_point on "accuracy problems." This is a good lesson in realizing that the results of computing ...


5

$x = 3^n, y = 2^n$ $$ 3 x^2 + x y - 4 y^2 = (3x + 4y)(x - y)$$


4

If you think negative numbers are weird, try imaginary numbers. There's not really a good way to think of positive numbers. You could say an imaginary number is positive if its real part is positive, but then you still get results like $(1+2i)(1+2i) = -3+4i$ so you immediately lose the fact that the product of two positive numbers is positive. That means ...


3

C150 is the product of the two primes, one with 55 (decimal) digits and one with 96 digits, $P55=1449299471738053389661827008867152641816024786660724327$ and $P96=3113530633988882752054263646036326764738143136281825948356832545797877\ 39684923341337929165875133$. I found these factors using the GMP-ECM algorithm incorporated in Sage, with the initial ...


2

I think n=43 works. The roots are 1, 6, 7, 36, 37, 42, 8, and 33. The first observation is that this polynomial factors over $\mathbb{Z}$ as $(x^6-1)(x^2+2x+6)$. We can actually go further and factor $(x^6-1)$ more, but I don't think it's necessary. The next simplification is to just look for $n$ prime, in particular then the multiplicative group ...


2

It can be shown by induction on $p$ that $f(a^p) = pf(a)$ for any integers $a,p \geq 1$. We will prove that $f(n) = \log_b n$ for some $b > 1$. After multiplying $f$ by a positive constant, we may assume $f(10) = 1$. Let $n$ be fixed. We will prove that $f(n) = \log n = \log_{10} n$. Otherwise, we have either $f(n) < \log n$ or $f(n) > \log n$. ...


1

I think this result will follow if you assume that given primes $p_1,p_2$ and $\epsilon>0$ there exists integers $n_1,n_2,n_1',n_2'$ such that \begin{equation} 1-\epsilon<\frac{n'_1\log p_1}{n'_2\log p_2}<1<\frac{n_1\log p_1}{n_2\log p_2}<1+\epsilon \end{equation} With this approximation, you can argue that $f(p_1)/f(p_2)$ is arbitrarily close ...


1

Here is Euler's Other Proof by Gerald Kimble \begin{align*} \frac{\pi^2}{6}&=\frac{4}{3}\frac{(\arcsin 1)^2}{2}\\ &=\frac{4}{3}\int_0^1\frac{\arcsin x}{\sqrt{1-x^2}}\,dx\\ &=\frac{4}{3}\int_0^1\frac{x+\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!}\frac{x^{2n+1}}{2n+1}}{\sqrt{1-x^2}}\,dx\\ &=\frac{4}{3}\int_0^1\frac{x}{\sqrt{1-x^2}}\,dx ...


1

Basically, naming $s=\sigma+it$, the idea is to use the function $$ \xi(s)=\Gamma(s/2)\pi^{-s/2}(s-1)\zeta(s) $$ because it's real-valued on the critical line $t=1/2$, hence you'll find a zero whenever $\xi(1/2+it)$ changes sign. There are various method to do that, a very nice introduction can be found in Edwards' book ``Riemann's Zeta Function", see for ...


1

Hint We can proceed naively but efficiently. It is plausibly useful to factor $f(x)$ over $\Bbb Q$: $$f(x) = (x^6 - 1)(x^2 + 2 x + 6),$$ and recalling our cyclotomic polynomials, we can easily factor the first factor, giving, respectively, $$f(x) = (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1)(x^2 + 2 x + 6).$$ Obviously, we need $n \geq 7$. On the other hand, ...


1

What you're missing is that the equation $d\alpha=0$ is taking place in $P_0/IP_0$, not in $P_0$. Backing up a bit, if $\alpha\in P_1/IP_1$ is a cycle, let's choose a lift $\beta\in P_1$ for $\alpha$. Then you can't necessarily say that $d\beta=0$ as an element of $P_0$, since all you know is that the image of $d\beta$ in the quotient $P_0/IP_0$ is $0$, ...


1

This is probably more complicated than you are hoping for, but it at least restates the problem by breaking it into two smaller problems. Suppose you are given some number $A$ and wish to know whether it is equal to the given formula for some $n$ and $s$. If so, then we would have $$\frac{s-2}{2}n^2 - \frac{s-4}{2}n = A$$ which, after some very slight ...


1

For $Re(s)> 1$ we have the Euler product $$ \frac{1}{\zeta(s)}=\prod_p (1-p^{-s}). $$ Taking the limit $s\to 1$ shows that the infinite product $\prod_p (1-\frac{1}{p})$ is $0$, since $\lim_{s\to 1}\zeta(s)=\infty$. A second possibility is to use that the product $\displaystyle\prod_{n=1}^{\infty} (1- a_n)$ is non-zero if and only if $\sum_{n=1}^{\infty} ...


2

Apart from $2$, $3$, $5$ and $7$, every prime number must be relatively prime to $210$. Between $1$ and $210$, there are $\varphi(210) = 48$ numbers prime to $210$. Whether a number is relatively prime to $210$ depends only on its remainder modulo $210$. Therefore for all $k \geq 1$, we have $\pi(210k) \leq 48k + 4$. For any number $n \geq 1$, there is some ...



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