9

If it is reducible over $\Bbb{Z}$ then it has a root in $\Bbb{Z}$, say $k\in\Bbb{Z}$. Then $k^3+nk+1=0$ so $$-1=k^3+nk=k(k^2+n),$$ which shows that $k$ divides $-1$, so $k=\pm1$. Solving the two equations $$1^3+n\cdot1+1=0\qquad\text{ and }\qquad (-1)^3+n\cdot(-1)+1=0,$$ yields $n=-2$ and $n=0$ as the only values for which the polynomial is reducible over $\...


5

Use the continued fraction expansion for $\frac {\sqrt{3}}2$. The first $10$ convergents are $$\left \{ 0, 1, \frac 6 {7}, \frac {13}{ 15},\frac {84}{97}, \frac {181}{ 209}, \frac {1170}{1351}, \frac {2521}{ 2911}, \frac {16296}{ 18817}, \frac {35113}{ 40545}\right \} $$ Note that your $13,15$ pair occurs early on. Using the final one, we note that $$\...


3

Your expression is $$\left(p + \frac{r}{a}\right)\left(p -1 + \frac{r}{a}\right) - \left(m + \frac{1}{a}\right)\left(m - 1+ \frac{1}{a}\right) = 0 \tag{1}\label{eq1}$$ If you let $$b = p + \frac{r}{a} \tag{2}\label{eq2}$$ $$c = m + \frac{1}{a} \tag{3}\label{eq3}$$ and moving the second term to the right, \eqref{eq1} then becomes $$b(b - 1) = c(c - 1) \...


3

One way to think of xor is as bitwise mod 2 addition. Thus it would be natural to take tritwise mod 3 addition for your new operation. To be more explicit. Define our new operation $n*m$ by writing both numbers in base 3, $$n=\sum_{i=0}^\infty a_i3^i,\text{ $a_i\in\{0,1,2\}$},$$ $$m=\sum_{i=0}^\infty b_i3^i,\text{ $b_i\in\{0,1,2\}$},$$ then define $c_i\...


2

If "$X \vee Y = Z$" means just that at least one of $X, Y$ is $Z$ then all you are really asking is that $$ A(p*q) = (q,p) $$ since then $$ (p,q) \vee (q,p) = (q,p) $$ will be true. There are lots of pairs of functions $(*, A)$ that will do. You just have to encode a pair as a single number that uniquely determines both $p$ and $q$. Then let $A$ be the ...


2

The statement is not correct . For example $10^{201}-1$ is within the range of $[10^{200},10^{20585}] $ but it does not contain any $0$ digit because all its digits are $9$.


2

These numbers are of the form $n = 1XXXXXXXXXX$, where $X$ can either be a $0$ or a $1$. So, the equivalent question here is, how many $10$-digit binary numbers have at least as many $1$'s as $0$'s? If and only if this is the case, the the number $n$ will have more $1$'s than $0$'s. Let's split this into two cases: 1) how many $10$-digit binary numbers have ...


1

Yes, you can get arbitrarily close. A good way is to look at the convergents of the continued fraction for $\frac {\sqrt 3}2$ In this case it is $[0;1,\overline{6,2}]$ or $$\frac 1{1+\frac 1{6+\frac 1{2+\frac 1{6+\frac 1{2+\ldots}}}}}$$ $\frac {13}{15}$ comes from taking just the first $6$ and first $2$, so $$\frac 1{1+\frac 1{6+\frac 12}}=\frac 1{1+\frac ...


1

$ax+by=n$ is a line, thus with the density of continuum. If you take out the double integral (diophantine) solutions, which may be none, or countable (finite or infinite, depending on the bounds), then you are left, at the minimum, with $\mathbb R \backslash \mathbb Z$ (eventually, within the given bounds).


1

Without any constraints on $\ast$ and $A$, this simply follows from cardinality arguments. Since $|\mathbb{Z}\times \mathbb{Z}|\le |\mathbb{R}|$ (via the well known fact that $\mathbb{Z}^2$ is countable and $\mathbb{R}$ is uncountable), there must exist an injective function $\ast : \mathbb{Z}\times\mathbb{Z}\to \mathbb{R}$. When restricted to it's image ...


1

This is a well known problem - perhaps, you may want to have a look at Stars & Bars. In fact, the number of $k$-uples of non-negative integers $a_1,\ldots,a_k$ summing up to $n$ is $\binom{n+k-1}{n}.$ Then, to find out the numbers of $k$-uples of non-negative integers satisfying $a_1+\ldots+a_k\leqslant n,$ you can either use this binomial identity ...


1

Yup, that looks good! To prove that these were the only classes, you'd appeal to the principle of mathematical induction, by taking $n=0,1,2,3,4$ as your base cases.


1

Suppose $S$ is the set of all $k$ that satisfy $p(k)=k^3$. Then we can write $$p(x)=x^3+\prod_{i\in S}(x-i)q(x)$$ But we also know that $p(100)=100$ so $$100-100^3=\prod_{i\in S}(100-i)q(100)$$ and since $100-100^3$ has $8$ prime factors (counting multiplicity), there can be at most $10$ integers in the product; thus the answer is $\fbox{10 }$ (which can be ...


1

Obviously, $(n+r)^3>n^3$ and since we want $r$ to be minimal, this occurs when $(n+r)^3=n^3+1$. Now note that $(n+r)^3=n^3+3n^2r+3nr^2+r^3\approx n^3+3n^2r$ so we want $3n^2r\approx1$ which is achieved when $n\approx 18$. Just by subbing in, $18$ is too small as $\sqrt[3]{18^3+1}>18.001$ and $\sqrt[3]{19^3+1}<19.001$ so $19$ is the answer.


1

It's not that cumbersome. We have $\beta - 1 = \frac{\sqrt[3]5}3$, which gives us $$ (\beta - 1)^3 = \frac 5{27}\\ 27(\beta - 1)^3 - 5 = 0\\ 27\beta^3 - 81\beta^2 + 81\beta - 32 = 0 $$ and we see that the minimal integer polynomial is not monic.


1

Using $\, ab\bmod ac\, =\, a\,(b\bmod c) $ = mod Distributive Law to factor out $\,a =4\,$ yields $\ \ \ \ 8^{\large 3+4K}\!\bmod 20\, =\, 4\,(\underbrace{2\cdot \color{#0a0}{8^{\large 2}}\,\color{#c00}8^{\large \color{#c00}{4}K}}_{\Large 2\, (\color{#0a0}{-1})\,\color{#c00}1^{\Large K}}\bmod 5) = 4(3)\ $


1

Although $8^{4\cdot 2^{n-1}}\equiv 16\pmod {20} $ for all positive integer $n $, it is actually much simpler: $8^{4k}\equiv 16\pmod {20} $ for all $k>0$. That is for any multiple of $4$, not just $4$times powers of $2$. And furthermore $8^{4k+1}\equiv 16*8\equiv -4*8\equiv-32\equiv 8\pmod {20} $ for $k>0$ And $8^{4k+2}\equiv 8*8\equiv 4\pmod {20} $...


1

You are almost there. You can reduce anything modulo $20$ so: $$16\times 8^3\equiv 16\times 8^2\times 8\equiv 16\times 4 \times 8\equiv64\times 8\equiv 4\times 8\equiv 12$$ You could also have used $16\equiv -4$ if it had helped - sometimes negative numbers make life easier, but it was not necessary here.


1

$8^n=(2^3)^n=2^{3n}$ As $(2^{3n},20)=4$ for $n\ge1$ Let's find $2^{3n-2}\pmod5$ Now $2^{3n-2}\equiv2^{(3n-2)\pmod4}\pmod5$ as $\phi(5)=4$ $\implies2^{3n}\equiv4\cdot2^{(3n-2)\pmod4}\pmod{20}$ Here $n=119\implies3n-2=3\pmod4$


1

Yes, you definitely need more axioms. The listed axioms are also true for the rational numbers, for instance, which have lots more units. You also can't prove $0\neq 1$, since a singleton set would also satisfy all the axioms (although in that case, you technically wouldn't be wrong in saying that $1$ and $-1$ are the only units). Also conspicuously ...


1

Note that we just want the thing under the sqrt to be non-negative. $$y = \sqrt{a^{x+0.5}+a^{4}\sqrt{x}-x^{0.5+x\log_x a}-a^{4.5}}$$ So make the thing under the square root non-negative. $$a^x\sqrt{a}+a^4\sqrt{x}-a^x\sqrt{x}-a^4\sqrt{a}\ge0$$ Thus $$(a^x-a^4)(\sqrt{a}-\sqrt{x})\ge0$$ Note that this means $x=4$ is always a solution. If $a\ge4$, then if ...


1

Given $(a,b)$ as lengths of two sides of triangle, we can calculate number of lattice points in it ($n$) as follows: Calculate slope: $m=\frac b a$ Calculate number of points of one of the two sides by adding $1$ to the length: $$c=b+1$$ Use the following formula: $$n=\sum^{a}_{k=0} \lfloor c-|km| \rfloor$$ Note that: we used the absolute value for $km$ ...


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