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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 $2a=km$. ...

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 ...

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 ...

2

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

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: $$a(b+c) + b(a + c) + c(a + b) = 2(bc + ac + ab)$$

2

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

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

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

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 ...

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 $0 \in A$.

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 ...

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 ...

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