This problem from Bulgarian MO( before some year )

The topic mentions the source. If you or someone could please provide the solution please.

Hint : By Cauchy-Schwarz inequality we have : 3((a^2)+(b^2)+(c^2)) ≥ (|a|+|b|+|c|)^2 ≥ (a+b+c)^2 ⇒ 3(a+b+c) ≥ (a+b+c)^2 ⇒ 3 ≥ (a+b+c) (note that (a+b+c) ≥ 0) ⇒ (a+b+c) ∈ {0,1,2,3} (Conditions (a+b+c) ∈ {0,3} can be easily checked) Now take : a=(m/q) , b=(n/q) , c=(p/q) gcd(m,n,p,q)=1 where m,n,p,q are integers... .

(The proof is from M.SH in this forum) take : a=(m/q) , b=(n/q) , c=(p/q) gcd(m,n,p,q)=1 where m,n,p,q are integers. We have four cases: 1) a+b+c = 0 ⇒ (a^2) + (b^2) + (c^2) = 0 ⇒ a=b=c=0 2) a+b+c=1 ⇒ (m+n+p)/q = ((m^2)+(n^2)+(p^2))/(q^2) = 1 ⇒ (Σmn) = 0 and (m+n+p)/q = ((m^2)+(n^2)+(p^2))/(q^2) Let r be a prime number. We have: r^α||m r^β||n r^γ||p r^δ||q (r^α).m' = m (r^β).n' = n (r^γ).p' = p (r^δ).q' = q WLOG assume that: 0 ≤ α ≤ β ≤ γ → (Σmn) = 0 ⇒ (r^(α+β))(m'.n' + (r^(γ-β)).m'.p' + (r^(γ-α)).p'.n') = 0 But gcd(r,m') = gcd(r,n') = 1 ⇒ α=β=γ or β=γ>α If there exist a prime r such that α=β=γ>0 then: (Σmn) = (r^(2α))(Σm'n') = 0 ⇒ (Σm'n')=0 ⇒ (Σ(m'^2)) = (Σm')^2 ⇒ r^θ || (Σm') and r^μ || (Σ(m'^2)) r^0||q →(r^α)(m'+n'+p')q = (r^(2α))((m'^2)+(n'^2)+(p'^2)) ⇒ θ+α = 2α+2θ ⇒ α=θ=0 So for any prime number r we have β=γ>α We know that : (m+n+p)q = ((m^2)+(n^2)+(p^2)) ⇒(r^α).(r^δ).q'.(m' + (r^(β-α))(n'+p'))=(r^(2α)).((m'^2)+(r^(2β-2α)).((n'^2)+(p'^2))) ⇒α+δ = 2α ⇒ α=δ But (m,n,p,q) = 1 ⇒ α=δ=0 So abc = (mnp)/(q^3) is obviously a ratio of a cube and a square. 3) a+b+c=2 ⇒ (m+n+p)/q = ((m^2)+(n^2)+(p^2))/(q^2) = 2 ⇒ (Σ(m^2)) - 2(Σmn) = 0 (*) and (m+n+p)q = ((m^2)+(n^2)+(p^2)) From (*) we get that : ±√m ± √n ± √p = 0 So we have two conditions: a) √p + √m + √n = 0 which we can easily get that m=n=p=0 b)WLOG assume that : √p = √m + √n ⇒ p = m + n + 2√(mn) So mn is a perfect square. (like above just by considering the power of prime r in ''p = m + n + 2√(mn)'' and ''(m+n+p)q = ((m^2)+(n^2)+(p^2))'' You can easily get that mnp is a perfect square(If I'm not wrong in calculations!) 4) a+b+c = 3 So the inequality holds and we can conclude that : a=b=c=1 or 0.

@wiseman can you $LATEX$ it please (else please learn it)

I don't know how to write in LATEX yet, but I promise to learn it soon!

Here is a LaTeX'ified and hopefully a better written solution. Note that, if $a^2+b^2+c^2=0$, we trivially have the condition. So, suppose $a+b+c=a^2+b^2+c^2=t>0$. By Cauchy-Schwarz, we have $3t\geq t^2$, hence, $t\leq 3$. For $t=3$, we must have $|a|=|b|=|c|=1$, from which the condition follows trivially. Hence, we focus on $t=1,2$ cases. Now, let $(a,b,c) = (m/k,n/k,r/k)$ with $(m,n,r,k)=1$. We start with $t=1$ case. If $t=1$, we have $m+n+r=k$ and $m^2+n^2+r^2=k^2$. Both, together yield that $mn+mr+nr=0$. Now, suppose $p$ be a prime such that $p^a \mid \mid m$. It follows that, $p^a \mid nr$. Notice now that $p$ cannot divide $n$ and $r$ simultaneously. Indeed, had this been the case, we would have had, $p\mid n,r$ and thus, $p\mid m+n+r=k$, contradicting with the fact that $(m,n,r,k)=1$. Hence, $p^a\mid n$. We now claim that, $p^a\mid \mid n$. To see this, let $m=p^a m'$ and $n=p^a n'$. We know that $p\nmid m'$, and we want to show $p\nmid n'$. Now, from the condition $mn+mr+nr=0$, we have $p^a m' n' + rm' +rn'=0$. Hence, $rm' + rn' \equiv 0\pmod{p}$. Since $p\nmid r$, it holds that $p\mid m'+n'$, and since $p\nmid m'$, it must be the case that, $p\nmid n'$. In particular, for the product, $mnr$, the exponent of $p$ is precisely $2a$. Repeat the same idea. We have, for any $q\mid mnr$, the exponent of $q$ in the prime decomposition of $mnr$ is even, and $(q,k)=1$. Finally, since $abc = \frac{mnr}{k^3}$, and for each $q\mid mnr$, $(q,k)=1$, we obtain that, $abc$ can be expressed, indeed, as a fraction of a perfect square and a perfect cube, that are coprime, as desired. Finally, we consider the case $t=2$. We have, $m+n+r=2k$, and $m^2+n^2+r^2=2k^2$. It holds, therefore, that $mn+nr+mr = k^2$. Now, if $p$ is a prime (that $p\neq 2$, the case $p=2$ is trivial, but a bit annoying, that I skip), dividing $m$, it follows that, $k^2\equiv nr\pmod{p}$. Now, we claim $p\nmid k$. If not, and $p\mid k$, then $p\mid nr$. So, $p$ divides either $n$, or $r$. If $p\mid n$, then from $m+n+r=2k$, it holds also that $p\mid r$, hence, $p\mid (m,n,r,k)=1$, again, a contradiction. This logic gives $(mnr,k^3)=1$. Now, we claim that $mnr$ is a perfect square. To see this, we again return back to $mn+mr+nr=k^2$. If $p\mid m$ and $p\mid n$, then it follows that, $p\mid k$, which in turn brings $p\mid r$. Continuing in this manner, we get $(m,n)=(m,k)=(n,k)=1$. Now, equipped with this, we have $m+n=2k-r$, which implies $4k^2-4kr+r^2 = 2k^2-r^2-2mn$, which gives $mn=(k-r)^2$. In particular, since $(m,n)=1$, it follows that $m$ and $n$ are both perfect squares, so do $mnr$, finishing the proof. (Note that, I've omitted the factor $2$ above, and I don't think it is too hard to fix that nuisance).