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2

Assume that $N$ is a prime larger than $3$ and not of the form $6n\pm1$: $N\equiv0\pmod6\implies N$ is divisible by $6\implies N$ is not a prime $N\equiv2\pmod6\implies N$ is divisible by $2\implies N$ is not a prime (or $N=2$) $N\equiv3\pmod6\implies N$ is divisible by $3\implies N$ is not a prime (or $N=3$) $N\equiv4\pmod6\implies N$ is divisible by $2$ ...


8

Can you divide $(6n+2)$ by anything? What about $(6n+3)$?


3

You have proved that $x^2 + x + 1 = 0 \implies x^3 = 1$. This is a nice fact, but it doesn't contradict the statement that the second equation has real solutions while the first one doesn't. One way to look at this is the factorization $x^3 - 1 = (x-1)(x^2 + x + 1)$. Clearly, the right-hand side vanishes when $x=1$, but that tells us nothing about $x^2 + ...


0

It is a problem of logic. "If it is raining then the sidewalk is wet", it does not follow that, "if the sidewalk is wet then it is raining". This is called fallacy of the converse. You showed that if $A(x)$ has a solution then $x=1$. This does not mean that $x=1$ must be a solution.


0

Your statement is not an equivalence but is just implication so the contradiction is because $A(1)=3\not=0$ and because $x\not\in(-\infty,-1]$


1

Your counterexample reveals a flawed implicit assumption in the proof. The contradiction the proof uses is based the fact that $A\subseteq\cup F$ and $A\subseteq\cup G$. It claims, based on this, that $\cup F$ and $\cup G$ are not disjoint. However, if $A=\emptyset$, then $A\subseteq S$ for any set $S$.


3

A good strategy is to to run through the proof with your counter-example. The issue is that while it is also true for $A = \emptyset$ that every element of $A$ is in both $\cup F$ and $\cup G$ this does not contradict the assertion that the sets are disjoint. Only if there is an element in $A$, there is a contradiction. Note that this also shows the the ...


0

Surely reaching a contradiction is acceptable, though it is usually not the most convenient path to follow. Anyway, here's a futile example. We have $1^2=0^2+2\cdot0+1$. Assume $(k+1)^2=k^2+2k+1$ but $(k+2)^2\ne k^2+4k+4.$ Then, $$\require{cancel} (k+2)^2-(k+1)^2\ne2k+3\\(\cancel{k}+2-\cancel{k}-1)(k+k+2+1)\ne2k+3\\ 2k+3\ne2k+3.$$ Hence, $(n+1)^2=n^2+2n+1$ ...


0

Check the options of $x^2$ and $3y^2$ $\text{mod }4$. $$\begin{array}{|cc|c|cc|} x & x^2 & & y & 3y^2\\ 0 & 0 & & 0 & 0 \\ 1 & 1 & & 1 & 3 \\ 2 & 0 & & 2 & 0 \\ 3 & 1 & & 3 & 3 \\ \end{array}$$ Therefore $x^2 + 3y^2 \equiv 0,1,\text{ or }3 \pmod{4}$ But $m \equiv 2 \pmod 4$. ...


1

If $x^2+3y^2=4a+2\iff(x+y)(x-y)=2(2a+1-2y)$ Now for integers $x,y; x\pm y$ have the same parity as $x+y-(x-y)$ is even If one is odd, the other & consequently the product must be odd If one is even, the other will be even & consequently the product must be divisible by $4$


0

We have $m\equiv2\pmod4$ Any integer $a\equiv0,1,2,3\pmod4\implies a^2\equiv0,1\pmod4$ then we can establish $x^2+3y^2\not\equiv2\pmod4$


4

Here is a proof that uses a bit of group theory. We show that, if $p$ is prime, then $2^p-1$ is divisible by a prime $q \gt p$. Suppose that $q$ divides $2^p - 1$. Consider the multipicative group $\mathbb Z_q \setminus \{0\}$. Since $q$ divides $2^p-1$, the element $2$ has order $p$ in $\mathbb Z_q \setminus \{0\}$. Since this group has order $q-1$, by ...


3

Let $F_n=2^{2^n}+1$ be the sequence of Fermat numbers. It is relatively easy to show that, for distinct $i,j$ we have $\gcd(F_i,F_j)=1$. Therefore, if we let $p_n$ be a sequence of primes such that $p_n$ divides $F_n$ (which is possible, since every number has at least one prime divisor), then each $p_n$ is distinct (since $p_n>1=\gcd(F_i,F_j)$), and ...


4

In the proof that is attributed to Euclid, what you are really doing is this: For any integer $n \geq 1$ you are producing a number, namely one plus the product of all primes less than or equal to $n$, whose prime factors have to be greater than the biggest prime less than $n$. Therefore, given the fundamental theorem of arithmetic, Euclid's proof can be ...


7

Actually, that proof can be phrased in a way to avoid contradiction. Let ${\cal P}=\{p_1,...,p_n\}$ be a non empty finite set of primes. Then let $a=1+\prod_{p\in\cal P}p$. By the usual argument there must be a prime $p\mid a$, $p\notin\cal P$. This proves that any finite set of primes cannot include all primes and so there must be infinitely many. EDIT: ...


6

The simplest direct proof I know is one which involves infinite series. One can show that $\sum\limits_{p\text{ prime}}\frac{1}{p}$ diverges. Particularly, it means that there cannot be finitely many primes since the sum of finitely many nonzero numbers is finite. (There is a very slight contradiction - or contrapositive - argument in here if you look ...


4

It is quite easy; suppose that there exist integers $a$ and $b$ such that $18a+6b=1$, then $6(3a+b)=1$, and that implies that $3a+b$ (which is an integer for any $a$ and $b$) is an multiplicative inverse of $6$. This is a contradiction, because the only integers with inverses are $1$ and $-1$.


3

Hint: $$18a+6b=6\cdot(3a+b)$$


0

Let $a=2n$ be even. $$4n^2-4b=3\implies n^2-b=\frac34\text{ !}$$ Let $a=2n+1$ be odd. $$4n^2+4n+1-4b=3\implies n^2+n-b=\frac12\text{ !}$$


3

$a^2=4b+3 \implies a^2\equiv 3\mod 4$ Contradiction!


1

Consider the remainder of $a^2 \pmod 4$. Which values can it take?


2

I think that Jorge Fernández' solution is the most effective. So, here we go for an overkill. Since $p(x)$ splits as: $$ (x^2+x-1)(x^3-x-1)$$ over $\mathbb{F}_3$, it has no rational root.


5

Assume you have a solution $x = \frac{p}{q}$ where $p$ and $q$ are relatively prime (so the fraction is reduced. Plug $x$ into the equation and then multiply both sides by $q^5$ to clear fractions. You get an equation that allows you conclude that one of $p$ or $q$ must divide the other, which you assumed was not the case, so you have a contradiction..


7

By the rational root theorem the only possible solutions would be $1$ and $-1$. Evaluate for both and conclude they are not roots.


0

I assume the equivalence relation $\sim$ is "there exists a bijective map between". Then neither does $A\sim X$ have anything to do with $a\sim x$ for $a\in A$, $x\in X$ (for example $\{\mathbb N\}$ and $\{\mathbb R\}$ are both one-element sests and yet theri only elements have different cardinalitty). Nor does the fact that some injective map $A\to X$ ...



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