New answers tagged

0

Since the roots are single digits they are also the roots in base 10. So work out your equation in base 10, remembering that the leading coefficient is to be 3 (also a single digit, therefore also the same in base 10) not 1. Then the actual base $b$ has to satisfy two equations: $-(2b+5)=$ your linear coefficient including the sign $6b+6=$ your constant ...


0

Notice that if you plug in the roots given, $x_1 = 4$, $x_2 = 9$ in the given quadratic $f(x) = 3x^2 - 25x + 66$, the values you get are $$y_1= f(x_1) = 3 *4^2 - 25 *4 + 66 = 14$$ $$y2 = f(x_2) = 3 * 9^2 - 25 * 9 + 66 = 84$$ Hence, if you need a number system where the roots $y_1$ and $y_2$ must be 0, it must take both $14$ and $84$ to $0$. What's the ...


0

For a more computational introduction and definitely manageable to work through on your own, I would suggest one of "A Friendly Introduction to Number Theory" or "Elementary Number Theory by Jones and Jones". I prefer the former, as it get's to some pretty complex stuff (elliptic curves by the end I think). The Jones and Jones is good too though, and manages ...


1

The denominator $d_n$ is divisible by every prime $p$ satisfying $p\leq n<2p$ (only one term in the sum $H_n$ has denominator divisible by $p$). Bertrand's Postulate implies such a prime exists if $n \geq 2$, so $d_n\geq n/2 \to\infty$ as $n\to\infty$.


2

Each of $E_\pm: x^4 \pm x^2 y^2 + y^4 = z^2$ is an elliptic curve. It turns out that in each case a Fermat-style "descent" suffices to find all solutions; in practice these days such questions are solved by reducing the curve to standard form and then either using software such as J.Cremona's mwrank, or consulting tables of elliptic curve if the curve is ...


0

You ask about a part of the theorem of Korselt of which I adjoint a copy. Unfortunately I don't remember the source of the publication however I give you a curious detail: Korselt has made his discovery before these numbers were called Carmichael and could not find any. It was Carmichael who discovered the first, 561,several years after this theorem.


0

We start as in your post, letting $n=p^tm$ where $t\ge 2$ and $m$ is not divisible by $p$. By the Chinese Remainder Theorem, the system of congruences $x\equiv 1+p\pmod{p^t}$, $x\equiv 1\pmod{m}$ has a solution $a$. Note that $\gcd(a,n)=1$. Since $n$ is Carmichael, we have $a^{n-1}\equiv 1\pmod{n}$. In particular, $a^{n-1}\equiv 1\pmod{p^t}$, and ...


0

The existence of local lifts is open in general, even for $F = \mathbf{Q}_p$ and general $n$. One expects the stronger result that there exist de Rham lifts. There is recent work of Gee-Hezig-Liu-Savitt on this problem (https://cms.math.ca/Events/winter15/abs/pdf/ant-fh.pdf). However, I do know some people who have an idea to answer the general case. I can ...


0

Note that $2\cdot 5-1=9$, $2\cdot13-1=25$, and $5\cdot15-1=64$. These are all perfect squares. Thus, there exists $a$ and $b$ such that $ab-1$ is a perfect square if and only if at least one of $$2d-1, 5d-1, \text{ or } 13d-1$$ is a perfect square. We will continue by contradiction. Assume temporarily that all of these are perfect squares. We then have ...


1

Given any polynomial $f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0\in\mathbb C[x]$ with complex roots $p_1,\dots,p_n$ (counted with multiplicity), we have $$ \sum_{i\neq j}p_ip_j=\frac{a_{n-2}}{a_n} $$ following from Vieta's formulas. Consider the cyclotomic polynomial $$\Phi_{83}(x)=\prod_{\zeta\text{ primitive $83^\text{th}$ root}}(x-\zeta).$$ Since $83$ is ...


1

Let $g$ be a primitive root of $83$. Then all the primitive roots are $g^k$ where $k$ is relatively prime to $82$, so all $g^k$ with odd $k$ from $k=1$ to $k=81$, with the exception of $k=41$. These are all the quadratic non-residues of $83$ except $-1$. Let $S$ be the sum of the primitive roots. Then the sum of the quadratic non-residues of $83$ is ...


1

I was surprised to learn, recently, that a simple idea ( made up, it appears, by Tito Piezas, without him knowing it was a neologism) allowed me to get the information in Conway's topograph with a fairly simple computer program, as long as what I wanted was to guarantee finding all solutions $(x,y)$ to $ax^2 + bxy+ cy^2 = n$ with integers $x,y > 0$ and ...


1

The square roots of $-1 \pmod 5$ are $2,3 \pmod 5$ but $154 \equiv 4 \pmod 5.$ The square roots of $-1 \pmod {13}$ are $5,8 \pmod {13}$ but $154 \equiv 11 \pmod {13}.$ The square roots of $-1 \pmod {17}$ are $4,13 \pmod {17}$ but $154 \equiv 1 \pmod {17}.$ The square roots of $-1 \pmod {29}$ are $12,17 \pmod {29}$ but $154 \equiv 9 \pmod {29}.$ The ...


4

Let $G=\sum_{a}\Big(\frac{a}{p}\Big)\zeta_p^a$. Here's a proof that $G^2=\Big(\frac{-1}{p}\Big)p$, which in particular shows that $|G|^2=p$. As $a$ runs over $(\mathbb{Z}/p\mathbb{Z})^{\times}$, so does $ab$ for fixed $b\neq0$, so we have: $$ ...


0

COMMENT.- I feel but I have no proof that there are only the trivial solutions $(x,y,z)= (t,0,t^2),(0,t,t^2)$. I give here an outline of what could perhaps lead to a proof. $$x^4-x^2y^2+y^4=z^2\iff (x^2-y^2)^2+x^2y^2=z^2\qquad (1)$$ Hence, as it is well known, $$\begin{cases}x^2-y^2=t^2-s^2\\xy=2ts\\z=t^2+s^2\end{cases}\qquad (2)$$ It follows in particular ...


1

My best guess would be that they just want you to do trial division by primes $p = 5,13,17,\ldots$ up to $36$ (there are much fewer primes to try with the $1$ mod $4$ restriction). For instance, to test whether $13$ divides $36^2+1$ you can compute $$ 36^2 + 1 \equiv (-3)^2 + 1 = 10 \not\equiv 0 \pmod{13}.$$ For $154^2+1$ you might in the worst case have ...


1

Yes, $d(n)$ can attain any positive integer for some $n$, consider $n=2^k$. Hence all ring numbers are of the form $d(n)$.


1

Take $R = \mathbb{F}_2[X]/(X^n)$. Then $R$ is finite and has exactly $n+1$ ideals. Indeed, ideals of $R$ are in canonical bijection with ideals of $\mathbb{F}_2[X]$ containing $X^n$, ie with polynomials dividing $X^n$ : these are the $X^k$ for $0\leqslant k\leqslant n$.


6

Yes, it is true, for odd $p$. The point is that a generator $g$ has order $p-1$, which is even. Yet, if $g= a^2$, then $g^{(p-1)/2} = a^{p-1} = 1$, a contradiction to $p-1$ being the order of $g$.


5

Note that this is $$ (x + y)^3 + y^3 = 37^3; $$ by Wiles' theorem the only integral solutions are $(37,0)$ and $(-37,37)$.


0

Basically the two equivalences \begin{align} x &\equiv a \pmod A \\ x &\equiv b \pmod B \end{align} have a common solution if and only if $$a \equiv b \pmod{\gcd(A,B)}$$ In the case of \begin{align} x &\equiv 2 \pmod 5\\ x &\equiv 1 \pmod {10}\\ x &\equiv 0 \pmod 3 \end{align} We notice that $1 \not \equiv 2 \pmod{\gcd(5,10)}$, ...


0

No there is not: the archimedean place is defined inherently by the ordering, which is not an algebraic property. In fact, this can easily be seen because equivalent absolute values generate the same topology, and you know that the standard topology on $\Bbb R$ is generated by it's total ordering (this is in general how one can define topologies on totally ...


2

Your proof is fine if $x$ and $y$ are rational, but your claim is that it works for $x$ and $y$ real. You assume that $y-x$ is rational. If not, $(y-x)\frac 1 {\sqrt 2}$ might be rational and the proof fails.


1

We are told that $a\varphi(a)=b\varphi(b)$, and want to conclude that $a=b$. Suppose the result is true whenever $\max(a,b)\lt n$. We show the result is true when $\max(a,b)=n$. Let $p$ be the largest prime that divides $a\varphi(a)$, or equivalently $b\varphi(b)$. Then $p$ is the largest prime that divides $a$, and $p$ is the largest prime that divides ...


1

This is good. Alternatively, for a direct proof, note $A$ is non-empty and bounded below. Every lower bound $c$ of $A$ must satisfy $c \notin \mathbb{R_+}$ since for all $\epsilon \in \mathbb{R_+}$ there is some $n$ such that $\epsilon > \frac{1}{n}$ and $\frac{1}{n} \in A$. Since $\max \mathbb{R} \setminus \mathbb{R_+} = 0$, and $0$ is a lower bound of ...


0

I'd like to point out that $$\frac b2<b\leq\text{inf}(A).$$ This being a contradiction implies that $b>\inf(A)$.


1

Your answer is fine. Just take some care with the assumption that "since $b \in A$, b can not be a lower bound of A". For example: $A = [0, 1]$. The infimum of this set is also $0$ and $0 \in A$.


0

On The Euler (Totient) Function Multiplicity Define theTotient function as Eu(.). To prove Eu(N∙M) = Eu(N)∙Eu(M). The set 1:N∙M is partitioned into set of columns. Each column is a sum of a base column: C0 = [ 0, M, .. (N-1)M]’; and an offset - k; k: 1,2,..M; A column Ck is: Ck= C0 + k. An element of a column Ck is the ...


3

Here's an outline of the proof: Show that there are $14$ primes under $\sqrt{2015}$. Let $R$ be a set of $14$ composite coprime numbers under $2015$. It is well-known that any composite number has a prime factor under its square root, so in this case, each of these composite numbers has one of the prime factors under $\sqrt {2015}$. Now, using that ...


1

In the 6th line of the text, $m$ is defined in such a way that $p_r|m$. Hence $\frac{m}{p_r} \in \mathbb{Z}$.


2

The number $5$ is a square modulo the prime $5$, though I would not call it a quadratic residue of $5$. Now assume that $p$ is an odd prime other than $5$. By Quadratic Reciprocity, since $5$ is of the form $4k+1$, we have $(5/p)=(p/5)$. Note that $(p/5)=(r/5)$, where $r$ is the remainder when we divide $p$ by $5$. It is easy to check that $1$ and $4$ are ...


3

If $p=2$ then $-1=1$ so $-1=1^2+0^2$. Now when $p>2$, let $S$ be the set of squares of $\Bbb Z/p$. Then $|S|={p+1\over 2}$. Now consider the set $-1-S = \{-1-s: S\in S\}$. This has the same cardinality, and since $$|S|+|-1-S|>p$$ it must be that there is an element in their intersection, call it $c$ then $a^2=c$ for some $a$ since $c\in S$. But ...


0


1

You can't take the square root like this. By Fermat's little theorem, we know that for $p$ prime we have that $5^{p-1} \equiv 1 \mod p$. This means that $p \mid 5^{p-1} -1$, and hence $$p \mid \left(5^{\tfrac{p-1}{2}} -1\right)\left(5^{\tfrac{p-1}{2}} +1\right)$$ Therefore $5^{\tfrac{p-1}{2}} \equiv 1 \mod p$ or $5^{\tfrac{p-1}{2}} \equiv -1 \mod p$. Note ...


0

Definition: $x\in\mathbb{N}$ is a Hilbert Prime if $x\equiv 1\pmod{4}$ and there does not exists a $y<x$ such that $y\equiv 1\pmod{4}$ and $y|x$. Let $\alpha=4k+1$. If $\alpha$ is not a Hilbert Prime, then by definition there exists a $\beta_0$ such that $\beta_0$ is a of the form $4k_0+1$ and $\beta_0|\alpha$. If $\beta_0$ is a Hilbert Prime we are ...


2

Nonsense. $a=105, b=7, k=98$. How did you get that idea? Or $a=3, b=1, k=2$.


0

In this case, if $k=5,$ then we would have $70$ as the sum of two integer squares, which it is not. $$ a^2 + b^2 = 5a + 5b + 5, $$ $$ 4a^2 + 4 b^2 = 20a + 20 b + 20, $$ $$ 4 a^2 - 20a + 4 b^2 - 20 b = 20, $$ $$ 4 a^2 - 20a + 25 + 4 b^2 - 20 b + 25 = 20 + 25 + 25 = 70, $$ $$ (2a-5)^2 + (2b-5)^2 = 70. $$ However, $70 = 2 \cdot 5 \cdot 7$ is not the sum of ...


1

Either $p$ or $q$ must have a factor of $r^2$. The other one can have either $r^0, r^1$ or $r^2$. That's five ways to the distribute $r$-factors. Similarily, the $s$-factors may be distributed in $9$ ways, and the $t$ factors can be distributed in $5$ ways. All in all we get $5.9.5 = 225$ different pairs.


1

If you look at C-S for just two terms and its use here becomes more obvious $$ (a_1b_1 + a_2b_2)^2 \le (a_1^2+a_2^2)(b_1^2+b_2^2) $$ Now look at inequality we want to prove. Square both sides and examine how the terms match with our C-S $$(a\sqrt{b+c} + b\sqrt{a+c} + c\sqrt{a+b})^2 \le 2(a+b+c)(ab+ac+bc)$$ We can see how our inequality will resemble C-S by ...


2

Use the Cauchy-Schwarz inequality on the two vectors $(\sqrt a, \sqrt b, \sqrt c)$ and $(\sqrt{a(b+c)}, \sqrt{b(a+c)}, \sqrt{c(a+b)})$ (then take the square root on both sides or not, depending on which version of the CS inequality you use), and lastly note that we have: $$ a(b+c) + b(a + c) + c(a + b) = 2(bc + ac + ab) $$


0

Case $a=1$ is trivial. Let's assume $a>1$, then this is equivalent to prove that $\gcd(a,b) \neq 1$. If $\gcd(a,b)=1$, it's easy to see that $\gcd(a,b^x)=1$, which is contradictory to $a \mid b^x$. Let $k=\gcd(a,b)$, then $k$ is factor of $a$ that divides $b$.


3

Once we take into account the zero solution. Next, we consider $a,b\ne0$. Number $b^3$ divided by $a^2$, which implies that $b^2$ divided by $a$. Then the left side is divided into $a^3$, then $b^3$ divided by $a^3$, and that's why $b$ divided by $a$. Let $b=ka$. The equation takes the form $$k^2a=4a^2+k^3.$$ Then $(2a)^2$ divided by $k^2$, and let $2a=km$. ...


1

Your proof looks good. It is not exactly how Euclid's proved it, but it works, and most importantly you understand it. A couple of minor points: Like lulu said in the comments, invoking the Fundamental Theorem of Arithmetic isn't necessary. Also in the end you state "No number, or prime, divides 1." This technically isn't true since 1 divides 1. It would ...


4

I would not add the complication of making this into a proof by contradiction. Euclid did not do it that way, despite many modern authors, dating back at least to Dirichlet in the middle of the 19th century, asserting that Euclid did it that way. Start with any finite set $S$ of prime numbers. (For example, we could have $S=\{2, 31, 97\}$) Let $p = 1 + ...


1

In case this was a question in good faith: since this problem is equivalent to the Riemann Hypothesis, don't expect to actually prove it. LAGARIAS For the curious, the best of the elementary RH versions to experiment with is the first one, by Jean-Louis Nicolas (1983). It says that RH is equivalent to the conjecture that, for all primorial numbers $P,$ ...


1

Since $r \in Q$ there exist $m_1,n_1\in Z$ with $n_1\ne 0$ and $r=m_1/n_1.$ So $r=m_2/n_2$ where $n_2=|n_1|\in N,$ and $m_2=(n_1/|n_1|)m_1=\pm m_1\in Z.$ So r is an integer divided by a natural number, so there is a LEAST $n\in N$ such that some $m\in Z$ satisfies $r=m/n.$ Now if $p\in Z$ and $|p|>1$ we cannot have $p|m$ and $p|n.....$ .... otherwise ...


0

Let $\frac{a}{b}\in\mathbb{Q}$ (with $(a,b)\in\mathbb{Z}\times\mathbb{Z_{>0}}$). If $a=0$, then $\frac{0}{1}$ does the job. Otherwise, $a\neq0$ and we consider the set $$ E:=\left\{c\in\mathbb{Z}:\exists d\in\mathbb{Z}_{>0}\text{ s.t. }\frac{c}{d}=\frac{a}{b}\right\} $$ First, $a\in E$, so $E$ is not empty. Also, remark that $\frac{a}{b}=\frac{c}{d}$ ...


0

If the denominator, $b<0$ then form the equivalent fraction by multiplying both numerator $a$ and denominator $b$ by $-1$. So begin by saying something like '...it is only necessary to consider $b>0$ since negative denominators can be made positive by forming an equivalent fraction etc, etc, ...' Hope this is clear.


2

So here is a way to show that these are all the numbers with the given property and which uses a small bit of machinery. What we want to show is that if $n = 2^km$ with $m$ odd has this property then $k\leq 3$ and $m=3$ or $m=1$. To see this note that by the Chinese remainder theorem, it suffices to find those prime powers $p^r$ for which the units in ...


0

In a cyclic group of even order, there are exactly $2$ solutions for $x^2=1$. The condition you seek is that $U(n)$ is not cyclic. This is equivalent to $n$ not being of the form $2,4,p^k,2p^k$, where $p$ is an odd prime. This is a consequence of the Chinese remainder theorem, which implies $$ U(n) \cong U(p_1^{e_1}) \times \cdots \times U(p_m^{e_m}) $$ ...



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