Suppose that there exist numbers $ a,b,c,k$ such that $ a^2+b^2+c^2=3k(ab+bc+ca)$ .Then $ (a+b+c)^2=(3k+2)(ab+bc+ca)$ . Now we choose a prime $ p\equiv 2(\mod 3)$ such that $ p^{2m-1}||3n+2$ . Such a prime exist because $ 3n+2$ cannot be a perfect square . Now because the exponent of $ p$ at $ (a+b+c)^2$ will be an even number we have that $ p|ab+bc+ca$ and subtituting the relation that $ c\equiv -a-b(\mod p)$ we have that $ p|a^2+ab+b^2$ or $ p|(2a+b)^2+3b^2$ from which we conclude that $ -3$ is a quadratic residue mod p which is a contradiction since $ p\equiv 2(\mod 3)$

2.Solution Suppose that $ x,y,z$ and $ n$ are positive integers such that $ \frac {x^2 + y^2 + z^2}{3(xy + yz + zx)} = n$ $ \therefore x^2 + y^2 + z^2 = 3n(xy + yz + zx).$ Therefore $ (x + y + z)^2 = (3n + 2)(xy + yz + zx)$ Choose prime number $ p \equiv 2 (mod 3)$ which divides $ 3n + 2$ with an odd exponent such that $ p^{2i - 1} | 3n + 2$ and $ p^{2i} \nmid 3n + 2 \exists i \in \mathbb{N}$.Then $ p^i | x + y + z$ Therefore $ p | xy + yz + zx$. Substituting $ z \equiv - x - y (mod p)$ $ \therefore p | x^2 + xy + y^2 \therefore p | (2x + y)^2 + 3y^2$ $ \therefore (\frac { - 3}{p}) = 1$, Which is impossible because $ p \equiv 2 (mod 3)$.

Consider the following identity: \[ (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+xz)\] Now assume that $ 3(xy+yz+xz)|x^2+y^2+z^2$. This means $ x^2+y^2+z^2=3t(xy+yz+xz)$ for some positive integer $ t$. It follows that by combining the left side of the identity, \[ (x+y+z)^2=(3t+2)(xy+yz+xz)\] Now we will use congruences modulo $ 3(xy+yz+xz)$ (this will be assumed and not written every line). $ (x+y+z)^2 \equiv (3t+2)(xy+yz+xz) \equiv 2(xy+yz+xz)$ $ \frac{(x+y+z)^2}{xy+yz+xz} \equiv 2$ Note $ \frac{(x+y+z)^2}{xy+yz+xz}$ must be an integer because if $ xy+yz+xz|x^2+y^2+z^2$, then it must also divide $ (x+y+z)^2$ because of the original identity. As a result of the modular algebra above, for some integer $ k$, \[ \frac{(x+y+z)^2}{xy+yz+xz} = 3k(xy+yz+xz) +2\] Now we will divide both sides of the original identity by $ 3(xy+yz+xz)$. This gives \[ \frac{(x+y+z)^2}{3(xy+yz+xz)} = t +\frac23\] \[ \frac{3k(xy+yz+xz) +2}{3} = t+\frac23\] \[ k(xy+yz+xz) +\frac23 = t+\frac23\] \[ k(xy+yz+xz) = t\] Since $ x^2+y^2+z^2 = 3t(xy+yz+xz)$, our previous conclusion implies that $ x^2+y^2+z^2 = 3k(xy+yz+xz)^2$, which implies $ \frac{x^2+y^2+z^2}{(xy+yz+xz)^2} = 3k$, an integer. This is an absurdity because $ (xy+yz+xz)^2$ is strictly larger than $ x^2+y^2+z^2$, as shown below: \[ (xy+yz+xz)^2>x^2y^2+y^2z^2+z^2x^2+xy^2z>x^2+y^2+z^2+xy^2z>x^2+y^2+z^2\] A contradiction has been reached, so $ 3(xy+yz+xz)$ never divides $ x^2+y^2+z^2$.

- assume the statement false. than there exists such a positive integer $ k$ that $ x^2 + y^2 + z^2 = 3k(xy + yz + xz)$. the expression is equivalent to $ (x + y + z)^2 = (3k + 2)(xy + yz + xz)$. - since $ x^2 = 0,1 \mod 3$ for integer $ x$ we can conclude that $ 3k + 2$ is not a perfect square. which imples that there exists such a prime number $ p$ of a form $ p = 3l + 2$ (for a positive integer $ l$) that $ p$ devides $ xy + yz + zx$ (otherwise $ (3k + 2)(xy + yz + xz)$ can't be a perfect square). this also implies that $ p$ devides $ x + y + z$ - if $ p$ devides $ x$ than it has to devide $ yz$ and $ y + z$ implying $ p|y$ and $ p|z$. because $ x,y,z$ cannot be infinitely devisable by $ p$ we can assume without loss of generality that $ p$ doesn't devide any of $ x,y,z$ and is therefore relatively prime to each of them. - obviously $ x = - (y + z) \mod p$ which by putting this into $ xy + yz + zx = 0 \mod p$ leads to $ x^2 = yz \mod p$. simmilary we conclude $ y^2 = xz \mod p$ and $ z^2 = xy \mod p$. - now $ y^2 \cdot yz = x^2 \cdot xz \mod p$ and since $ p,z$ are relatively prime we conclude that $ x^3 = y^3 \mod p$ - if that is true than also $ x^{3l} = y^{3l} \mod p$. on the other hand since $ x,y$ are relatively prime to $ p$ ,${ x^{p - 1} = y^{p - 1} = 1 \mod p}$ or in other words $ x^{3l + 1} = y^{3l + 1} \mod p$ which combined with the former implies $ x = y \mod p$. - using the same method as above but for a different pair of $ x,y,z$ we obtain $ x = y = z \mod p$. than $ x + y + z = 3x = 0 \mod p$ which implies $ x = 0 \mod p$ which is in contradiction with our assumption that $ x$ is relatively prime to $ p$ - our starting assumption must have been wrong, therefore such a $ k$ does not exist.

suppose that there are $ x,y,z$ that satisfy $ 3(xy + yz + zx)$ divides $ x^2 + y^2 + z^2$. let $ x^2 + y^2 + z^2 = 3k(xy + yz + zx)$. if $ x, y, z$ are all even numbers, divides them all with 2. then this case moves to other cases. if two numbers of them are even, $ x^2 + y^2 + z^2$ is odd and $ xy + yz + zx$ is even so contradiction. in the other hand, $ (x + y + z)^2 = (3k + 2)(xy + yz + zx)$, so there exists a prime number $ p$ so that is $ p \equiv 2(mod3)$, and the order of $ 3k + 2$ for $ p$ is an odd number.. if $ p$ only can be $ 2$, then $ 3k + 2 = 2^{(2l - 1)}$ for some $ l$. then $ 2$ divides $ xy + yz + zx$ because $ (x + y + z)^2$ is a square. but $ xy + yz + zx \equiv 1(mod2)$ so contradiction. so there is $ p$ that is not $ 2$. also, because $ (x + y + z)^2$ is a square number, $ p$ also divides $ xy + yz + zx$. (the order of $ (x + y + z)^2$ for $ p$ is an even number) meanwhile, because $ p$ divides $ x + y + z$, $ x \equiv - y - z(mod p)$, so $ x^2 \equiv y^2 + z^2 + 2yz(mod p)$. therefore $ x^2 + y^2 + z^2 \equiv 2(y^2 + z^2 + yz) \equiv 0(mod p)$, so, $ 4y^2 + 4yz + z^2 \equiv - 3z^2(mod p)$. this means $ - 3$ is quadratic residues of $ p$.(because $ 4y^2 + 4yz + z^2 = (2y + z)^2$). but because $ p \equiv 2(mod 3)$, this is contradiction. this means there are no $ x,y,z$ that satisfy $ 3(xy + yz + zx)$ divides $ x^2 + y^2 + z^2$.

Sol: Lemma 1: If $ x,y,z$ are sides of one triangle.Then: $ x^2 + y^2 + z^2 < 2(xy + yz + zx)$. dem: If $ 2P$: perimeter of the triangle $ ABC$ whit sides $ x,y,z$. Then: $ x = (P - y) + (P - z)$, $ y = (P - z) + (p - x)$, $ z = (P - x) + (P - y)$, If. $ a = P - x$,$ b = P - y$,$ c = P - z$. then : $ x^2 + y^2 + z^2 = 2(a^2 + b^2 + c^2 + ab + bc + ca)$ , $ 2(xy + yz + zx) = 2(a^2 + ab + bc + ca + b^2 + bc + ba + ca + c^2 + ac + ab + cb) = 2(a^2 + b^2 + c^2 + ab + bc + ca) + 4(ab + bc + ca)$$ Q.E.D$ Lemma 2: If $ x,y,z \in R^{ + }$ Then $ x^2 + y^2 + z^2 > = xy + yz + zx$.dem: $ (x - y)^2 + (y - z)^2 + (z - x)^2>=0$ where equality is given if only if $ x = y = z$ . $ Q.E.D$ Now: If $ K = (x^2 + y^2 + z^2)/(xy + yz + zx)$.Where $ x,y,z$ /$ K \in Z^{ + }$, WLG $ x > = y > = z$. $ I$ $ case$: $ x < y + z$ Then for lemma 1, $ K < 2$, But, for lemma 2. $ K > = 1$. how $ K$ is integer , $ K = 1$, then $ x = y = z$, $ II$$ case$: $ x > = y + z$, give the ecuation in $ x$. $ x^2 - k(y + z)x + y^2 + z^2 - kyz = 0$ if $ x_0$ the other solution.For vietta. $ x + x_0 = k(y + z)$, $ x.x_0 = y^2 + z^2 - kyz$ ,Suppose that $ x + y + z$ is minimum. $ a)$ $ x_0 > = 0$. Then: $ k(y + z) = x + x_0 > = 2x > = 2(y + z)$,-> $ k > = 2$..(1) $ y^2 + z^2 - kyz = x.x_0 > = x^2 > = (y + z)^2 = y^2 + z^2 + 2yz$ -> $ k < = 2$..(2). of (1) and (2). K=2. $ b)$ $ x_0 < 0$.

Suppose that a integer triple $ (a,b,c)$ satisfy the condition $ a^{2} + b^{2} + c^{2} = k(ab + bc + ca)$ when $ k$ is multiple of 3. Then $ (a + b + c)^{2} = (k + 2)(ab + bc + ca)$. (We can assume $ a,b,c$ are pairwise coprime. If $ p|a,b$ for prime $ p$, $ p|ab + bc + ca \Rightarrow p|a + b + c \Rightarrow p|a,b,c$. Now we can replace $ (a,b,c)$ into $ (\frac {a}{p},\frac {b}{p},\frac {c}{p})$.) $ k + 2\equiv 2\pmod 3$. So it must has a prime factor in the form of $ 3t + 2$ which appears odd times in the prime factorization of $ k + 2$. If there is a odd prime factor of $ k + 2$, $ p$ such that $ p\equiv 2\pmod 3$, $ p|(a + b + c)^{2} \Rightarrow p|a + b + c$. The exponent of $ p$ in $ (a + b + c)^{2} = (k + 2)(ab + bc + ca)$ is even. Since that of $ k + 2$ is odd, $ p|ab + bc + ca$. Applying $ p|a + b + c$, we get $ 0\equiv ab + bc + ca \equiv - (b + c)(b + c) + bc = - b^{2} - bc - c^{2}\pmod p$. Hence, $ (2b + c)^{2}\equiv - 3c^{2}\pmod p$. $ 2b + c, c$ can't be a $ 0\pmod p$. ($ (b,c) = 1$) Therefore $ \left( \frac { - 3}{p}\right) = 1$. But $ \left(\frac { - 3}{p}\right) = \left(\frac { - 1}{p}\right)\left(\frac {3}{p}\right) = ( - 1)^{\frac {p - 1}{2}}( - 1)^{\frac {p - 1}{2}\frac {3 - 1}{2}}\left(\frac {p}{3}\right) = - 1$. Contradiction! If exponent of $ 2$ in $ k + 2$ is odd, by the same way, $ a + b + c$ and $ ab + bc + ca$ are even. Both to be a even, $ a,b,c$ all should be a even. Contradiction! Therefore $ 3(xy + yz + zx) \not| x^{2} + y^{2} + z^{2}$.