Clearly,$gcd(a,b)gcd(b,c)*gcd(a,c)=1$.We assume statement is wrong.WLOG$ c>b>a.$ We have $0<a+b-c<a \implies ab>a^2>(a+b-c)^2 .$But $ab|(a+b-c)^2$.Contradiction.

Misha57 wrote: Clearly, $gcd(a,b)gcd(b,c)*gcd(a,c)=1$... I think you misunterstood something: $\textnormal{gcd}(a,b,c)=1$ means that there is no integer $d > 1$ that divides all of $a,b,c$, but they needn´t be pairwise coprime. Consequently, your solution is not completely correct since $a \mid (a+b-c)^2$ and $b \mid (a+b-c)^2$ together do not imply $ab \mid (a+b-c)^2$.

From conditions gcd (a,b,c)=1 and a|(b-c)^2 it follows that gcd (a,b)=1.

Meh. WLOG, $a>b>c$. Also note that $\gcd(a,b)=\gcd(b,c)=\gcd(a,c)=1$. For the sake of contradiction, assume that there are $a,b,c$ such that they form a non-degenerate triangle. This means that $b+c>a$. We firstly consider small cases for $c$. Observe that if $a=b+c-1$, then $b\mid (a-c)^2=(b-1)^2\implies b=1$, which is impossible as $c\geq 1$. Also observe that if $c\geq 2$, then $a=b+1$ is impossible as $c\mid (a-b)^2=1$. Note also that $b+c>a\geq b+1$. Here, $c=1$ obviously impossible, thus assume $c\geq 2$. Now, we get a better bound, i.e. $b+c-1>a\geq b+2$. This nukes $c=2,3$. Consider also $c=4,5$, $c=4$: $a=b+2$, meaning $b\mid (a-c)^2=(b-2)^2\implies b\mid 4$, impossible as $b>c=4$. $c=5$: If $a=b+2$, then $c\mid (a-b)^2=4$, which is absurd. If $a=b+3$, then $c\mid (a-b)^2=9$, which is absurd. Hereinafter, we consider $c\geq 6$. Now, note that $$abc+a^2+b^2+c^2> 3ab+a^2+b^2+c^2>2ab+2ac+2bc$$and $$abc+2ab+2ac+2bc>3b^2+3c^2+6bc>3a^2>a^2+b^2+c^2,$$which is true as $abc>6b^2>3b^2+3c^2\Longleftrightarrow b^2(c-6)+bc>0$ as $c\geq 6$. This shows that $abc> \vert a^2+b^2+c^2-2ab-2ac-2bc \vert$. However, divisibilities yield, \begin{align*} a\mid a^2+b^2+c^2-2ab-2ac-2bc,\\ b\mid a^2+b^2+c^2-2ab-2ac-2bc,\\ c\mid a^2+b^2+c^2-2ab-2ac-2bc.\end{align*}And as $\gcd(a,b)=\gcd(b,c)=\gcd(a,c)=1$, $$abc\mid a^2+b^2+c^2-2ab-2ac-2bc,$$hence $abc\leq \vert a^2+b^2+c^2-2ab-2ac-2bc\vert$, unless $a^2+b^2+c^2-2ab-2ac-2bc=0$. We now consider equation $$a^2+b^2+c^2=2ab+2ac+2bc\Longleftrightarrow a^2-a(2b+2c)+(b-c)^2=0,$$thus $$a=\frac{2b+2c\pm \sqrt{(2b+2c)^2-4(b-c)^2}}{2}=b+c\pm 2\sqrt{bc}.$$As $b+c>a$, we have $a=(\sqrt{b}-\sqrt{c})^2$, but then $\sqrt{b}<\sqrt{a}=\sqrt{b}-\sqrt{c}$, which is absurd. Therefore, we conclude that $a,b,c$ never form a non-degenerate triangle.

Wlog $a\ge b \ge c$. We have $bc|(a-b-c)^2$ from the $2.$ and $3.$ condition. Assume that $b+c>a$. We have $bc \le (b+c-a)^2 \le c^2 \implies b=c$, contradicts with $gcd=1$. Thus $a\ge b+c$ which shows there does not exist a triangle with sides $a,b,c$.

Assume $a,b,c$ are triangle sides then, $a=x+y$; $b=y+z$ ; $c=z+x$ $(x+y) | (y-x)^2$ also $(x+y) |(y+x)^2$ $(x+y)|4xy$ $\gcd(xy,x+y)=1$ $x+y|4$ contradicition.

ismayilzadei1387 wrote: Assume $a,b,c$ are triangle sides then, $a=x+y$; $b=y+z$ ; $c=z+x$ $(x+y) | (y-x)^2$ also $(x+y) |(y+x)^2$ $(x+y)|4xy$ $\gcd(xy,x+y)=1$ $x+y|4$ contradicition. why is it a contradiction? also how did you get $(x+y) | (y-x)^2$

ismayilzadei1387 wrote: Assume $a,b,c$ are triangle sides then, $a=x+y$; $b=y+z$ ; $c=z+x$ $(x+y) | (y-x)^2$ also $(x+y) |(y+x)^2$ $(x+y)|4xy$ $\gcd(xy,x+y)=1$ $x+y|4$ contradicition. contradiction of what. I have same question as @above. Gan xie.

RenheMiResembleRice wrote: ismayilzadei1387 wrote: Assume $a,b,c$ are triangle sides then, $a=x+y$; $b=y+z$ ; $c=z+x$ $(x+y) | (y-x)^2$ also $(x+y) |(y+x)^2$ $(x+y)|4xy$ $\gcd(xy,x+y)=1$ $x+y|4$ contradicition. why is it a contradiction? also how did you get $(x+y) | (y-x)^2$ $(x+y) | (y-x) ^2 $ $(x+y) | (y+x) ^2 $ So, $(x+y) | 4xy$ Then, $(x+y) |4$ $x+y={1,2,4}$ $y+z={1,2,4}$ $z+x={1,2,4}$ Wlog $a>b>c$ $a=4;b=2;c=1$ $a<b+c$ but $a>b+c$ Remark: Ravi substituting $a|(b-c)^2$ $x+y |(y+z-z-x)^2=(y-x)^2$