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16

If $n$ is not prime, then at least one of the factors of $n$ is at most as large as $\sqrt n$. To see why, let's suppose not. Since $n$ is not prime, $n = ab$ for some $a,b \neq 1$. If both $a$ and $b$ are larger than $\sqrt n$, then $a\cdot b > \sqrt n \cdot \sqrt n = n$. This clearly cannot be! So you only need to check for factors up to $\sqrt n$.


11

$$15x^2+19x\equiv 5 \mod 11$$ is equivalent to $$15x^2+19x +6 \equiv 0 \mod 11$$ by the fact that we can subtract $5$ from both sides of the congruence (and $-5\equiv 6 \mod 11$). From here, we factor $15x^2+19x+6$ to give $(3x+2)(5x+3)$. This means that our original quadratic congruence is equivalent to $$(3x+2)(5x+3)\equiv 0 \mod 11.$$ This will ...


10

Write your equation as $m^2 - m = n^5 - n$. You want $m = (1 + \sqrt{4 n^5 - 4 n + 1})/2$ to be an integer. Trying the first $10^6$ values of $n$, we find that $4n^5 - 4n + 1$ is a square for $n = 1, 2, 3, 30$, corresponding to $m = 1, 6, 16, 4930$. The curve $x^2 - x - y^5 + y$ has genus $2$ (according to Maple), so by Faltings's theorem there are only ...


9

John, in cyclotomic extensions of $\mathbf Q$ the Galois group acts like a power map only on the roots of unity, not on general elements, so don't think about the Galois group as "exponential." For comparison, complex conjugation is an element of ${\rm Gal}(\mathbf C/\mathbf R)$ and it acts on $i$ by sending it to $-i$, but that doesn't mean the Galois ...


8

Choose any positive integer $a > n$ such that $\gcd(a,n)=1$ Since $d|n \implies \gcd(a,d)=\gcd(a,n)=1$ By Euler's Theorem, $a^{\phi (n)} \equiv 1 \pmod n$ $\implies n|(a^{\phi (n)}-1)$ $\implies d|(a^{\phi (n)}-1)$ $\implies a^{\phi (n)} \equiv 1 \pmod d $ Again, by Euler's Theorem, since $\gcd(a,d)=1$, $a^{\phi (d)} ...


7

I'd say you're on a good track, but more should be said. As you noted the sum of all elements of $A$ is $55$. You should argue (using the $55$ fact) that for any subset $B$ of $A$, exactly one of $B$ and $B^c$ (complement of $B$) have a sum greater than or equal to $28$. Then pairing up subsets of $A$ by complements, we see that exactly half the subsets ...


6

HINT: I expect that you know that a number is rational if and only if its decimal expansion is eventually periodic. Suppose that the expansion eventually repeats with period $p$. Show that there are two consecutive powers of $2$ whose after the initial aperiodic segment whose lengths (when written in the usual way in base ten) are the same multiple of $p$. ...


6

Assume $x=3^{100} a$ to balance all those large powers of 3: $$ y^7 = 2\cdot 3^{700} \left( 7a^6 + 35a^4 + 21a^2 + 1 \right) $$ Due to the common factor, $y$ should be of the form $y=2\cdot 3^{100} b$: \begin{align} 64 b^7 &= 7a^6 + 35a^4 + 21a^2 + 1 \\ &= \sum_{k=0}^3 \binom7{2k} a^{2k} \\ &= \frac{(1+a)^7 + (1-a)^7}2 \\ (2b)^7 &= (1+a)^7 ...


5

Using the characterisation of these triples, it suffices to show that any such number can be written as $m^2-n^2$, $2mn$ or $m^2+n^2$ with some numbers $m>n$. The case $m^2-n^2$ covers "the most" numbers (only those $\equiv 2 \mod 4$ remain), the rest is covered by $2mn$.


5

Oftentimes in mathematics we find that by expanding our universe we can find out more about a subject that has its roots in something simpler. Look at polynomials over the real numbers, some of them don't have roots, but all of them have roots in $\Bbb C$, the complex numbers. Quadratic fields grew out of the early study of quadratic forms in two variables, ...


5

Since $\sigma(n)>0$ we have $d<2N$, together with $(2/3)N^2<d$ we get $(2/3)N^2<2N$ which is verified only for $0< N< 3$, which leaves only 2 cases to be checked, namely $1$ and $2$. edit:After the question being asked has been changed the first part of my answer is irrelevant. I can't completely answer the new question, but I can show ...


5

$(1+\sqrt{71})^{71} + (\sqrt{71}-1)^{71}$ is not an integer. However, if you meant $(1+\sqrt{71})^{71} + (1-\sqrt{71})^{71}$, then it is indeed an integer. If $x_n = (1+\sqrt{71})^n + (1-\sqrt{71})^n$, we then have $x_n$ to satisfy the recurrence $$x_{n+1} = 2x_n + 70x_{n-1}$$ where $x_0 = 2$ and $x_1=2$. Hence, we see that $$x_{n+1} = 2x_n \pmod{10}$$ ...


4

I think you refer to a theorem of Kummer: Fermat's Last Theorem is true for an odd prime $p$ if and only if $p$ doe not divide the class number of the cyclotomic extension $\mathbf Q(\zeta_p)$, i. e. the order the the group of fractionary ideals of this field modulo principal ideals. Such a prime number is called a regular prime. Kummer criterion: An odd ...


4

I'm not sure which level to write this at. Here is a completely elementary proof: As you say, if $p \equiv 1$, $3$, $7$ or $9 \bmod 20$, then $\left( \frac{-5}{p} \right) = 1$. Let $p$ be in one of these residue classes and let $a$ be a square root of $-5$ modulo $p$. Let $\Lambda \subset \mathbb{Z}^2$ be the lattice of pairs $(x,y)$ such that $x \equiv ay ...


4

HINT Note that $m^3 \equiv 0, \pm1 \pmod9$.


4

Here's how I would start (don't know if I will finish): If $m^2-n^5 = m - n$, then $m^2-m = n^5-n$ or $m(m-1) =n(n^4-1) =n(n^2-1)(n^2+1) =n(n-1)(n+1)(n^2+1) $. The RHS needs to factor into two terms that differ by $1$. This seems important, but I don't know how to make use of this. Also, multiplying by 4 and adding $1$ (to make the left side a square), ...


4

Note that if $n$ is composite, then $n$ has a non-trivial factor $i$ such that $i \le \sqrt{n}$. Otherwise, any nontrivial factorization $n = ab$ would force $a, b > \sqrt{n}$, so that $n = ab > \sqrt{n}\times \sqrt{n} = n$, a contradiction.


4

Hint: What does 'primitive' mean? So the statement we are trying to prove here is if $(x,y,z)$ is a primitive Pytagorian triple, that is to say $hcf(x,y,z)=1$ and $x^2+y^2=z^2$, then $5$ divides at most one of them. If 5 divides all three of them then 5 divides the highest common factor of $x,y$ and $z$ and this is a contradiction. So suppose 5 divides ...


3

The proof is by infinite descent. Let $n,a,b,c$ be the smallest possible solution to $W(n,a,b,c) = 0$, where $n>3$. Since he is almost sure that there exists a polynomial, discoverable by computer, with positive coefficients such that: $$W(n,a,b,c) = P\left(W(n,a-1,b,c), W(n,a,b-1,c), \ldots W(n -1,a,b,c), \ldots\right)$$ for $n>3$. This would mean ...


3

Kummer's Theorem, a proof of which is given in this answer, in the case $p=2$ says If, when you add $m$ and $n$ in binary, there are no carries, then $\binom{m+n}{n}$ will be odd. Thus, the largest $n\le m$, will be the binary ones'-complement of $m$ (truncated to the size of $m$). For example, if $m=5=101_{\text{two}}$, then $n=10_{\text{two}}=2$. ...


3

Roth's Theorem will give you an upper bound: http://wiki.math.toronto.edu/TorontoMathWiki/images/2/2d/Expo_paper.pdf In that paper it is Theorem 1.3 (bottom of first page). It says: Theorem (Roth, 1953) For any $\delta>0$ if $k>\exp(\exp(c\delta^{-1}))$ (for some absolute constant $c>0$) and $A\subseteq \{1, 2, 3, ..., k\}$ and $|A|\geq \delta k$ ...


3

As for an upper bound, we get into work related to Erdos' conjecture on arithmetic progressions (which has not been proven even in the case of arithmetic progression of length 3). I believe the strongest known result is Sander's proof that, if $f(k)$ is the size of the largest subset of a $k$-element progression free from arithmetic progression of length ...


3

[Edit: I didn't read the post properly, so this answers a quite different though related question: what is the count of $(n_1,n_2)$ such that $n_1^2+n^2 \leq N$. Gerry Myerson's response answers the question that was actually asked.] This is a nice question that was first studied thoroughly by Gauss and goes by the name 'Gauss's circle problem.' We know ...


3

Assume you have the Pythagorean relation $u^2 + v^2 = c^2$ Then $$ \begin{align} (u^2 + v^2) + (u^2 + v^2) & = 2c^2\\ (u^2 + v^2 + 2uv) + (u^2 + v^2 - 2uv) & = 2c^2\\ (u + v)^2 + (u - v)^2 & = 2c^2\\ \end{align} $$ Thus if $a = u + v$ and $b = |u - v|$ $a^2 + b^2 = 2c^2$


3

HINT: Induction is much easier, but if you really must use Binet’s formula, note that your summations are geometric, with ratios $\alpha^2$ and $\beta^2$, respectively; you can use the familiar formula for the sum of a finite geometric series. You also know that $\alpha$ and $\beta$ satisfy the equation $x^2-x-1=0$, so $\alpha^2-1=\alpha$ and ...


3

We have $$ \sum_{n\le x}\tau(n)\sim \sum_{n\le x}\log(n), $$ so that the average value of $\tau(n)$ is indeed $\log(n)$, but from Hardy and Ramanujan we know, since $2^{\omega(n)}\le \tau(n)\le 2­^{\Omega(n)}$, that for most numbers $n$, $$ \tau(n)=\log(n)^{\log(2)+o(1)}, $$ where $\log (2)$ is around $0.693$. The normal value of $\omega (n)$ resp. ...


3

Given $x_1,x_2,\ldots,x_7$, set $x_i = \tan(t_i)$, where $t_i \in \left(-\dfrac{\pi}2, \dfrac{\pi}2\right)$. Divide $\left(-\dfrac{\pi}2, \dfrac{\pi}2\right)$ into $6$ equal intervals of length $\dfrac{\pi}6$. By PHP, we have two of the $t_i$'s to lie in the same interval, i.e., we have $$-\dfrac{\pi}6 < t_i - t_j < \dfrac{\pi}6$$ Now taking $\tan$ on ...


3

No induction necessary. The equation $x^2 -10x +22 = 0$ has the roots $5-\sqrt 3$ and $5+\sqrt 3$. We know that this function is positive outside the roots, which means on $(-\infty, 5-\sqrt 3) \cup (5+\sqrt 3, \infty)$. The first interval contains the natural numbers $0,1,2,3$, while the second one all the natural numbers starting from $7$ (use that $\sqrt ...


3

A number is rational if and only if it has an eventually repeating decimal expansion. So you should show that this decimal expansion does not have an eventually repeating expansion.



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