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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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