One can prove that $\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}$ is never divisible by 3.

Assume $a,b,c$ do not have some common divisor and write \[ (2013k+2)(a^2+b^2+c^2) = 2013k \cdot (a+b+c)^2. \] Since $2013k + 2 \equiv 2 \pmod{3}$, there is a prime $p \equiv 2 \pmod{3}$ with $v_p(2013k+2)$ odd (in particular, $p \mid 2013k+2$). =============================== Claim: $p \neq 2$. Let us assume on the contrary that $v_2(2013k+2)$ is odd (in particular, $k$ is even). Remark that since $a,b,c$ are not all even, $v_2(a^2+b^2+c^2) \le 1$. Furthermore, \[ v_2(a+b+c) = 0 \iff v_2(a^2+b^2+c^2)=0. \] Now we consider two cases. Case 1: $v_2(k) \ge 2$. Then $v_2(2013k+2)=1$, $v_2(a^2+b^2+c^2) \le 1$. Therefore \[ v_2(k) + 2v_2(a+b+c) \ge 2 \ge v_2(2013k+2) + v_2(a^2+b^2+c^2) \] but equality cannot occur since the relations \[ v_2(a+b+c) = 0 \quad \text{and} \quad v_2(a^2+b^2+c^2) = 1 \] cannot hold simultaneously. Case 2: $v_2(k) = 1$. Then $v_2(2013k+2) > 1$ and is odd. Now \[ v_2(2013k+2) + v_2(a^2+b^2+c^2) = 1 + 2v_2(a+b+c). \] Considering mod 2, we see $v_2(a^2+b^2+c^2)$ must be even, so it is zero; consequently $v_2(a+b+c) = 0$ as well and we obtain $1 < v_2(2013k+2) = 1$. =============================== Now for the interesting part. Remark $p \mid a+b+c$ and $p \mid a^2+b^2+c^2$. Without loss of generality $b \not\equiv 0 \pmod{p}$, so that \[ a^2+b^2+(a+b)^2 \equiv 0 \pmod{p} \implies a^2+ab+b^2 \equiv 0 \pmod{p}. \] Then if $x = ab^{-1}$, we get that $x^2+x+1 \equiv 0 \pmod{p}$. But the LHS is the third cyclotomic polynomial, so either $p=3$ or $3 \mid p-1$, but neither is the case. Therefore, such a triple $(a,b,c)$ does not exist.

Let $a^2 + b^2 + c^2 = k(ab + ac + bc)$. Looking at this as a quadratic in $c$, we have $c^2 - c(ka + kb) + (a^2 + b^2 - kab) = 0$. Thus we can vieta-jump from a solution $(a,b,c)$ to $\left (a,b, \frac{a^2 + b^2 - kab}{c} \right )$. Thus we can WLOG $c \le \sqrt{a^2 + b^2}$ and $a \le b \le c$ (just take the minimum solution) unless $c = ka + kb$. We'll deal with that case in a bit. Now note that: \begin{eqnarray*} \frac{a^2 + b^2 + c^2}{ab + ac + bc} & \le & \frac{2a^2 + 2b^2}{ab + ac + bc} \\ & \le & \frac{2a^2 + 2b^2}{ab + a^2 + b^2} \\ & \le & 2 \end{eqnarray*} So obviously in this case $a^2 + b^2 + c^2$ is not divisible by $2013(ab + ac + bc)$. Now suppose $c = ka + kb$. We have: \begin{eqnarray*} \frac{a^2 + b^2 + c^2}{ab + ac + bc} & = & \frac{a^2 + b^2 + k^2(a+b)^2}{ab + k(a+b)^2} \\ & = & k + \frac{a^2 + b^2 - kab}{ab + k(a+b)^2} \\ & = & k \end{eqnarray*} due to $a^2 + b^2 - kab = 0$ as well. Thus it suffices to show $a^2 + b^2 = kab$ has no solutions for $k = 2013$. However, to do this just vieta jump again to get the minimum solution has $a=b$ and thus $k=2$. Therefore in fact if $(ab+ac+bc)|(a^2 + b^2 + c^2)$, then $a^2 + b^2 + c^2$ is either double $ab + ac + bc$ or equal to it which shows the result. EDIT : oops I forgot that $c$ can be greater than $ka + kb$ darn... disregard this solution. Apparently $\frac{a^2 + b^2+c^2}{ab + ac + bc} =1,2,5,10,14$ based on a program. I think some trickier vieta jumping can make this solution work.

@dinoboy, I did almost exactly as you did but I got stuck at this point. dinoboy wrote: Thus we can vieta-jump from a solution $(a,b,c)$ to $\left (a,b, \frac{a^2 + b^2 - kab}{c} \right )$. What happens if $a^2+b^2 < kab$? It happens for $(a, b, c) = (3, 5, 41)$.

Indeed, dinoboy's solution claims that $ab + bc + ca \mid a^2 + b^2 + c^2 \implies k=2$, but this is false for many triples. evan@ArchAir ~/Downloads $ python Python 2.7.4 (default, Apr  6 2013, 19:20:36) [GCC 4.8.0] on linux2 Type "help", "copyright", "credits" or "license" for more information. >>> import itertools >>> for a,b,c in itertools.product(range(1,100), repeat=3): ...     m = a*a + b*b + c*c ...     n = a*b + b*c + c*a ...     if m % n != 0: continue ...     if m/n > 2: ...             print a,b,c ... 2 3 71 2 5 71 2 71 3 2 71 5 3 2 71 3 5 41 3 41 5 3 71 2 5 2 71 5 3 41 5 41 3 5 71 2 6 10 82 6 82 10 10 6 82 10 82 6 41 3 5 41 5 3 71 2 3 71 2 5 71 3 2 71 5 2 82 6 10 82 10 6

Ok, so I messed up. darn. However, one thing I find that is it appears if $k > 2$ then usually $c = ka + kb + 1$ in the minimum solution (. Can anybody prove this? As this would lead to a solution. Or better, if somebody could prove apluscactus's claim that would be nice.

The solution proposed by V_ENhance prouves my claim : just replace 2013 by any multiple of 3. This was my solution to.

But the problem is that his solution is incomplete... Also, one thing to note is that $v_3 \left ( \frac{a^2 + b^2 + c^2}{ab + ac + bc} \right ) = 1$ is possible if you remove the restriction it must be an integer (just see $a=1, b = 8, c = 88$) so arguments looking purely at the $3$-adic evalution fail.

This is similar to v_enhance's solutions. First note that if $p$ divides 2 of $a,b,c$, the $p$ divides the last one. So we let $a^2 + b^2 + c^2 = k(ab+bc+ac)$, where $k$ is a multiple of 3. Assume $gcd(a, b, c) = 1$. This rearranges as $a^2 - a\left(k(b+c)\right) + b^2-kbc+c^2 = 0$. So the discriminant of the quadratic is $(k^2-4)b^2 + (2k^2+4k)bc + (k^2-4)c^2 = (k+2)\left((k-2)b^2 + 2kbc + (k-2)c^2\right)$. Since $k+2 \equiv 2 \pmod{3}$, there exists a prime $p$ dividing $k+2$ with odd multiplicity. Distinguish 2 cases: Case 1: $p$ is odd. Then $p \mid \left((k-2)b^2 + 2kbc + (k-2)c^2\right)$, and since $p \mid k+2$, \[ (k-2)b^2 + 2kbc + (k-2)c^2 \equiv -4b^2-4bc-4c^2 \pmod{p} \] So $p \mid b^2+bc+c^2 \mid b^3-c^3$. Since $p \equiv 2 \pmod{3}$, all residues $x^3 \pmod{p}$ are distinct, so $p \mid b-c$. If $p \mid b$ and $p \mid c$, then we can divide $a, b, c$ by $p$ and get a smaller triple. Otherwise, $\upsilon_p(b^2+bc+c^2) = \upsilon_p(b^3-c^3) - \upsilon_p(b-c) = \upsilon_p(b-c) + \upsilon_p(3) -\upsilon_p(b-c) = 0$, by the Lifting Exponent Lemma. So this case is impossible. Case 2: $p = 2$. Since we know that the multiplicity of 2 in $(k+2)$ is odd, we can write $k = j\cdot 2^{2m+1} - 2$, where $j$ is odd and $m \ge 0$. Assume $m = 0$. Then clearly $4 \vert k$. But going back to the beginning, we get that $4 \mid a^2+b^2+c^2$, so $a,b,c$ are all even, contradiction. Now let $m \ge 1$. Then note that \[ ((k-2)b^2+2kbc+(k-2)c^2) \] \[= (j\cdot 2^{2m+1} - 4)b^2+(j\cdot 2^{2m+2} - 4)bc+(j\cdot 2^{2m+1} - 4)c^2 \] \[= 4((j\cdot 2^{2m-1} - 1)b^2 + (j\cdot 2^{2m} - 1)bc + (j\cdot 2^{2m-1} - 1)c^2) \] So we still need $2 \mid \left((j\cdot 2^{2m-1} - 1)b^2 + (j\cdot 2^{2m} - 1)bc + (j\cdot 2^{2m-1}-1)c^2\right)$. But the coefficients of $b^2, bc, c^2$ are all odd, so it is easy to check that it is even iff $b, c$ are both even. Then $a, b, c$ are all even again, contradiction. Remark: In fact this shows that $3 \not\vert k$ and $\upsilon_2(k+2) \not\equiv 1 \pmod{2}$.

$11|a^2+b^2+c^2\Rightarrow 11|\gcd(a,b,c)$ ???!!!

Finished my solution. During chemistry I unfortunately did not realize that $v_2$ was in fact sufficient. By the way, thanks to whichever moderator fixed a typo for me siddigss wrote: $11|a^2+b^2+c^2\Rightarrow 11|\gcd(a,b,c)$ ???!!! Sorry, what is this referring to? Anyway, it is true in the context of this problem (namely, $11 \mid a+b+c$ and $11 \mid a^2+b^2+c^2$ implies $a,b,c \equiv 0 \pmod{11}$. See the last paragraph of my solution.)

v_Enhance wrote: Sorry, what is this referring to? Anyway, it is true in the context of this problem (namely, $11 \mid a+b+c$ and $11 \mid a^2+b^2+c^2$ implies $a,b,c \equiv 0 \pmod{11}$. See the last paragraph of my solution.) I exactly meant what I've written $11|a^2+b^2+c^2$ implies $a,b,c\equiv0(\mod 11) $ even if $11\not|a+b+c$ -using it the problem answer is NO , because there is no 3 coprimes such that the sum of their squares is divisible by 11 ( $11|2013$ ) -( but we didn't use $ab+bc+ca|a^2+b^2+c^2$ !! ) thanks v_Enhance

siddigss wrote: I exactly meant what I've written $11|a^2+b^2+c^2$ implies $a,b,c\equiv0(\mod 11) $ even if $11\not|a+b+c$ Oh, sorry, I thought you were asking a question about the solution above yours. But I don't think this claim is true: $11 \mid 3^2 + 1^2 + 1^2$, for instance.

Yeah yeah sorry unforgivable mistake .. Thanks

Suppose $\sum ab=\prod_{i=1}^{k} p_i^{2\beta_i+1}.x^2,\sum a=\prod_{i=1}^{k} p_i^{\alpha_i+\beta_i}xy$. Where $gcd(p_i,x)=gcd(p_i,y)=1$.Now suppose there for all $i$ we've $\alpha_i+\beta_i\neq 2beta_i+1$. So we get eliminating $c$ we get $a^2+b^2+ab=\prod p_i^{min(\alpha_i+\beta_i, 2beta_i+1)}(m(a+b)xy-nx^2)$ for some $m,n$ and $gcd(m,n)=1$ and $\{S(m),S(n)\}=\{p_1,p_2,.....,p_k\}$ where $S(x)$ denotes set of prime divisors of $x$. Now if for some $i$ we get $v_p(a^2+b^2+ab)>min(\alpha_i+\beta_i, 2\beta_i+1)$ then $p|RHS$ while $p$ divides exactly one of them , since we can take $gcd(p,a+b)=1$ since all of $p_i$'s must be greater than two.Now so basically we've $v_p(a^2+b^2+ab)=(\alpha_i+\beta_i, 2beta_i+1)$. Thus we note for all prime $p\in\{p_1,p_2,....,p_k\}$ the for a fixed $p$ and the $v_p(a^2+b^2+ab)=v_p(b^2+c^2+bc)=v_p(a^2+c^2+ac)=\min(\alpha_i+\beta_i,2\beta_i+1)$. Now it's easy to see for $p>2$ we can't have $v_p(a^2+b^2+ab)=v_p(b^2+c^2+bc)=v_p(a^2+c^2+ac)=v_p(a+b+c)<v_p(ab+bc+ac)$ now if $v_p(a^2+b^2+ab)=v_p(b^2+c^2+bc)=v_p(a^2+c^2+ac)=v_p(ab+bc+ac)<v_p(a+b+c)$ then certainly we're getting $a\equiv b\equiv c =k (mod p)$ but also that implies $p|3k,p|3k^2\implies p|k$ since $p\neq 3$, but also it's absurd. So now we've $v_p(a^2+b^2+ab)=v_p(b^2+c^2+bc)=v_p(a^2+c^2+ac)=v_p(ab+bc+ac)=v_p(a+b+c)$. Now so we take like $a^2+b^2+ab=mk_1,b^2+c^2+bc=mk_2,a^2+c^2+ac=mk_3$ and also $\sum ab=mx^2,\sum=mxy$. Where $(x,m)=(y,m)=1=(k_i,k_j)=1$. Now as we had $k=my^2-2$ thus there must exist a prime factor of form $3k-1$ dividing $m$ but that directly implies from $p|gcd(a^3-b^3,a^{p-1}-b^{p-1})$ that $m$ must be even. Now so all of $\sum a,\sum ab,a^2+ab+b^2,b^2+bc+c^2,a^2+c^2+ac$ are even and that directly implies all of $a,b,c$ are even, and that's the contra.

suppose that $(a^2+b^2+c^2/(ab+ac+bc))=2013t_{1}=d$ so $a^2+b^2+c^2-d(ab+ac+bc)=0\Rightarrow \Delta =[(b+c)d]^2-4(b^2+c^2-dbc)=n^2$ so we have $ \Delta =(d+2)[(d-4)(b^2+c^2)+2bcd]=m^2$ . Its not hard to prove that $GCD((d+2),((d-4)(b^2+c^2)+2bcd))=1$ so $2013t_{1}+2=x^2\Rightarrow 2\equiv x^2(mod 3)$ which is a contraction .

shekast-istadegi wrote: $a^2+b^2+c^2-d(ab+ac+bc)=0\Rightarrow \Delta =[(b+c)d]^2-4(b^2+c^2-dbc)=n^2$ so we have $ \Delta =(d+2)[(d-4)(b^2+c^2)+2bcd]=m^2$. Where does the $d+2$ come from? I have $n^2 = \left[ (d-4)(b^2+c^2) + 2bcd \right]$ and $d+2 = \tfrac{(a+b+c)^2}{ab+bc+ca}$. shekast-istadegi wrote: Its not hard to prove that $GCD((d+2),((d-4)(b^2+c^2)+2bcd))=1$. Try $(a,b,c) = (2,3,71)$ and $d=14$, or $(a,b,c) = (4,1,1)$ and $d=2$. This solution looks wrong to me because, like the other solutions, it really just shows that $3 \nmid \tfrac{a^2+b^2+c^2}{ab+bc+ca}$. However, you does not use any properties of $3$ other than $\left( \tfrac{2}{3} \right) = -1$. Since $\left( \tfrac{2}{5} \right) = -1$ as well, by simply replacing $3$ with $5$ your solution can be extended to show $5 \nmid \tfrac{a^2+b^2+c^2}{ab+bc+ca}$. But this is false for $(a,b,c) = (3,5,41)$, among others.

Sorry , but how can I prove $3 \nmid \frac{a^2+b^2+c^2}{ab+bc+ca}$

sorry i don't understand! Is it true $a^2+b^2+c^2<2013(ab+bc+ca)?$

We will infact show that there exist no positive integers $a, b, c$ such that $\frac{a^2+b^2+c^2}{3(ab+bc+ca)} \in \mathbb{Z}$. This will show that the given divisibility has no solutions for $3$ divides $2013$. Let us proceed by contradiction and suppose there exist positive integers $a, b, c, n$ such that $a^2+b^2+c^2 = 3n(a+b+c)$. Dividing $a, b, c$ by $\gcd(a, b, c)$ if necessary, we may assume $\gcd(a, b, c) = 1$. We have \[(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) = (3n+2)(ab+bc+ca)\]Clearly there exists a prime $p$ congruent to $2$ modulo $3$ dividing $3n+2$ with $\nu_p(n)$ odd; otherwise $3n+2 \equiv 1 \pmod 3$, absurd. Take such a prime. Then $p \mid a+b+c$ and since $\nu_p((a+b+c)^2)$ is even, it follows that $\nu_p(ab+bc+ca)$ is odd, in particular $p \mid ab+bc+ca$. We obtain \[ab+bc+ca \equiv ab + c(a+b) \equiv ab-(a+b)^2 \equiv -a^2-ab-b^2 \pmod p\]Thus $p \mid a^2+ab+b^2$. Since $p \equiv 2 \pmod 3$ simple quadratic reciprocity computations yield that $p \mid a, b$ following which $p \mid c$ as well. But we assumed that $\gcd(a, b, c)$ is $1$ and this is a contradiction. We're done.

goodar2006 wrote: Do there exist natural numbers $a, b$ and $c$ such that $a^2+b^2+c^2$ is divisible by $2013(ab+bc+ca)$? Proposed by Mahan Malihi We show that the quantity \[\frac{a^2+b^2+c^2}{3(ab+bc+ca)}\]is never an integer. Then we are done. Assume that \[a^2+b^2+c^2=3n(ab+bc+ca)\]then \[(a+b+c)^2=(3n+2)(ab+bc+ca)\]Now choose a prime \(p\) congruent to \(2\pmod3\) so that \(p\mid3n+2\). Then, we have that for some \(i\), \[p^i\mid a+b+c\]and \(p\mid ab+bc+ca\) (simply choose an odd power of \(p\) dividing \(3n+2\)). So, \(c\equiv-a-b\pmod p\) and substituting in the other relation gives \[p\mid a^2+ab+b^2\]or \[p\mid (2a+b)^2+3b^2\]so by Quadratic Residue, \[\left(-3\over p\right)=1\]which is a contradiction since \(p\equiv2\pmod3\).

The answer is NO. Assume contrary. WLOG $\gcd(a,b,c) =1 $ and $a \ge b \ge c$. Let $x=a-b,y=b-c$. Write \begin{align*} a^2 + b^2 + c^2 &= 2013k(ab + bc + ca) \\ \implies a^2 + b^2 + c^2 - (ab + bc + ca) &= (2013k-1)(ab + bc + ca) \\ \implies (2013k-1)(ab+bc+ca) &= \frac{1}{2} \left( (a-b)^2 + (b-c)^2 + (c-a)^2 \right) \\ &= \frac{1}{2} \left( x^2 + y^2 + (x+y)^2 \right) = x^2 + y^2 + xy \\ \implies x^2 + y^2 + xy &= (2013k-1)(ab + bc + ca) \qquad \qquad (1) \end{align*}Pick a prime $p \equiv 2 \pmod{3}$ such that $v_p(2013k-1)$ is even (such a $p$ exists as $2013k-1 \equiv 2 \pmod{3}$). Write $x = dm,y=dn$ with $\gcd(m,n) = 1$. Note $p$ cannot divide both $m,n$. Claim: $p \mid m^2 + n^2 + mn$. Proof: If not, then $p \mid x,y$ and also $p \mid ab + bc + ca$. From $p \mid x,y$ we get $a \equiv b \equiv c \pmod{p}$. But as $p \mid ab + bc + ca$, so this forces $p \mid a,b,c$, which contradicts $\gcd(a,b,c) = 1$. $\square$ Now $4(m^2 + n^2 + mn) = (2m+n)^2 + 3n^2$. Hence, $$ (2m+n)^2 + 3n^2 \equiv 0 \pmod{p} $$As $p \nmid m,n$, so $-3$ is a QR mod $p$. But as $p \equiv 2 \pmod{3}$, so that's a contradiction. $\blacksquare$

Can someone explain why if $p \equiv 2 \pmod 3$, then $-3$ cannot be a is a QR mod $p$? I'm looking at $p=2$, and $-3 \equiv 1$ seems to be a QR.

bump bump

bump bump

eduD_looC wrote: Can someone explain why if $p \equiv 2 \pmod 3$, then $-3$ cannot be a is a QR mod $p$? I'm looking at $p=2$, and $-3 \equiv 1$ seems to be a QR. That statement only holds for odd primes $p$. It follows by quadratic reciprocity theorem.

v_Enhance wrote: eduD_looC wrote: Can someone explain why if $p \equiv 2 \pmod 3$, then $-3$ cannot be a is a QR mod $p$? I'm looking at $p=2$, and $-3 \equiv 1$ seems to be a QR. That statement only holds for odd primes $p$. It follows by quadratic reciprocity theorem. Yes, but then how is the $p=2$ case dealt with? The last two solutions seem to disregard that case, so I was wondering if there were conditions I was missing out that would eliminate this extra case.

$p=2$ can be solved directly by noting $2\mid x^2+xy+y^2$ implies $2\mid x,y$

We prove the following general problem: if $a^2+b^2+c^2=k(ab+bc+ca)$ for some integer $k$, then $3 \nmid k$. Assume otherwise. WLOG we may assume $\gcd(a, b, c) = 1$. Then, $$(a+b+c)^2=(3n+2)(ab+bc+ca),$$so there exists a $3k+2$ form prime $p \mid 3n+2$. Now, note that $\nu_p((a+b+c)^2)$ is even and $\nu_p(3n+2)$ is odd, thus $p \mid ab+bc+ca$. But now $$a^2+ab+b^2 \equiv ab+bc+ca \equiv 0 \pmod p,$$and thus we must have $p \mid a, b, c$, which contradicts $\gcd(a, b, c) = 1$.

We claim there are no solutions to this equation. Assume the contrary. See that we can rewrite the problem statement as \[{(a+b+c)}^2=(2013n+2)(ab+bc+ca)\]where $n \in \mathbb{Z}$ and WLOG assume $\gcd(a,b,c)=1$. Claim We cannot have $2 \mid a+b+c$ and $2 \mid ab+bc+ca$. Subproof. See that since of our assumption, we must have two of $a$, $b$ and $c$ to be odd and last one to be even but this means $ab+bc+ca$ must be odd. $\blacksquare$ See that $2013n+2 \equiv 2 \pmod 3$, and hence there exists a prime $p \equiv 2 \pmod 3$ such that $v_p(2013n+2)$ is odd which means $p \mid a+b+c$ but $v_p \left({(a+b+c)}^2 \right)$ is even and hence we get $p \mid ab+bc+ca$ (see that $p \geq 5$). See that if $p$ divides any of $a$, $b$ or $c$ then it divides all of them, a contradiction. And now considering $c \equiv -a-b \pmod p$ and let $x \equiv \frac{a}b \pmod p$, we get \[p \mid x^2+x+1 \Longleftrightarrow {(2x+1)}^2 \equiv -3 \pmod p \Longleftrightarrow \left( \frac{-3}p \right) \Longleftrightarrow p \equiv 1 \pmod 3\]which is a contradiction.

Nice Problem. Assume $(a,b,c)=1$ Let $a^2+b^2+c^2 = 3k(ab+bc+ca)$ as $3|2013$ then $(a+b+c)^2 = (3k+2)(ab+bc+ca)$, now there must exist a prime $p$ dividing $3k+2$ such that $v_p(3k+2)$ is odd. $p|(a+b+c)^2$ $\rightarrow$ $p|a+b+c$ and as $v_p(3k+2)$ is odd so $p|ab+bc+ca$ $\rightarrow p|(a+b)(-a-b)+ab$ and hence $p|a^2+b^2+ab$ if $p$ does not divide $a,b$ then $(2a+b)^2 \equiv -3b^2 (\mod p)$ $(2a \cdot b^{-1}+1)^2 \equiv -3 (\mod p)$ $$1=\left(\frac{-3}{p}\right)= (-1)^{\frac{p-1}{2}} \cdot \left(\frac{3}{p}\right)=(-1)^{\frac{p-1}{2}} \cdot \left(\frac{p}{3}\right) \cdot (-1)^{\frac{p-1}{2}}=-1$$This yields a contradiction as desired.

Assume gcd(a,b,c) = 1. Now let $n = \frac{a^2 + b^2 + c^2}{2013(ab + bc + ac)}$, which is equivalent to $(a + b + c)^2 = (2013n + 2)(ab + bc + ca)$. Since $2013n + 2 \equiv 2 \pmod 3$, then $p \mid 2013n + 2$, where $p \equiv 2 \pmod 3$ and $\nu_p(2013n + 2) = odd$. Since $(a + b + c)^2$ is a square and $\nu_p(2013n + 2) = odd$, then we must have $p \mid ab + bc + ca$. Now $p \mid a + b + c$ $\Rightarrow$ $p \mid a^2 + b^2 + c^2$ and $p \mid (a+b)^2 - c^2$ $\Rightarrow$ $p \mid a^2 + b^2 + c^2 + (a+b)^2 - c^2$ $\Rightarrow$ $p \mid a^2 + b^2 + (a+b)^2$ $\Rightarrow$ $p \mid a^2 + ab + b^2$. Now by lemma for $p \equiv 2 \pmod 3$ we have that $p \mid a$ and $p \mid b$. Since $p \mid a + b + c$, $p \mid c$ too $\Rightarrow$ gcd(a,b,c) = p and since we assumed gcd(a,b,c) = 1 we have a contradiction. There don't exist such a,b,c that work.

No. Suppose there did exist such $a,b,c$. Choose $a,b,c$ such that $a + b + c$ is minimal. Claim: $\gcd(a,b,c) = 1$ Proof: If $d>1$ divided $a,b,c$, we find that $\frac ad, \frac bd, \frac cd$ satisfies the divisibility, contradiction to the minimality of $a+b+c$. $\square$ Now let $a^2 + b^2 + c^2 = 2013k (ab + bc + ca)$. We find that \[ (2013k + 2) (ab + bc + ca) = (a+b+c)^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \] Since $2013k + 2\equiv 2\pmod 3$, there exists a prime $p\equiv 2\pmod 3$ with $\nu_p(2013k + 2)$ is odd. Choose $p$ minimal. Case 1: $p = 2$ Then notice the LHS of $(1)$ must be a perfect square, so it has an even $\nu_2$, meaning that $ab + bc + ca$ is even and $a + b + c$ is also even. Since $\gcd(a,b,c) = 1$, at least one of $a,b,c$ is odd. Hence two of $a,b,c$ are odd and the other is even, but this is absurd since it means $ab + bc + ca$ is odd. Case 2: $p > 2$ Then notice the LHS of $(1)$ must have an even $\nu_p$, so $p$ divides $ab + bc + ca$ and $a + b + c$. But then $p$ also divides $(a+b+c)^2 - 2(ab + bc + ca) = a^2 + b^2 + c^2$. If we write $c \equiv -(a+b)\pmod p$, we see that $p$ divides $a^2 + b^2 + (a+b)^2 = 2(a^2 + ab + b^2)$. Since $p$ is odd, $p$ also divides $a^2 + ab + b^2$. If one of $a,b$ is divisible by $p$, then all three of $a,b,c$ are, contradicting $\gcd(a,b,c) = 1$. Now, we find that $p$ divides $(a-b)(a^2 + ab + b^2) = a^3 - b^3$, so this also means that $p$ divides $(a/b)^3 - 1$. If $a/b \not \equiv 1\pmod p$, then the order of $\frac ab$ modulo $p$ equals $3$, so $p\equiv 1\pmod 3$, contradiction. Hence $\frac ab \equiv 1\pmod p$, so $a\equiv b \pmod p$. But then $a^2 + ab + b^2 \equiv 3a^2 \pmod p$, so $p\mid 3a^2$. Since $p \equiv 2\pmod 3$, $p$ cannot equal $3$, and since $p\nmid a$, this is absurd.

I used way too many hints after I got stuck at the $p \equiv 3 \mod 2...$ Suppose $a^2+b^2+c^2=k(2013)(ab+bc+ca).$ Choose $a,b,c$ with minimal $a.$ We have that \[ (a+b+c)^2=(2013k+2)(ab+bc+ca) \]Observe $2013k+2 \equiv 2 \mod 3$ so $\exists p \equiv 3 \mod 2$ such that $p \mid 2013k+2,$ and observe there must exist such a $p$ to an odd power or else $2013k+2$ would $\equiv 1 \mod 3$ forming a contradiction. Thus $p \equiv 2 \mod 3$ divides $(a+b+c)$ and $(ab+bc+ca).$ Next observe \[ p \mid (a+b+c)(a+b)-ab-ac-bc=a^2+b^2+ab \]Next, observe \[p \equiv 3 \mod 2 \iff a^2+ab+b^2 \equiv 0 \mod p \implies a,b \equiv 0 \mod p\]as if we let $x=ab^{-1}$ we know $x^2+x+1 \equiv 0 \mod p$ and as $(x-1)(x^2+x+1)=x^3-1$ this is a cyclotomic polynomial and solutions $p$ must either be $3$ or $\equiv 1 \mod 3.$ As $p \equiv 3 \mod 2$ we know that $p \mid a,b$ and thus $p \mid a+b+c-a-b=c,$ so taking $\left( \dfrac{a}{p}, \dfrac{b}{p}, \dfrac{c}{p} \right)$ contradicts the minimality of $a$ and we are done.