Everybody loves integers, right?

We shall show that there are no solutions.

Since $(a, b, c)$ is a solution if and only if so is $(na, nb, nc)$ for some $n \in \mathbb{Z},$ we can assume WLOG that $a,b,c$ are coprime. Say $\dfrac{a^2+b^2+c^2}{ab+bc+ca}= 3n.$ Rewrite it as $3n(a+b+c)^2 = (3n+2)(a^2+b^2+c^2).$ Since $3n+2 \equiv -1 \pmod{3},$ there must exist a prime $p \equiv 2 \pmod{3}$ which divides $a^2+b^2+c^2$ an odd number of times (if all primes dividing $3n+2$ were $1 \pmod{3},$ we would have $3n+2 \equiv 1 \pmod{3};$ if all primes $p \equiv 2 \pmod{3}$ divided $3n+2$ an even number of times, we would have $3n+2 \equiv (-1)^{v_p(3n+2)} \equiv 1 \pmod{3}$ ).

If $p \neq 2,$ $\gcd (p, 2)= 1.$ Note that $p$ divides $(a+b+c)^2$ an even number of times but $3k+2$ an odd number of times, so $p$ must divide $a^2+b^2+c^2$ an odd number of times. We simultaneously have the congruencies $\begin{cases} a+b+c \equiv 0 & \pmod{p} \\ a^2+b^2+c^2 \equiv 0 & \pmod{p} \end{cases}$ which implies $a^2+b^2+(a+b)^2 \equiv 0 \pmod{p} \implies 2(a^2+ab+b^2) \equiv 0 \pmod{p}.$ Since $\gcd (p,2)= 1,$ $a^2 + b^2 + ab \equiv 0 \pmod{p} \implies \left( ab^{-1} \right ) ^ 3 \equiv 0 \pmod{p}.$ Thus, the order of $ab^{-1} \pmod{p}$ is either $1$ or $3.$ Since $\varphi (p) = p-1 \equiv 1 \not \equiv 0 \pmod{3},$ the order of $ab^{-1} \pmod{p}$ must be $1,$ so $a \equiv b \pmod{p}.$ This again gives $3a^2 \equiv 3b^2 \equiv 0 \pmod{p},$ and since $p \neq 3,$ $a^2 \equiv b^2 \equiv 0 \pmod{p},$ which is a contradiction since we have assumed $a$ and $b$ to be coprime.

If $p=2,$ we divide into two cases. The highest power of $2$ dividing $n$ will be denoted by $v_2 (n),$ and we shall use the trivial properties $v_2 (\prod a_i ) = \sum v_2 (a_i)$ and $v_2 (x^n) = n v_2 (x).$

Case 1: $4 \mid n$

Then, $3n+2 \equiv 2 \not \equiv 0 \pmod{4},$ and $3n+2 \equiv 0 \pmod{2},$ so $v_2 (3n+2)= 1.$ Also since $a,b,c$ are coprime, $4 \not \mid a^2+b^2+c^2,$ so $v_2 (a^2+b^2+c^2) \leq 1.$ We have that $\begin{array}{rcl} v_2 ((3n+2)(a^2+b^2+c^2)^2 & = & 1 + v_2 (a^2+b^2+c^2) \leq 2 \\ v_2 (3n(a+b+c)^2) = v_2 (3) + v_2 (k) + v_2((a+b+c))^2 & = & v_2 (n) + 2 v_2 (a+b+c) \geq 2. \end{array}$ For equality to hold, we must have $v_2 (n) = 2, v_2 (a+b+c)= 0,$ and $v_2 (a^2+b^2+c^2)= 1,$ which is not possible since $2 \mid a^2+b^2+c^2$ implies that there are an even number of even numbers in the set $\{a,b,c\},$ which in turn implies $2 \mid a+b+c .$

Case 2: $4 \not \mid n$

Then, $v_2 (n) = 1$ and $3n+2 \equiv 0 \pmod{4},$ so $v_2 (3n+2) \geq 2.$ Now, $\begin{array}{lcl} v_2 ((3n+2)(a^2+b^2+c^2)) & = & v_2 (3n(a+b+c)^2) \\ \implies v_2 (3n+2) + v_2 (a^2+b^2+c^2) & = & v_2 (n) + 2 v_2 (a+b+c). \end{array}$

• Subcase 1: $v_2 (3n+2)$ is even.

Since $2$ is the only prime divisor of $3n+2$ which is $\equiv -1 \pmod{3},$ modulo $3,$ we have $3n+2 \equiv(-1)^{v_2 (3n+2)} \equiv 1 \pmod{3},$ which is a contradiction.

• Subcase 2: $v_2 (3n+2)$ is odd.

Then, $v_2 (a^2+b^2+c^2)$ is even. But since $4 \not \mid a^2+b^2+c^2,$ we have that $v_2 (a^2+b^2+c^2)= 0.$ This again implies that $a+b+c$ is also odd, so $v_2 (a+b+c)=0.$ Then, we have $v_2 (3n+2) = 1,$ but we have already shown that $4 \mid 3n+2.$ Hence, contradiction.

Therefore, there don't exist any integers $a,b,c$ satisfying the problem conditions.

The given expression can be rewritten: $\frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 3(ab+bc+ac) } =\frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2ab+2bc+2ac-2(ab+bc+ac) }{ 3(ab+bc+ac) } =\frac { { (a+b+c) }^{ 2 } }{ 3(ab+bc+ac) } -\frac { 2 }{ 3 }$

In order for this to be an integer ${ (a+b+c) }^{ 2 }$ must be congruent to $2\quad mod\quad 3$ since the denominator of the first part of the expression is a multiple of 3 and the numerator of the second part of the expression is congruent to $2\quad mod\quad 3$ .

It is impossible for a square to be congruent to $2\quad mod\quad 3$ , so there is indeed no ordered triple that satisfies the conditions.

Proof of above statement: $({ 0\quad mod\quad 3) }^{ 2 }\equiv \quad 0\quad mod\quad 3$ and ${ 1\quad mod\quad 3) }^{ 2 }\equiv \quad 1\quad mod\quad 3$ and $({ 2\quad mod\quad 3) }^{ 2 }\equiv \quad 1\quad mod\quad 3$