For (b) just do https://artofproblemsolving.com/community/c6h530379p3026714

(a) $n=k^2+1$ is always silesian. Just take \begin{align*} a&=k^4+k^3+3k^2+2k+1 \\ b&=k^2+1 \\ c&=k+1 \end{align*}

For $(b)$; any $n=6k+3$ cannot be expressed in this form. To see this, observe that, $$ 6k+3 = \frac{a^2+b^2+c^2}{ab+bc+ca}\iff (a+b+c)^2 = (6k+5)(ab+bc+ca). $$Let $q \mid 6k+5$ be a prime number, such that $q\equiv 2\pmod{3}$, and the exponent of $q$ in $6k+5$ is odd (such a prime must exist, since the sum of exponents of primes of form $3\ell +2$, dividing $6k+5$ is odd). Then, since $q\mid a+b+c$; we deduce that the exponent of $q$, on the right hand side, is even. In particular, $q\mid ab+bc+ca$. Now, using $a+b+c\equiv 0 \pmod{q}\iff c\equiv -(a+b)\pmod{q}$, we get $ab+bc+ca \equiv -(a+b)^2+ab \pmod{q}$. In particular, $$ q\mid ab+bc+ca \iff q\mid a^2-ab+b^2 \iff (2a-b)^2+3b^2 \equiv 0\pmod{q}. $$Hence, either $q\mid a,b$; or $-3$ is a quadratic residue in modulo $q$. By dividing both sides with $q$ as many times as necessary, we may suppose $q\nmid a,b,c$; which will yield $\left(\frac{-3}{q}\right)=1$, contradicting with $q\equiv 2\pmod{3}$. We are done.

Sorry for reviving the topic, but how did #3 obtain these values? Randomly, or by using some methods?

I can't find a way to do (a), but (b) is easy: We can assume $gcd(a,b,c)=1$, consider $n+2=p>2$ is a prime, now we begin to search the condition for $p$. Clearly $(a+b+c)^2=p(ab+bc+ca)$, then $p \mid a+b+c$, $p \mid ab+bc+ca$, $$0=ab+bc+ca=a(b+c)+bc=-(b+c)^2+bc(modp),$$then $$(2b+c)^2+3c^2=0(modp),$$if $p \mid c$, then $p \mid b$, and $p \mid a$, a contradiction. So $(\frac{-3}{p})=1$, by Gauss' quadratic reciprocity law, we know $p =1(mod3)$. So, at the begining, we can select $p=-1(mod3)$ to get a contradiction. We did it.

DerJan wrote: (a) $n=k^2+1$ is always silesian. Just take \begin{align*} a&=k^4+k^3+3k^2+2k+1 \\ b&=k^2+1 \\ c&=k+1 \end{align*} How do you find it, it's very hard.

(b)$n=p-2$ $p \equiv -1(mod3)$ $$(ab+bc+ca)p=(a+b+c)^2$$$$p|a+b+c$$$$p|ab+bc+ca$$$$p|a^2+b^2+ab$$Exist $x$ $$x^2 \equiv -3 (modp)$$contradiction

creativehand09 wrote: Sorry for reviving the topic, but how did #3 obtain these values? Randomly, or by using some methods? WypHxr wrote: How do you find it, it's very hard. One very natural way to come up with one parametrisation, although slightly different, is the following: A natural way to look for solutions is to fix $b,c$ and then look for solutions in $a$. This means that we need to find $a$ with \[ab+bc+ca \mid a^2+b^2+c^2.\]Now, the LHS is linear in $a$ and the RHS is quadratic so we can try to reduce it to something independent of $a$, which will mean that we can check finitely many cases for $a$ (if $b,c$ are fixed). Doing this is straightforward, we find that \[ab+bc+ca \mid (b+c)(a^2+b^2+c^2)-a(ab+bc+ca)=(b+c)(b^2+c^2)-abc\]and then \[ab+bc+ca \mid (b+c)^2(b^2+c^2)-abc(b+c)+bc(ab+bc+ca)=(b+c)^2(b^2+c^2)+b^2c^2=(b^2+bc+c^2)^2.\]Moreover, we have essentially down equivalence manipulations, at least when $a,b,c$ are coprime. So nothing is lost. This reformulation is nice to check for solutions, but it also helps to actually solve the problem. We just need to find divisors of the RHS that are $\equiv bc \equiv -c^2 \pmod{b+c}$. But the RHS is $\equiv b^2c^2 \equiv c^4 \pmod{b+c}$. So hey, let's just choose $c^2 \equiv -1 \pmod{b+c}$. Then we can find $a$ with $ab+bc+ca=(b^2+bc+c^2)^2$ and will get a solution. So we can just take $b+c=c^2+1$ i.e. $\boxed{b=c^2-c+1}$ and then a short computation shows that $\boxed{a=c^6-2c^5+6c^4-6c^3+7c^2-3c+1}$ and then $\boxed{n=c^4-2c^3+5c^2-4c+2}$ is Silesian. Done. (Note that we even don't have to really do the computations, since it is clear that we get some polynomial expression for $n$ of degree $4$, which already suffices to solve the problem!)

Sorry to revive this thread once more, but I think I found solutions to both parts worth showing, especially since they differ from the solutions above. For (a): We look at the expression $n=\frac{a^2+b^2+c^2}{ab+bc+ca}$ and want to find integer values for $a,b,c$ that result in an integer value of $n$. This is not so easy as it is difficult to control this divisibility in terms of $a,b,c$. What we would like to have is a easier denominator we can control. Now, the magic begins. What would be really nice is if it just was $1$ - then the divisibility clearly holds. So, we want to choose $c$ such that $$n=\frac{a^2+b^2+c^2}{c(a+b)+ab}=\frac{a^2+b^2+c^2}{1},$$in other words, $c(a+b)+ab=1$. This is rather easy: Let's write $ab=k(a+b)+1$ where $k$ is an arbitrary positive integer. We want $a,b$ so that this equation holds, so by SFFT we find $(a-k)(b-k)=k^2+1$. As we really may choose whatever we like to, let's just say $a=k+1$ and $b=k^2+k+1$. So now, $ab=k(a+b)+1$ clearly holds. The trick now is to set $c=-k$. This is not allowed as $c$ must be positive, but if we follow this indeed $c(a+b)+ab=-k(a+b)+k(a+b)+1=1$, so $n=a^2+b^2+c^2=(k+1)^2+(k^2+k+1)^2+(-k)^2$ is indeed a positive integer solution. It is easy to check that $n$ is strictly increasing in terms of $k$, so we find infinitely many values. The only problem left is that $c$ is not positive by now. So, we want to manipulate $c$ so that it indeed is positive. But the initial condition is equivalent to $n(ab+bc+ca)=a^2+b^2+c^2$ or $c^2-c(a+b)n+a^2+b^2-nab=0$. This is a quadratic equation in $c$, so we can try Vieta Jumping! Indeed, with the $c$ found, $c^*=(a+b)n-c$ is also a solution for the same positive $n,a,b$. But $c^*$ is obviously positive! So we indeed found a solution $(a,b,c)$ for infinitely many $n$ - all these are silesian. For (b): We look modulo $4$. The quadratic residues are $0,1$, so the numerator is precisely divisible by $4$ iff all of $a,b,c$ are divisible by two. But as the expression is homogenous of degree $0$, then $(\frac{a}{2},\frac{b}{2},\frac{c}{2})$ would also be a solution for the same $n$. We find a contradiction by infinite descent. Therefore, $n$ cannot be divisible by $4$, so that e.g. $n=4$ is not silesian.