# Tag Info

13

If we prove that for every x there exists a prime number between $x^2$ and $x^2+2x+1$, we are done. This is Legendre's conjecture, which remains unsolved. Hence the big smile on your teacher's face.

12

Consider the cases $n=10^k$. Then we get that the $n$th digit of $a$ is the $k$th digit of $\sqrt{2}$. Now, if $a$ is rational, then it repeats with some frequency, $f$. But then we can find $d$ so that $f\mid 10^{k+d}-10^k$ for $k$ large enough. Therefore, for large enough $k$, the $10^{k+d}$th digit of $a$ and the $10^k$th digit of $a$ must be the same. ...

7

$$\begin{array}{rll} (-a)b &= (-a)b + 0 &0 \text{ is identity for addition}\\ &= (-a)b + (ab + (-ab))&-ab \text{ is additive inverse of } ab\\ &= ((-a)b+ab) + (-ab)&\text{addition is associative}\\ &= ((-a+a)b) + (-ab)&\text{multiplication is distributive}\\ &= \color{blue}{(0b)} + (-ab)&-a \text{ is additive inverse ... 6 We look first at the simpler problem of Pythagorean triples (x,y,z) where z=y+1. Such a triple must be primitive, and therefore z=u^2+v^2, y=2uv, for some u and v. Then u^2+v^2=2uv+1, giving u=v+1. That gives the triple x=2v+1, y=2v(v+1), z=2v^2+2v+1. Multiply each entry by 3 to get infinitely many triples (a,b,c) where c=b+3. ... 4 Daniel Shiu has proved (in "Strings of congruent primes", JLMS 2000) that there are arbitrarily long strings of consecutive primes in any given residue class. In particular, the number described above has arbitrarily long strings 11111..., 33333..., 77777..., and 99999... in its decimal expansion. This is enough to show that it is irrational. The same proof ... 4 This is Legendre's theorem on the sum of four squares. As the squares are congruent to 0, 1 or 4 (mod 8), anything integer congruent to 7 (mod 8) cannot be written as the sum of three squares. Perhaps the easiest way to prove this theorem is using Quaternions, an extension of the complex numbers in R^4. 4 Any of the accepted conjectures on sieves and random-like behavior of primes would predict that the chance of finding counterexamples to the conjecture in (x^2, (x+1)^2) decrease rapidly with x, since they correspond to random events that are (up to logarithmic factors) x standard deviations from the mean, and probabilities of those are suppressed very ... 4 As (17,100)=1, and using Carmichael function \displaystyle\lambda(100)=20 \displaystyle\implies17^{20}\equiv1\pmod{100} 16^{100}=(2^4)^{100}=2^{400} As (2^{400},100)=4=2^2\ne1 let us find 2^{400-2}\pmod{25} As \displaystyle\lambda(25)=\phi(25)=20, 2^{20}\equiv1\pmod{25} and \displaystyle400-2\equiv18\pmod{20}\implies ... 3 The derivative \Phi_n'(t) is useful not only for t=0 or t=1, but for t=\zeta_{p^r}, a p^r-th primitive root of unity. To show that of all prime ideals in \mathbb{Z}, only P=(p) is ramified in the cyclotomic field K=\mathbb{Q}(\zeta_{p^r}), one needs that the discriminant is given by$$ D(1,\zeta_{p^r},\ldots ,\zeta_{p^r}^{\phi(p^r)})=\pm ...

3

Consider $a\in \mathbb{Z}_m$ with $a\ne0$ and the map $\mu: x \mapsto ax$. Then $\mathbb{Z}_m$ is a field iff $\mu$ is surjective. Since $\mathbb{Z}_m$ is finite, this is equivalent to $\mu$ is injective. Now $\mu$ is injective iff ($\mu(x)=0$ implies $x=0$) iff ($m \mid ax$ implies $m \mid a$ or $m \mid x$). This proves that $\mu$ is injective when $m$ ...

3

Let $$\mathbb N_n=\{1,2,\ldots,n\}$$ and $$A_d=\{k\in \mathbb N_n\;|\; \gcd(k,n)=d\}$$ Hint Prove that $(A_d)_{d|n}$ is a partition of $\mathbb N_n$ and then $$(x^n-1)=\prod_{k\in \mathbb N_n}(x-e^{2ik\pi/n})=\prod_{d|n}\prod_{k\in A_d}(x-e^{2ik\pi/n})$$ finally notice that $$k\in A_d\iff\gcd(k,n)=d\\\iff \gcd(k/d=k',n/d=d')=1\iff k'\in\{k\in \mathbb ... 3 This is a straightforward application of the law of quadratic reciprocity. The Legendre symbol, \left(\frac pq\right), is set to be equal to 1 if p is a QR \pmod q and -1 is p is not a QR \pmod q:$$\left(\frac{p}{q} \right) \cdot \left(\frac{q}{p} \right) = -1^{\frac{p-1}{2} \cdot \frac{q-1}{2}}$$Now, plugging 5 in for p: ... 3 If you examine the inductive step in Apostol's proof you will see that it does not generally lift-up any positivity of the coefficients. Namely, to get a linear common divisor of \rm\,a,b\, it first obtains, by induction, a common divisor \rm\,d\, of \rm\, a-b,b\, of sought linear form \rm\, d = (a-b)x+by.\, Since \rm\,d\mid a-b,b\,\Rightarrow d\mid ... 3 Look for palindromic numbers of the most obvious format which also have a simple algebraic formula - the numbers 10^n+1 are palindromic and are easy to deal with mathematically. The equation$$10^n+1 = 2013 +317m$$is equivalent to the congruence$$10^n \equiv 2012 \equiv 110 \pmod{317}$$It is easy to see that this has solutions$$n = 76, 155, 234, ...

3

I'm not entirely sure why one would ask such a question, but I suspect it might be pretty hard to solve it completely (or maybe not!). On the other hand, it is not difficult to show, for example, that there is only the one known solution with, say, $n > m^2$, by noting that $n!$ divides the left-hand-side and that each prime between $n/2$ and $n$ divides ...

3

As your intuition might tell, $$2^{3^{4^{5^{\cdots^{1000^{1001^1}}}}}}$$ will give the largest result. First of all note that setting $a_i=1$ for $i<1001$ can never give the maximal value. Replacing $1$ to the end of the sequence will give a number larger then the one you had before. So let's just ignore the $1$ from now on. Now for a proof, we need the ...

3

I am afraid it does depend! Take $b=2$ and $x=8$, in which case $x=1\cdot 2^3+0\cdot 2^2+0\cdot 2+0\cdot 1$, i.e., $a_0=a_1=a_2=0$ and $a_3=1$. Clearly the condition $a_3\ge b/4$ is satisfied since $a_3=1> b/4=1/2$ and $\lfloor\sqrt{x}\rfloor=2$. On the other hand, if $a_0$ becomes equal to $1$, then $x$ becomes $9$ and $\lfloor\sqrt{9}\rfloor=3$. ...

3

Yes, it really is a prime. It took about a week to get it checked by Primo on my spare, not-exactly-cutting-edge machine. You can confirm its primality certificate by Primo yourself; verification is considerably faster than searching for the certificate in the first place.

3

What you are trying to prove is false, I'm afraid, since $$7^3 \mid 10^{147}+1,$$ and there are many other examples, such as: $$13^3\mid 10^{507}+1.$$ It is quite easy to find these things even with a limited computer using software such as pari/gp, which is free and is designed for number-theoretic investigations.

2

Your property is $(2)$ in the list below of properties of domains that are equivalent to uniqueness of factorizations into atoms (irreducibles). Nonunits satisfying $(2)$ are called primal. One easily checks that atoms are primal $\iff$ prime. Products of primes are also primal. So "primal" may be viewed as a generalization of the notion "prime" $(10)$ from ...

2

I can help you anwer the first question: This has something to do with solutions to Pell's equation. I let mathematica solve it, and it gave the following solution: $$\frac{1}{4} \left(\sqrt{2} \left(3+2 \sqrt{2}\right)^{c_1}-\left(3+2 \sqrt{2}\right)^{c_1}-\sqrt{2} \left(3-2 \sqrt{2}\right)^{c_1}-\left(3-2 \sqrt{2}\right)^{c_1}-2\right)$$ where $c_1$ is ...

2

Yes. Let $x$ be the solution to $x+\arctan(x) = \pi$, then $$\arctan(x)=\pi-x \\ \Rightarrow x=tan(\pi-x) \\ \Rightarrow x=-tan(x).$$ Thus if $x$ would be rational, also $tan(x)$ would be rational. This is impossible: You can use the statement you gave for showing that $x+arctan(x)=1$ is irrational. Here is another reference: Prove that if $x$ is a ...

2

Another way is to specialize the ascent in the ternary tree of Pythagorean triples (see below). Specializing $\,z = y+3\,$ in the $\rm\color{#c00}{\,formula\,}$below yields the following generation rule $$x^2 + y^2 = (y+3)^2\ \Rightarrow\ (6-x)^2 + Y^2 = (Y+3)^2,\quad Y = y-2x+6$$ This yields $\ (-3,0,3)\to (-9, 12, 15)\to (-15, 36, 39)\to (-21, 72, ... 2 The generalized pentagonal numbers begin, in order,$1, 2, 5, 7, 12, 15, ...$. The recurrence for the partition number is $$p(n) = p(n-1) + p(n-2) -p(n-5) -p(n-7) + p(n-12)+p(n-15) - \cdots$$ with the sign alternating in pairs. For any given$n$, only a finite number of terms are non-zero, since$p(m)=0$if$m<0$. For example,$$p(2) = ... 2 For your follow-up question, only linear polynomials don't work : Suppose$G$is a subgroup of$S_n$such that$G$acts transitively on$\{1\ldots n\}$, and let$H_i^j = \{\sigma \in G \mid \sigma(i)=j \}$. If$\tau(i)=j$then$H_i^k = H_j^k \tau$, and$H_k^j = \tau H_k^i$. Since$G$is transitive, every$H_i^j$has the same cardinal. Since every element ... 2 The proof in T. Bongers answer is perhaps the most direct way to proceed. There is however another viewpoint that it is insightful, since it relates to our prior intuition about the number of roots a polynomial may have. Recall that if$x=1$is a root of a polynomial$f(x)$then, by the Factor Theorem, we infer$f(x) = (x-1) g(x)$. Now suppose that$x = r ...

2

Somehow it turns out to not be reasonable to try to prove this by direct substitution and manipulation. Proofs were known in the 19th century (see also things about "Dedekind's $\eta$-function", which is the $24$-th root of Ramanujan's $\Delta$). Succinct, but subtle, proofs were given by Siegel and Weil: these are recalled, with citations to the original ...

2

Here is a collection of links for this question (a bit too long for the comment field). These numbers are studied since Euler in $1769$, who conjectured that for all integers $n$ and $k$ greater than $1$, if the sum of $n$ $k$-th powers of positive integers is itself a $k$-th power, then $n$ is greater than or equal to $k$. There are several discussions on ...

2

This can even be extended to higher powers: Edward Waring, says that for every natural number $n$ there exists an associated positive integer s such that every natural number is the sum of at most $s$ $k$th powers of natural numbers (for example, every number is the sum of at most 4 squares, or 9 cubes, or 19 fourth powers, etc.) from here.

2

Just for completion, here is an elementary proof using modular arithmetic, which I wrote up previously for a number theory course. Claim For any $n \in \mathbf{N}$ there exists $w,x,y,z \in \mathbf{Z}$ such that $n=w^2+x^2+y^2+z^2$. Proof: First, we observe that sums of four squares are closed under multiplication. That is given $a,b,c,d,e,f,g,h$ integers, ...

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