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8

Suppose to the contrary that $p^2-4qr=x^2$, where $x$ is an integer. Then $p^2-x^2=4qr$. It is clear that $x$ must be odd. So $p^2\equiv 1\pmod{8}$ and $x^2\equiv 1\pmod{8}$, and therefore $p^2-x^2$ is divisible by $8$. But $4qr$ is not divisible by $8$.


6

$$a^{\sqrt{b}}=\sqrt{a^b}$$ Squaring both sides, $$a^{2\sqrt{b}}=a^b$$ Case 1: $a=1$. It holds regardless of $b$. ($100$ cases) Case 2: If $a \neq 1$, $2\sqrt{b}=b \implies b=4$ ($99$ cases)


4

Note that $\sqrt{a^b}=(a^b)^{1/2}=a^{b/2}$. If $a>1$, then $a^{\sqrt{b}}=a^{b/2}$ iff $\sqrt{b}=b/2$, which happens only for $b=4$. On the other hand, if $a=1$, then $b$ can be anything and both sides will be $1$. So the solutions are $(1,b)$ for any $b$ and $(a,4)$ for any $a$. There are $100$ solutions in each of these cases, but both cases include $...


4

Let $A=(0,1)$, $B=(1,1)$, and $C=(-x,x)$ as in the picture. Then $$AC=\sqrt{x^2+(1-x)^2}, BC=\sqrt{(1+x)^2+(1-x)^2}.$$ Let $D$ be symmetric to $A$ about $x+y=0$ (trajectory of $C$). Apparently $$AC+BC\ge BD=AE+BE.$$ Minimal is attained at $C=E$. I leave you figure the coordinates of $E$.


4

Randomly colour the members of the set black and white, independently with probabilities $1/2$ and $1/2$. The probability that any given $18$-term a.p. in the set is monochromatic is $2^{-17}$. There are $117587$ such a.p.'s, and this is less than $2^{17}$. Thus the expected number of monochromatic a.p.'s is less than one, which means that it must be ...


4

This is equivalent to proving that for odd $m$, if $4|n^2-m^2$ then $8|n^2-m^2$. This is easy if we notice $n^2-m^2=(n+m)(n-m)$, and if one of them is even, both are even, and also, one is a multiple of $4$.


3

And now a strange solution. Assuming that $p^2-4qr$ is a square, the polynomial $$s(x)= qx^2 + px + r $$ has to be reducible over $\mathbb{Q}$. However, $x^2+x+1$ is irreducible over $\mathbb{F}_2$, hence that cannot happen.


3

A little culture. If $p^2 - 4 q r$ were a perfect square, then we would be able to factor $$ qx^2 + p x + r $$ over the integers as $$ qx^2 + p x + r = ( q_1x + r_1) (q_2 x + r_2); $$ see Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable or Formula for factorization of a Quadratic Equation? However, $$ ( q_1x + r_1) (q_2 x + r_2) = q_1 q_2 x^...


2

If $p,q,r$ are odd then $p^2-4qr$ is odd. So if $p^2-4qr=s^2$ then $s$ must be odd. But $4qr=(p-s)(p+s)$ The LHS is divisible by $4$ and not $8$, the RHS is divisible by $8$ because one of $p+s$ or $p-s$ is divisible by $2$ and the other by $4$.


2

If it's square, $ x^2\! + p x\! +\! qr\,$ has odd integer roots (factors of odd $qr),\,$ contra root sum $ {-}p\,$ is odd


2

Using Minkowski Inequality $$\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\geq \sqrt{(a+c)^2+(b+d)^2}$$ and equality hold when $\displaystyle \frac{a}{b} = \frac{c}{d}$ So $$\sqrt{x^2+(1-x)^2}+\sqrt{(1-x)^2+(1+x)^2}\geq \sqrt{[x+(1-x)]^2+[(1-x)+(1+x)]^2}=\sqrt{5}$$ and Equality hold when $$\frac{x}{1-x}=\frac{1-x}{1+x}\Rightarrow x^2-2x+1=x^2+x\Rightarrow x=\frac{1}{3}$$...


1

A Sketch of Proof Let $W$ be the space of polynomials over $\mathbb{C}$ in variable $X$ of degree at most $8$. Define an inner product $\langle\_,\_\rangle$ on $W$ via $$\langle AB\rangle:=V\left(A\bar{B}\right)\,,$$ where $\bar{B}$ is the complex conjugate of $B$. Prove that $\langle\_,\_\rangle$ is indeed an inner product. Now, $W$ has a basis ...


1

I try a solution. Let $E=\mathbb{R}_{8}[x]$, and define on $E$ $$<A,B>=\sum_{k=1}^{16}A(\alpha_k)B(\alpha_k)$$ We see easily that $<.,.>$ is a scalar product on $E$. Let $\{x^j, j=0,\cdots, 8\}$ the standard basis of $E$, and $A_j$, $j=0,\cdots,8$, the orthonormal basis deduced from this standard basis by Gram-Schmidt. Then the usual ...


1

Write $f(x)=g(x)+h(x)$. Now $g(x)$ has minima at $x=1/2$ and $h(x)$ is increasing function with minima at $x=0$.Then $f(x)$ has minima at ?


1

Let's use this inequality: for two positive numbers $a$ and $b$, we have $\sqrt{a} + \sqrt {b} \geq 2 (ab)^{1/4}.$ Say $a= \sqrt{x^2 +(1-x)}$ and $b = \sqrt{(1-x)^2+(1+x)^2}$ and then we can compute the maximum of $(ab)^{1/4}$ easily, which is the minimum of $f.$


1

A posible way: Let $A,B,C$ a triangle and $AD$ the height. The problem is $AD=1-x$, $BD=x$, $CD=1+x$ and you want minimize $BA+AC$, but note that the area is fixed ($(1-x)(2x+1)/2$), then this sum is minime if the triangle is right in A.


1

$p^2$ is congruent to $1$ mod $8$ (it is equal to $8k+1$ for some $k$) but $4qr$ is congruent to $4$ mod $8$ (it is $8m+4$ for some $m$). Therefore the difference of those two expressions is $(1 - 4) = 5$ mod $8$ which is not the value of any perfect square mod $8$.


1

Well, the first problem is the circle is an uncountable set, so the usual way of defining infinite sums with sequences and partial sums doesn't really work. There's a way to extend sums to uncountable sets, but the sum will diverge unless all but a countable subset of the terms are zero. There are also alternative summation methods which are interesting ...


1

For olympiads Number Theory this is a must-"Number Theory-Andrescu Titu"-https://blngcc.files.wordpress.com/2008/11/andreescu-andrica-problems-on-number-theory.pdf And for Geometry this one-"Coxeter-Geometry Revisited"-http://www.aproged.pt/biblioteca/geometryrevisited_coxetergreitzer.pdf I have another good book,but I have no idea if it's available ...


1

Problem solving strategy by Engel. Pretty advanced though


1

At the beginning we transform the original inequality: $$\dfrac{8x^4}{8x^3+5y^3}+\dfrac{8y^4}{8y^3+5z^3}+\dfrac{8z^4}{8z^3+5x^3}\geq \dfrac8{13}(x+y+z),$$ $$x-\dfrac{5xy^3}{8x^3+5y^3}+y-\dfrac{5yz^3}{8y^3+5z^3}+z-\dfrac{5zx^3}{8z^3+5x^3}\geq \dfrac8{13}(x+y+z),$$ $$\dfrac{xy^3}{8x^3+5y^3}+\dfrac{yz^3}{8y^3+5z^3}+\dfrac{zx^3}{8z^3+5x^3}\leq \dfrac1{13}(x+y+z)....



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